{"paper_meta":{"paper_id":"arxiv:0708.1529","title":"0708.1529","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0708.1529v1 [cs.CC] 10 Aug 2007\nRESOLUTION OVER LINEAR EQUATIONS\nAND MULTILINEAR PROOFS\nRAN RAZ AND IDDO TZAMERET\nAbstract. We develop and study the complexity of propositional proof systems of varying strength\nextending resolution by allowing it to operate with disjunctions of linear equations instead of\nclauses. We demonstrate polynomial-size refutations for hard tautologies like the pigeonhole prin-\nciple, Tseitin graph tautologies and the clique-coloring tautologies in these proof systems. Using\nthe (monotone) interpolation by a communication game technique we establish an exponential-size\nlower bound on refutations in a certain, considerably strong, fragment of resolution over linear\nequations, as well as a general polynomial upper bound on (non-monotone) interpolants in this\nfragment.\nWe then apply these results to extend and improve previous results on multilinear proofs (over\nfields of characteristic 0), as studied in [RT06]. Specifically, we show the following:\n• Proofs operating with depth-3 multilinear formulas polynomially simulate a certain, consid-\nerably strong, fragment of resolution over linear equations.\n• Proofs operating with depth-3 multilinear formulas admit polynomial-size refutations of the\npigeonhole principle and Tseitin graph tautologies. The former improve over a previous result\nthat established small multilinear proofs only for the functional pigeonhole principle. The\nlatter are different than previous proofs, and apply to multilinear proofs of Tseitin mod p\ngraph tautologies over any field of characteristic 0.\nWe conclude by connecting resolution over linear equations with extensions of the cutting planes\nproof system.\nContents\n1.\nIntroduction\n2\n1.1.\nComparison to Earlier Work\n4\n1.2.\nSummary of Results\n5\n2.\nNotation and Background on Propositional Proof Systems\n7\n3.\nResolution over Linear Equations and its Subsystems\n8\n3.1.\nDisjunctions of Linear Equations\n8\n3.2.\nResolution over Linear Equations – R(lin)\n9\n3.3.\nFragment of Resolution over Linear Equations – R0(lin)\n10\n4.\nReasoning and Counting inside R(lin) and its Subsystems\n11\n4.1.\nBasic Reasoning inside R(lin) and its Subsystems\n12\n4.2.\nBasic Counting inside R(lin) and R0(lin)\n13\n5.\nImplicational Completeness of R(lin) and its Subsystems\n16\n6.\nShort Proofs for Hard Tautologies\n17\n6.1.\nThe Pigeonhole Principle Tautologies in R0(lin)\n17\n6.2.\nTseitin mod p Tautologies in R0(lin)\n19\n2000 Mathematics Subject Classification.\n03F20, 68Q17, 68Q15.\nKey words and phrases. proof complexity, resolution, algebraic proof systems, multilinear proofs, cutting planes,\nfeasible monotone interpolation.\nThe first author was supported by The Israel Science Foundation and The Minerva Foundation.\nThe second author was supported by The Israel Science Foundation (grant no. 250/05).\n1\n\n6.3.\nThe Clique-Coloring Principle in R(lin)\n23\n7.\nInterpolation Results for R0(lin)\n26\n7.1.\nInterpolation for Semantic Refutations\n26\n7.2.\nPolynomial Upper Bounds on Interpolants for R0(lin)\n28\n8.\nSize Lower Bounds\n30\n9.\nApplications to Multilinear Proofs\n32\n9.1.\nBackground on Algebraic and Multilinear Proofs\n32\n9.2.\nFrom R(lin) Proofs to PCR Proofs\n34\n9.3.\nFrom PCR Proofs to Multilinear Proofs\n35\n9.4.\nSmall Depth-3 Multilinear Proofs\n38\n10.\nRelations with Extensions of Cutting Planes\n38\nAppendix A.\nFeasible Monotone Interpolation\n41\nAcknowledgments\n42\nReferences\n42\n1. Introduction\nThis paper considers two kinds of proof systems. The first kind are extensions of resolution\nthat operate with disjunctions of linear equations with integral coefficients instead of clauses. The\nsecond kind are algebraic proof systems operating with multilinear arithmetic formulas. Proofs in\nboth kinds of systems establish the unsatisfiability of formulas in conjunctive normal form (CNF).\nWe are primarily concerned with connections between these two families of proof systems and with\nextending and improving previous results on multilinear proofs.\nThe resolution system is a popular propositional proof system that establishes the unsatisfiability\nof CNF formulas (or equivalently, the truth of tautologies in disjunctive normal form) by operating\nwith clauses (a clause is a disjunction of propositional variables and their negations). It is well\nknown that resolution cannot provide small (that is, polynomial-size) proofs for many basic count-\ning arguments. The most notable example of this are the strong exponential lower bounds on the\nresolution refutation size of the pigeonhole principle and its different variants (Haken [Hak85] was\nthe first to establish such a lower bound; see also [Razb02] for a survey on the proof complexity of\nthe pigeonhole principle). Due to the popularity of resolution both in practice, as the core of many\nautomated theorem provers, and as a theoretical case-study in propositional proof complexity, it\nis natural to consider weak extensions of resolution that can overcome its inefficiency in provid-\ning proofs of counting arguments. The proof systems we present in this paper are extensions of\nresolution, of various strength, that are suited for this purpose.\nPropositional proof systems of a different nature that also attracted much attention in proof\ncomplexity theory are algebraic proof systems, which are proof systems operating with (multivariate)\npolynomials over a field. In this paper, we are particularly interested in algebraic proof systems\nthat operate with multilinear polynomials represented as multilinear arithmetic formulas, called by\nthe generic name multilinear proofs (a polynomial is multilinear if the power of each variable in its\nmonomials is at most one). The investigation into such proof systems was initiated in [RT06], and\nhere we continue this line of research. This research is motivated on the one hand by the apparent\nconsiderable strength of such systems; and on the other hand, by the known super-polynomial\nsize lower bounds on multilinear formulas computing certain important functions [Raz04, Raz06],\ncombined with the general working assumption that establishing lower bounds on the size of objects\na proof system manipulates (in this case, multilinear formulas) is close to establishing lower bounds\non the size of the proofs themselves.\n2\n\nThe basic proof system we shall study is denoted R(lin). The proof-lines1 in R(lin) proofs are\ndisjunctions of linear equations with integral coefficients over the variables ⃗x = x1, . . . , xn.\nIt\nturns out that (already proper subsystems of) R(lin) can handle very elegantly basic counting\narguments.\nThe following defines the R(lin) proof system. Given an initial CNF, we translate\nevery clause W\ni∈I xi ∨W\nj∈J ¬xj (where I are the indices of variables with positive polarities and\nJ are the indices of variables with negative polarities) pertaining to the CNF, into the disjunction\nW\ni∈I(xi = 1)∨W\nj∈J(xj = 0). Let A and B be two disjunctions of linear equations, and let ⃗a·⃗x = a0\nand ⃗b · ⃗x = b0 be two linear equations (where ⃗a,⃗b are two vectors of n integral coefficients, and\n⃗a · ⃗x is the scalar product Pn\ni=1 aixi; and similarly for ⃗b · ⃗x). The rules of inference belonging to\nR(lin) allow to derive A ∨B ∨((⃗a +⃗b) · ⃗x = a0 + b0) from A ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0) (or\nsimilarly, to derive A ∨B ∨((⃗a −⃗b) ·⃗x = a0 −b0) from A ∨(⃗a ·⃗x = a0) and B ∨(⃗b ·⃗x = b0)). We can\nalso simplify disjunctions by discarding (unsatisfiable) equations of the form (0 = k), for k ̸= 0.\nIn addition, for every variable xi, we shall add an axiom (xi = 0) ∨(xi = 1), which forces xi to\ntake on only Boolean values. A derivation of the empty disjunction (which stands for false) from\nthe (translated) clauses of a CNF is called an R(lin) refutation of the given CNF. This way, every\nunsatisfiable CNF has an R(lin) refutation (this can be proved by a straightforward simulation of\nresolution by R(lin)).\nThe basic idea connecting resolution operating with disjunctions of linear equations and multilin-\near proofs is this: Whenever a disjunction of linear equations is simple enough — and specifically,\nwhen it is close to a symmetric function, in a manner made precise — then it can be represented\nby a small size and small depth multilinear arithmetic formula over fields of characteristic 0. This\nidea was already used (somewhat implicitly) in [RT06] to obtain polynomial-size multilinear proofs\noperating with depth-3 multilinear formulas of the functional pigeonhole principle (this principle\nis weaker than the pigeonhole principle). In the current paper we generalize previous results on\nmultilinear proofs by fully using this idea: We show how to polynomially simulate with multilinear\nproofs, operating with small depth multilinear formulas, certain short proofs carried inside resolu-\ntion over linear equations. This enables us to provide new polynomial-size multilinear proofs for\ncertain hard tautologies, improving results from [RT06].\nMore specifically, we introduce a certain fragment of R(lin), which can be polynomially simu-\nlated by depth-3 multilinear proofs (that is, multilinear proofs operating with depth-3 multilinear\nformulas). On the one hand this fragment of resolution over linear equations already is sufficient\nto formalize in a transparent way basic counting arguments, and so it admits small proofs of the\npigeonhole principle and the Tseitin mod p formulas (which yields some new upper bounds on\nmultilinear proofs); and on the other hand we can use the (monotone) interpolation technique to\nestablish an exponential-size lower bound on refutations in this fragment as well as demonstrating a\ngeneral (non-monotone) polynomial upper bound on interpolants for this fragment. The possibility\nthat multilinear proofs (possibly, operating with depth-3 multilinear formulas) possess the feasible\nmonotone interpolation property (and hence, admit exponential-size lower bounds) remains open.\nAnother family of propositional proof systems we discuss in relation to the systems mentioned\nabove are the cutting planes system and its extensions. The cutting planes proof system operates\nwith linear inequalities with integral coefficients, and this system is very close to the extensions\nof resolution we present in this paper. In particular, the following simple observation can be used\nto polynomially simulate cutting planes proofs with polynomially bounded coefficients (and some\nof its extensions) inside resolution over linear equations: The truth value of a linear inequality\n⃗a · ⃗x ≥a0 (where ⃗a is a vector of n integral coefficients and ⃗x is a vector of n Boolean variables) is\n1Each element (usually a formula) of a proof-sequence is referred to as a proof-line.\n3\n\nequivalent to the truth value of the following disjunction of linear equalities:\n(⃗a · ⃗x = a0) ∨(⃗a · ⃗x = a0 + 1) ∨· · · ∨(⃗a · ⃗x = a0 + k) ,\nwhere a0 + k equals the sum of all positive coefficients in ⃗a (that is, a0 + k =\nmax\n⃗x∈{0,1}n (⃗a · ⃗x)).\nNote on terminology. All the proof systems considered in this paper intend to prove the unsatis-\nfiability over 0, 1 values of collections of clauses (possibly, of translation of the clauses to disjunctions\nof linear equations). In other words, proofs in such proof systems intend to refute the collections\nof clauses, which is to validate their negation. Therefore, throughout this paper we shall sometime\nspeak about refutations and proofs interchangeably, always intending refutations, unless otherwise\nstated.\n1.1. Comparison to Earlier Work. To the best of our knowledge this paper is the first that\nconsiders resolution proofs operating with disjunctions of linear equations. Previous works consid-\nered extensions of resolution over linear inequalities augmented with the cutting planes inference\nrules (the resulting proof system denoted R(CP)). In full generality, we show that resolution over\nlinear equations can polynomially simulate R(CP) when the coefficients in all the inequalities are\npolynomially bounded (however, the converse is not known to hold). On the other hand, we shall\nconsider a certain fragment of resolution over linear equations, in which we do not even know how to\npolynomially simulate cutting planes proofs with polynomially bounded coefficients in inequalities\n(let alone R(CP) with polynomially bounded coefficients in inequalities). We now shortly discuss\nthe previous work on R(CP) and related proof systems.\nExtensions of resolution to disjunctions of linear inequalities were first considered by Kraj ́ıˇcek\n[Kra98] who developed the proof systems LK(CP) and R(CP). The LK(CP) system is a first-order\n(Gentzen-style) sequent calculus that operates with linear inequalities instead of atomic formulas\nand augments the standard first-order sequent calculus inference rules with the cutting planes\ninference rules. The R(CP) proof system is essentially resolution over linear inequalities, that is,\nresolution that operates with disjunctions of linear inequalities instead of clauses.\nThe main motivation of [Kra98] is to extend the feasible interpolation technique and consequently\nthe lower bounds results, from cutting planes and resolution to stronger proof systems. That paper\nestablishes an exponential-size lower bound on a restricted version of R(CP) proofs, namely, when\nthe number of inequalities in each proof-line is O(nε), where n is the number of variables of the\ninitial formulas, ε is a small enough constant and the coefficients in the cutting planes inequalities\nare polynomially bounded.\nOther papers considering extensions of resolution over linear inequalities are the more recent\npapers by Hirsch & Kojevnikov [HK06] and Kojevnikov [Koj07]. The first paper [HK06] considers\na combination of resolution with LP (an incomplete subsystem of cutting planes based on simple\nlinear programming reasoning), with the ‘lift and project’ proof system (L&P), and with the cutting\nplanes proof system. The second paper [Koj07] deals with improving the parameters of the tree-like\nR(CP) lower-bounds obtained in [Kra98].\nWhereas previous results concerned primarily with extending the cutting planes proof system,\nour foremost motivation is to extend and improve previous results on algebraic proof systems\noperating with multilinear formulas obtained in [RT06]. In that paper the concept of multilinear\nproofs was introduced and several basic results concerning multilinear proofs were proved.\nIn\nparticular, polynomial-size proofs of two important combinatorial principles were demonstrated:\nthe functional pigeonhole principle and the Tseitin (mod p) graph tautologies. In the current paper\nwe improve both these results.\nAs mentioned above, motivated by relations with multilinear proofs operating with depth-3 mul-\ntilinear formulas, we shall consider a certain subsystem of resolution over linear equations. For\nthis subsystem we apply twice the interpolation by a communication game technique. The first\n4\n\napplication is of the non-monotone version of the technique, and the second application is of the\nmonotone version. Namely, the first application provides a general (non-monotone) interpolation\ntheorem that demonstrates a polynomial (in the size of refutations) upper bound on interpolants;\nThe proof uses the general method of transforming a refutation into a Karchmer-Wigderson com-\nmunication game for two players, from which a Boolean circuit is then attainable. In particular,\nwe shall apply the interpolation theorem of Kraj ́ıˇcek from [Kra97]. The second application of the\n(monotone) interpolation by a communication game technique is implicit and proceeds by using\nthe lower bound criterion of Bonet, Pitassi & Raz in [BPR97]. This criterion states that (semantic)\nproof systems (of a certain natural and standard kind) whose proof-lines (considered as Boolean\nfunctions) have low communication complexity cannot prove efficiently a certain tautology (namely,\nthe clique-coloring tautologies).\n1.2. Summary of Results. This paper introduces and connects several new concepts and ideas\nwith some known ones. It identifies new extensions of resolution operating with linear equations,\nand relates (a certain) such extension to multilinear proofs. The upper bounds for the pigeonhole\nprinciple and Tseitin mod p formulas in fragments of resolution over linear equations are new. By\ngeneralizing the machinery developed in [RT06], these upper bounds yield new and improved re-\nsults concerning multilinear proofs. The lower bound for the clique-coloring formulas in a fragment\nof resolution over linear equations employs the standard monotone interpolation by a communica-\ntion game technique, and specifically utilizes the theorem of Bonet, Pitassi & Raz from [BPR97].\nThe general (non-monotone) interpolation result for a fragment of resolution over linear equations\nemploys the theorem of Kraj ́ıˇcek from [Kra97]. The upper bound in (the stronger variant of –\nas described in the introduction) resolution over linear equations of the clique-coloring formulas\nfollows that of Atserias, Bonet & Esteban [ABE02]. We now give a detailed outline of the results\nin this paper.\nThe proof systems. In Section 3 we formally define two extensions of resolution of decreasing\nstrength allowing resolution to operate with disjunctions of linear equations. The size of a linear\nequation a1x1 + . . . + anxn = a0 is the sum of all a0, . . . , an written in unary notation. The size of\na disjunction of linear equations is the total size of all linear equations in the disjunction. The size\nof a proof operating with disjunctions of linear equations is the total size of all the disjunctions in\nit.\nR(lin): This is the stronger proof system (described in the introduction) that operates with\ndisjunctions of linear equations with integer coefficients.\nR0(lin): This is a (provably proper) fragment of R(lin). It operates with disjunctions of (arbi-\ntrarily many) linear equations whose variables have constant coefficients, under the restriction that\nevery disjunction can be partitioned into a constant number of sub-disjunctions, where each sub-\ndisjunction either consists of linear equations that differ only in their free-terms or is a (translation\nof a) clause.\nNote that any single linear inequality with Boolean variables can be represented by a disjunction\nof linear equations that differ only in their free-terms (see the example in the introduction section).\nSo the R0(lin) proof system is close to a proof system operating with disjunctions of constant\nnumber of linear inequalities (with constant integral coefficients). In fact, disjunctions of linear\nequations varying only in their free-terms, have more (expressive) strength than a single inequality.\nFor instance, the parity function can be easily represented by a disjunction of linear equations,\nwhile it cannot be represented by a single linear inequality (or even by a disjunction of linear\ninequalities).\nAs already mentioned, the motivation to consider the restricted proof system R0(lin) comes from\nits relation to multilinear proofs operating with depth-3 multilinear formulas (in short, depth-3\n5\n\nmultilinear proofs): R0(lin) corresponds roughly to the subsystem of R(lin) that we know how\nto simulate by depth-3 multilinear proofs via the technique in [RT06] (the technique is based on\nconverting disjunctions of linear forms into symmetric polynomials, which are known to have small\ndepth-3 multilinear formulas). This simulation is then applied in order to improve over known\nupper bounds for depth-3 multilinear proofs, as R0(lin) is already sufficient to efficiently prove\ncertain “hard tautologies”.\nMoreover, we are able to establish an exponential lower bound on\nR0(lin) refutations size (see below for both upper and lower bounds on R0(lin) proofs). We also\nestablish a super-polynomial separation of R(lin) from R0(lin) (via the clique-coloring principle, for\na certain choice of parameters; see below).\nShort refutations. We demonstrate the following short refutations in R0(lin) and R(lin):\n(1) Polynomial-size refutations of the pigeonhole principle in R0(lin);\n(2) Polynomial-size refutations of Tseitin mod p graph formulas in R0(lin);\n(3) Polynomial-size refutations of the clique-coloring formulas in R(lin) (for certain parameters).\nThe refutations here follow by direct simulation of the Res(2) refutations of clique-coloring\nformulas from [ABE02].\nAll the three families of formulas above are prominent “hard tautologies” in proof complexity\nliterature, which means that strong size lower bounds on proofs in various proof systems are known\nfor them (for the exact formulation of these families of formulas see Section 6).\nInterpolation results. We provide a polynomial upper-bound on (non-monotone) interpolants\ncorresponding to R0(lin) refutations; Namely, we show that any R0(lin)-refutation of a given formula\ncan be transformed into a (non-monotone) Boolean circuit computing the corresponding interpolant\nfunction of the formula (if there exists such a function), with at most a polynomial increase in size.\nWe employ the general interpolation theorem of Kraj ́ıˇcek [Kra97] for semantic proof systems.\nLower bounds. We provide the following exponential lower bound:\nTheorem 1. R0(lin) does not have sub-exponential refutations for the clique-coloring formulas.\nThis result is proved by applying a result of Bonet, Pitassi & Raz [BPR97], that (implicitly) use\nthe monotone interpolation by a communication game technique for establishing an exponential-\nsize lower bound on refutations of general semantic proof systems operating with proof-lines of low\ncommunication complexity.\nApplications to multilinear proofs. Multilinear proof systems are (semantic) refutation sys-\ntems operating with multilinear polynomials over a fixed field, where every multilinear polynomial\nis represented by a multilinear arithmetic formula.\nIn this paper we shall consider multilinear\nformulas over fields of characteristic 0 only. The size of a multilinear proof (that is, a proof in\na multilinear proof system) is the total size of all multilinear formulas in the proof (for formal\ndefinitions concerning multilinear proofs see Section 9).\nWe shall first connect multilinear proofs with resolution over linear equations by the following\nresult:\nTheorem 2. Multilinear proofs operating with depth-3 multilinear formulas over characteristic 0\npolynomially-simulate R0(lin).\nAn immediate corollary of this theorem and the upper bounds in R0(lin) described above are\npolynomial-size multilinear proofs for the pigeonhole principle and the Tseitin mod p formulas.\n(1) Polynomial-size depth-3 multilinear refutations for the pigeonhole principle over fields of\ncharacteristic 0. This improves over [RT06] that shows a similar upper bound for a weaker\nprinciple, namely, the functional pigeonhole principle.\n(2) Polynomial-size depth-3 multilinear refutations for the Tseitin mod p graph formulas over\nfields of characteristic 0. These refutations are different than those demonstrated in [RT06],\n6\n\nand further they establish short multilinear refutations of the Tseitin mod p graph formulas\nover any field of characteristic 0 (the proof in [RT06] showed how to refute the Tseitin mod\np formulas by multilinear refutations only over fields that contain a primitive pth root of\nunity).\nRelations with cutting planes proofs. As mentioned in the introduction, a proof system com-\nbining resolution with cutting planes was presented by Kraj ́ıˇcek in [Kra98]. The resulting system\nis denoted R(CP) (see Section 10 for a definition). When the coefficients in the linear inequalities\ninside R(CP) proofs are polynomially bounded, the resulting proof system is denoted R(CP*). We\nestablish the following simulation result:\nTheorem 3. R(lin) polynomially simulates resolution over cutting planes inequalities with polyno-\nmially bounded coefficients R(CP*).\nWe do not know if the converse also holds.\n2. Notation and Background on Propositional Proof Systems\nFor a natural number n, we use [n] to denote {1, . . . , n}. For a vector of n (integral) coefficients\n⃗a and a vector of n variables ⃗x, we denote by ⃗a · ⃗x the scalar product Pn\ni=1 aixi. If ⃗b is another\nvector (of length n), then ⃗a +⃗b denotes the addition of ⃗a and ⃗b as vectors, and c⃗a (for an integer\nc) denotes the product of the scalar c with ⃗a (where, −⃗a denotes −1⃗a). For two linear equations\nL1 : ⃗a · ⃗x = a0 and L2 : ⃗b · ⃗x = b0, their addition (⃗a +⃗b) · ⃗x = a0 + b0 is denoted L1 + L2 (and their\nsubtraction (⃗a −⃗b) · ⃗x = a0 −b0 is denoted L1 −L2). For two Boolean assignments (identified as\n0, 1 strings) α, α′ ∈{0, 1}n we write α′ ≥α if α′\ni ≥αi, for all i ∈[n] (where αi, α′\ni are the ith bits\nof α and α′, respectively).\nWe now recall some basic concepts on propositional proof systems. For background on algebraic\nproof systems (and specifically multilinear proofs) see Section 9.\nResolution. In order to put our work in context, we need to define the resolution refutation system.\nA CNF formula over the variables x1, . . . , xn is defined as follows. A literal is a variable xi or\nits negation ¬xi. A clause is a disjunction of literals. A CNF formula is a conjunction of clauses.\nThe size of a clause is the number of literals in it.\nResolution is a complete and sound proof system for unsatisfiable CNF formulas. Let C and D\nbe two clauses containing neither xi nor ¬xi, the resolution rule allows one to derive C ∨D from\nC ∨xi and D ∨¬xi. The clause C ∨D is called the resolvent of the clauses C ∨xi and D ∨¬xi on\nthe variable xi, and we also say that C ∨xi and D ∨¬xi were resolved over xi. The weakening rule\nallows to derive the clause C ∨D from the clause C, for any two clauses C, D.\nDefinition 2.1 (Resolution). A resolution proof of the clause D from a CNF formula K is a\nsequence of clauses D1, D2, . . . , Dl, such that: (1) each clause Dj is either a clause of K or a\nresolvent of two previous clauses in the sequence or derived by the weakening rule from a previous\nclause in the sequence; (2) the last clause Dl= D. The size of a resolution proof is the sum of all\nthe sizes of the clauses in it. A resolution refutation of a CNF formula K is a resolution proof of\nthe empty clause ✷from K (the empty clause stands for false; that is, the empty clause has no\nsatisfying assignments).\nA proof in resolution (or any of its extensions) is called also a derivation or a proof-sequence.\nEach sequence-element in a proof-sequence is called also a proof-line. A proof-sequence containing\nthe proof-lines D1, . . . , Dlis also said to be a derivation of D1, . . . , Dl.\n7\n\nCook-Reckhow proof systems. Following [CR79], a Cook-Reckhow proof system is a polynomial-\ntime algorithm A that receives a Boolean formula F (for instance, a CNF) and a string π over some\nfinite alphabet (“the (proposed) refutation” of F), such that there exists a π with A(F, π) = 1 if\nand only if F is unsatisfiable. The completeness of a (Cook-Reckhow) proof system (with respect\nto the set of all unsatisfiable Boolean formulas; or for a subset of it, e.g. the set of unsatisfiable\nCNF formulas) stands for the fact that every unsatisfiable formula F has a string π (“the refutation\nof F”) so that A(F, π) = 1. The soundness of a (Cook-Reckhow) proof system stands for the fact\nthat every formula F so that A(F, π) = 1 for some string π is unsatisfiable (in other words, no\nsatisfiable formula has a refutation).\nFor instance, resolution is a Cook-Reckhow proof system, since it is complete and sound for the\nset of unsatisfiable CNF formulas, and given a CNF formula F and a string π it is easy to check in\npolynomial-time (in both F and π) whether π constitutes a resolution refutation of F.\nWe shall also consider proof systems that are not necessarily (that is, not known to be) Cook-\nReckhow proof systems. Specifically, multilinear proof systems (over large enough fields) meet the\nrequirements in the definition of Cook-Reckhow proof systems, except that the condition on A\nabove is relaxed: we allow A to be in probabilistic polynomial-time BPP (which is not known to\nbe equal to deterministic polynomial-time).\nPolynomial simulations of proof systems. When comparing the strength of different proof\nsystems we shall confine ourselves to CNF formulas only. That is, we consider propositional proof\nsystems as proof systems for the set of unsatisfiable CNF formulas. For that purpose, if a proof\nsystem does not operate with clauses directly, then we fix a (direct) translation from clauses to\nthe objects operated by the proof system. This is done for both resolution over linear equations\n(which operate with disjunctions of linear equations) and its fragments, and also for multilinear\nproofs (which operate with multilinear polynomials, represented as multilinear formulas); see for\nexample Subsection 3.1 for such a direct translation.\nDefinition 2.2. Let P1, P2 be two proof systems for the set of unsatisfiable CNF formulas (we\nidentify a CNF formula with its corresponding translation, as discussed above). We say that P2\npolynomially simulates P1 if given a P1 refutation π of a CNF formula F, then there exists a\nrefutation of F in P2 of size polynomial in the size of π. In case P2 polynomially simulates P1 while\nP1 does not polynomially simulates P2 we say that P2 is strictly stronger than P1.\n3. Resolution over Linear Equations and its Subsystems\nThe proof systems we consider in this section are extensions of resolution. Proof-lines in res-\nolution are clauses.\nInstead of this, the extensions of resolution we consider here operate with\ndisjunctions of linear equations with integral coefficients. For this section we use the convention\nthat all the formal variables in the propositional proof systems considered are taken from the set\nX := {x1, . . . , xn}.\n3.1. Disjunctions of Linear Equations. For L a linear equation a1x1 + . . . + anxn = a0, the\nright hand side a0 is called the free-term of L and the left hand side a1x1 + . . . + anxn is called the\nlinear form of L (the linear form can be 0). A disjunction of linear equations is of the following\ngeneral form:\n \na(1)\n1 x1 + . . . + a(1)\nn xn = a(1)\n0\n \n∨· · · ∨\n \na(t)\n1 x1 + . . . + a(t)\nn xn = a(t)\n0\n \n,\n(1)\nwhere t ≥0 and the coefficients a(j)\ni\nare integers (for all 0 ≤i ≤n, 1 ≤j ≤t). We discard duplicate\nlinear equations from a disjunction of linear equations. The semantics of such a disjunction is the\nnatural one: We say that an assignment of integral values to the variables x1, ..., xn satisfies (1)\n8\n\nif and only if there exists j ∈[t] so that the equation a(j)\n1 x1 + . . . + a(j)\nn xn = a(j)\n0\nholds under the\ngiven assignment.\nThe symbol |= denotes the semantic implication relation, that is, for every collection D1, . . . , Dm\nof disjunctions of linear equations,\nD1, . . . , Dm |= D0\nmeans that every assignment of 0, 1 values that satisfies all D1, . . . , Dm also satisfies D0.2 In this\ncase we also say that D1, . . . , Dm semantically imply D0.\nThe size of a linear equation a1x1 + . . . + anxn = a0 is Pn\ni=0 |ai|, i.e., the sum of the bit sizes\nof all ai written in unary notation. Accordingly, the size of the linear form a1x1 + . . . + anxn is\nPn\ni=1 |ai|. The size of a disjunction of linear equations is the total size of all linear equations in it.\nSince all linear equations considered in this paper are of integral coefficients, we shall speak\nof linear equations when we actually mean linear equations with integral coefficients. Similar to\nresolution, the empty disjunction is unsatisfiable and stands for the truth value false.\nTranslation of clauses. As described in the introduction, we can translate any CNF formula to\na collection of disjunctions of linear equations in a direct manner: Every clause W\ni∈I xi ∨W\nj∈J ¬xj\n(where I and J are sets of indices of variables) pertaining to the CNF is translated into the\ndisjunction W\ni∈I(xi = 1) ∨W\nj∈J(xj = 0). For a clause D we denote by eD its translation into a\ndisjunction of linear equations. It is easy to verify that any Boolean assignment to the variables\nx1, . . . , xn satisfies a clause D if and only if it satisfies eD (where true is treated as 1 and false as\n0).\n3.2. Resolution over Linear Equations – R(lin). Defined below is our basic proof system\nR(lin) that enables resolution to reason with disjunctions of linear equations. As we wish to reason\nabout Boolean variables we augment the system with the axioms (xi = 0)∨(xi = 1), for all i ∈[n],\ncalled the Boolean axioms.\nDefinition 3.1 (R(lin)). Let K := {K1, . . . , Km} be a collection of disjunctions of linear equations.\nAn R(lin)-proof from K of a disjunction of linear equations D is a finite sequence π = (D1, ..., Dl)\nof disjunctions of linear equations, such that Dl= D and for every i ∈[l], either Di = Kj for\nsome j ∈[m], or Di is a Boolean axiom (xh = 0) ∨(xh = 1) for some h ∈[n], or Di was deduced\nby one of the following R(lin)-inference rules, using Dj, Dk for some j, k < i:\nResolution: Let A, B be two disjunctions3of linear equations and let L1, L2 be two linear\nequations.\nFrom A ∨L1 and B ∨L2 derive A ∨B ∨(L1 + L2).\nSimilarly, from A ∨L1 and B ∨L2 derive A ∨B ∨(L1 −L2).\nWeakening: From a disjunction of linear equations A derive A∨L , where L is an arbitrary\nlinear equation over X.\nSimplification: From A ∨(0 = k) derive A, where A is a disjunction of linear equations\nand k ̸= 0.\nAn R(lin) refutation of a collection of disjunctions of linear equations K is a proof of the empty\ndisjunction from K. The size of an R(lin)-proof π is the total size of all the disjunctions of linear\nequations in π, denoted |π|.\nSimilar to resolution, in case A ∨B ∨(L1 + L2) is derived from A ∨L1 and B ∨L2 by the\nresolution rule, we say that A ∨L1 and B ∨L2 were resolved over L1 and L2, respectively, and we\n2Alternatively, we can consider assignments of any integral values (instead of only Boolean values) to the variables\nin D1, . . . , Dm, stipulating that the collection D1, . . . , Dm contains all disjunctions of the form (xj = 0) ∨(xj = 1)\nfor all the variables xj ∈X (these formulas force any satisfying assignment to give only 0, 1 values to the variables).\n3Possibly the empty disjunction. This remark also applies to the inference rules below.\n9\n\ncall A ∨B ∨(L1 + L2) the resolvent of A ∨L1 and B ∨L2 (and similarly, when A ∨B ∨(L1 −L2)\nis derived from A ∨L1 and B ∨L2 by the resolution rule; we use the same terminology for both\naddition and subtraction, and it should be clear from the context which operation is actually\napplied). We also describe such an application of the resolution rule by saying that L1 was added\n(resp., subtracted) to (resp. from) L2 in A ∨L1 and B ∨L2.\nIn light of the direct translation between CNF formulas and collections of disjunctions of linear\nequations (described in the previous subsection), we can consider R(lin) to be a proof system for\nthe set of unsatisfiable CNF formulas:\nProposition 1. The R(lin) refutation system is a sound and complete Cook-Reckhow (see Sec-\ntion 2) refutation system for unsatisfiable CNF formulas (translated into unsatisfiable collection of\ndisjunctions of linear equations).\nProof: Completeness of R(lin) (for the set of unsatisfiable CNF formulas) stems from a straight-\nforward simulation of resolution, as we now show.\nClaim 1. R(lin) polynomially simulates resolution.\nProof of claim: Proceed by induction on the length of the resolution refutation to show that any\nresolution derivation of a clause A can be translated with only a linear increase in size into an R(lin)\nderivation of the corresponding disjunction of linear equations eA (see the previous subsection for\nthe definition of eA).\nThe base case: An initial clause A is translated into its corresponding disjunction of linear\nequations eA.\nThe induction step: If a resolution clause A ∨B was derived by the resolution rule from A ∨xi\nand B ∨¬xi, then in R(lin) we subtract (xi = 0) from (xi = 1) in eB ∨(xi = 0) and eA ∨(xi = 1),\nrespectively, to obtain eA ∨eB ∨(0 = 1). Then, using the Simplification rule, we can cut-off(0 = 1)\nfrom eA ∨eB ∨(0 = 1), and arrive at eA ∨eB.\nIf a clause A ∨B was derived in resolution from A by the Weakening rule, then we derive eA ∨eB\nfrom eA by the Weakening rule in R(lin).\nSoundness of R(lin) stems from the soundness of the inference rules (which means that: If D\nwas derived from C, B by the R(lin) resolution rule then any assignment that satisfies both C and\nB also satisfies D; and if D was derived from C by either the Weakening rule or the Simplification\nrule, then any assignment that satisfies C also satisfies D).\nThe R(lin) proof system is a Cook-Reckhow proof system, as it is easy to verify in polynomial-\ntime whether an R(lin) proof-line is inferred, by an application of one of R(lin)’s inference rules,\nfrom a previous proof-line (or proof-lines). Thus, any sequence of disjunctions of linear equations,\ncan be checked in polynomial-time (in the size of the sequence) to decide whether or not it is a\nlegitimate R(lin) proof-sequence.\nIn Section 5 we shall see that a stronger notion of completeness (that is, implicational complete-\nness) holds for R(lin) and its subsystems.\n3.3. Fragment of Resolution over Linear Equations – R0(lin). Here we consider a restriction\nof R(lin), denoted R0(lin). As discussed in the introduction section, R0(lin) is roughly the fragment\nof R(lin) we know how to polynomially simulate with depth-3 multilinear proofs.\nBy results established in the sequel (Sections 6.3 and 8) R(lin) is strictly stronger than R0(lin),\nwhich means that R(lin) polynomially simulates R0(lin), while the converse does not hold.\nR0(lin) operates with disjunctions of (arbitrarily many) linear equations with constant coefficients\n(excluding the free terms), under the following restriction: Every disjunction can be partitioned\n10\n\ninto a constant number of sub-disjunctions, where each sub-disjunction either consists of linear\nequations that differ only in their free-terms or is a (translation of a) clause.\nAs mentioned in the introduction, every linear inequality with Boolean variables can be rep-\nresented by a disjunction of linear equations that differ only in their free-terms. So the R0(lin)\nproof system resembles, to some extent, a proof system operating with disjunctions of constant\nnumber of linear inequalities with constant integral coefficients (on the other hand, it is probable\nthat R0(lin) is stronger than such a proof system, as a disjunction of linear equations that differ\nonly in their free terms is [expressively] stronger than a linear inequality [or even a disjunction of\nlinear inequalities]: the former can define the parity function while the latter cannot).\nExample of an R0(lin)-line:\n(x1 + . . . + xl= 1) ∨· · · ∨(x1 + . . . + xl= l) ∨(xl+1 = 1) ∨· · · ∨(xn = 1),\nfor some 1 ≤l≤n. The next section contains other concrete (and natural) examples of R0(lin)-\nlines.\nLet us define formally what it means to be an R0(lin) proof-line, that is, a proof-line inside an\nR0(lin) proof, called R0(lin)-line:\nDefinition 3.2 (R0(lin)-line). Let D be a disjunction of linear equations whose variables have\nconstant integer coefficients (the free-terms are unbounded). Assume D can be partitioned into a\nconstant number k of sub-disjunctions D1, . . . , Dk, where each Di either consists of (an unbounded)\ndisjunction of linear equations that differ only in their free-terms, or is a translation of a clause (as\ndefined in Subsection 3.1). Then the disjunction D is called an R0(lin)-line.\nThus, any R0(lin)-line is of the following general form:\n_\ni∈I1\n \n⃗a(1) · ⃗x = l(1)\ni\n \n∨· · · ∨\n_\ni∈Ik\n \n⃗a(k) · ⃗x = l(k)\ni\n \n∨\n_\nj∈J\n(xj = bj) ,\n(2)\nwhere k and all at\nr (for r ∈[n] and t ∈[k]) are integer constants and bj ∈{0, 1} (for all j ∈J) (and\nI1, . . . , Ik, J are unbounded sets of indices). Note that a disjunction of clauses can be combined\ninto a single clause. Hence, without loss of generality we can assume that in any R0(lin)-line only\na single (translation of a) clause occurs. This is depicted in (2) (where in addition we have ignored\nin (2) the possibility that the single clause obtained by combining several clauses contains xj ∨¬xj,\nfor some j ∈[n]).\nDefinition 3.3 (R0(lin)). The R0(lin) proof system is a restriction of the R(lin) proof system in\nwhich each proof-line is an R0(lin)-line (as in Definition 3.2).\nFor a completeness proof of R0(lin) see Section 5.4\n4. Reasoning and Counting inside R(lin) and its Subsystems\nIn this section we illustrate a simple way to reason by case-analysis inside R(lin) and its subsys-\ntems. This kind of reasoning will simplify the presentation of proofs inside R(lin) (and R0(lin)) in\nthe sequel (essentially, a similar – though weaker – kind of reasoning is applicable already in reso-\nlution). We will then demonstrate efficient and transparent proofs for simple counting arguments\nthat will also facilitate us in the sequel.\n4The simulation of resolution inside R(lin) (in the proof of Proposition 1) is carried on with each R(lin) proof-line\nbeing in fact a translation of a clause, and hence, an R0(lin)-line (notice that the Boolean axioms of R(lin) are\nR0(lin)-lines). This already implies that R0(lin) is a complete refutation system for the set of unsatisfiable CNF\nformulas. In section 5 we give a proof of a stronger notion of completeness for R0(lin).\n11\n\n4.1. Basic Reasoning inside R(lin) and its Subsystems. Given K a collection of disjunc-\ntions of linear equations {K1, . . . , Km} and C a disjunction of linear equations, denote by K ∨C\nthe collection {K1 ∨C, . . . , Km ∨C}.\nRecall that the formal variables in our proof system are\nx1, . . . , xn.\nLemma 4. Let K be a collection of disjunctions of linear equations, and let z abbreviate some linear\nform with integer coefficients. Let E1, . . . , Elbe ldisjunctions of linear equations. Assume that for\nall i ∈[l] there is an R(lin) derivation of Ei from z = ai and K with size at most s where a1, . . . , al\nare distinct integers. Then, there is an R(lin) proof of Wl\ni=1 Ei from K and (z = a1)∨· · ·∨(z = al),\nwith size polynomial in s and l.\nProof: Denote by D the disjunction (z = a1) ∨· · · ∨(z = al) and by πi the R(lin) proof of Ei from\nK and z = ai (with size at most s), for all i ∈[l]. It is easy to verify that for all i ∈[l] the sequence\nπi ∨W\nj∈[l]\\{i}(z = aj) is an R(lin) proof of Ei ∨W\nj∈[l]\\{i}(z = aj) from K and D. So overall, given\nD and K as premises, there is an R(lin) derivation of size polynomial in s and lof the following\ncollection of disjunctions of linear equations:\nE1 ∨\n_\nj∈[l]\\{1}\n(z = aj), . . . , El∨\n_\nj∈[l]\\{l}\n(z = aj) .\n(3)\nWe now use the Resolution rule to cut-offall the equations (z = ai) inside all the disjunctions\nin (3). Formally, we prove that for every 1 ≤k ≤lthere is a polynomial-size (in s and l) R(lin)\nderivation from (3) of\nE1 ∨· · · ∨Ek ∨\n_\nj∈[l]\\[k]\n(z = aj) ,\n(4)\nand so putting k = l, will conclude the proof of the lemma.\nWe proceed by induction on k.\nThe base case for k = 1 is immediate (from (3)).\nFor the\ninduction case, assume that for some 1 ≤k < lwe already have an R(lin) proof of (4), with size\npolynomial in s and l.\nConsider the line\nEk+1 ∨\n_\nj∈[l]\\{k+1}\n(z = aj) .\n(5)\nWe can now cut-offthe disjunctions W\nj∈[l]\\[k](z = aj) and W\nj∈[l]\\{k+1}(z = aj) from (4) and (5),\nrespectively, using the Resolution rule (since the aj’s in (4) and in (5) are disjoint).\nWe will\ndemonstrate this derivation in some detail now, in order to exemplify a proof carried inside R(lin).\nWe shall be less formal sometime in the sequel.\nResolve (4) with (5) over (z = ak+1) and (z = a1), respectively, to obtain\n(0 = a1 −ak+1) ∨E1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,k+1}\n(z = aj) .\n(6)\nSince a1 ̸= ak+1, we can use the Simplification rule to cut-off(0 = a1 −ak+1) from (6), and we\narrive at\nE1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,k+1}\n(z = aj) .\n(7)\nNow, similarly, resolve (4) with (7) over (z = ak+1) and (z = a2), respectively, and use Simplification\nto obtain\nE1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,2,k+1}\n(z = aj) .\n12\n\nContinue in a similar manner until you arrive at\nE1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,2,...,k,k+1}\n(z = aj) ,\nwhich is precisely what we need.\nUnder the appropriate conditions, Lemma 4 also holds for R0(lin) proofs. This is stated in the\nfollowing lemma.\nLemma 5. Let K be a collection of disjunctions of linear equations, and let z abbreviate a linear\nform with integer coefficients. Let E1, . . . , Elbe ldisjunctions of linear equations. Assume that for\nall i ∈[l] there is an R0(lin) derivation of Ei from z = ai and K with size at most s, where the\nai’s are distinct integers. Then, assuming Wl\ni=1 Ei is an R0(lin)-line, there is an R0(lin) proof of\nWl\ni=1 Ei from K and (z = a1) ∨· · · ∨(z = al), with size polynomial in s and l.\nProof: It can be verified by simple inspection that, under the conditions spelled out in the state-\nment of the lemma, each proof-line in the R(lin) derivations in the proof of Lemma 4 is actually\nan R0(lin)-line.5\nAbbreviations. Lemmas 4 and 5 will sometime facilitate us to proceed inside R(lin) and R0(lin)\nwith a slightly less formal manner. For example, the situation in Lemma 4 above can be depicted\nby saying that “if z = ai implies Ei (with a polynomial-size proof) for all i ∈[l], then Wl\ni=1(z = ai)\nimplies Wl\ni=1 Ei (with a polynomial-size proof)”.\nIn case Wl\ni=1(z = ai) above is just the Boolean axiom (xi = 0) ∨(xi = 1), for some i ∈[n], and\nxi = 0 implies E0 and xi = 1 implies E1 (both with polynomial-size proofs), then to simplify the\nwriting we shall sometime not mention the Boolean axiom at all. For example, the latter situation\ncan be depicted by saying that “if xi = 0 implies E0 with a polynomial-size proof and xi = 1 implies\nE1 with a polynomial-size proof, then we can derive E0 ∨E1 with a polynomial-size proof”.\n4.2. Basic Counting inside R(lin) and R0(lin). In this subsection we illustrate how to effi-\nciently prove several basic counting arguments inside R(lin) and R0(lin). This will facilitate us in\nshowing short proofs for hard tautologies in the sequel. In accordance with the last paragraph in\nthe previous subsection, we shall carry the proofs inside R(lin) and R0(lin) with a slightly less rigor.\nLemma 6. Let z1 abbreviate ⃗a · ⃗x and z2 abbreviate ⃗b · ⃗x. Let D1 be W\nα∈A(z1 = α) and let D2 be\nW\nβ∈B (z2 = β), where A, B are two (finite) sets of integers. Then there is a polynomial-size (in the\nsize of D1, D2) R(lin) proof from D1, D2 of:\n_\nα∈A,β∈B\n(z1 + z2 = α + β) .\n(8)\nMoreover, if ⃗a and ⃗b consist of constant integers (which means that D1, D2 are R0(lin)-lines), then\nthere is a polynomial-size (in the size of D1, D2) R0(lin) proof of (8) from D1, D2.\nProof: Denote the elements of A by α1, . . . , αk. In case z1 = αi, for some i ∈[k] then we can add\nz1 = αi to every equation in W\nβ∈B (z2 = β) to get W\nβ∈B(z1 + z2 = αi + β). Therefore, there exist\nk R(lin) proofs, each with polynomial-size (in |D1| and |D2|), of\n_\nβ∈B\n(z1 + z2 = α1 + β) ,\n_\nβ∈B\n(z1 + z2 = α2 + β) ,\n. . .\n,\n_\nβ∈B\n(z1 + z2 = αk + β)\n5Note that when the proofs of Ei from z = ai, for all i ∈[l], are all done inside R0(lin), then the linear form z\nought to have constant coefficients.\n13\n\nfrom z1 = α1, z1 = α2 ,. . . ,z1 = αk, respectively.\nThus, by Lemma 4, we can derive\n_\nα∈A,β∈B\n(z1 + z2 = α + β)\n(9)\nfrom D1 and D2 in a polynomial-size (in |D1| and |D2|) R(lin)-proof. This concludes the first part\nof the lemma.\nAssume that ⃗a and ⃗b consist of constant coefficients only. Then by inspecting the R(lin)-proof\nof (9) from D1 and D2 demonstrated above (and by using Lemma 5 instead of Lemma 4), one can\nverify that this proof is in fact carried inside R0(lin).\nAn immediate corollary of Lemma 6 is the efficient formalization in R(lin) of the following obvious\ncounting argument: If a linear form equals some value in the interval (of integer numbers) [a0, a1]\nand another linear form equals some value in [b0, b1] (for some a0 ≤a1 and b0 ≤b1), then their\naddition equals some value in [a0 + b0, a1 + b1]. More formally:\nCorollary 7. Let z1 abbreviate ⃗a·⃗x and z2 abbreviate ⃗b·⃗x. Let D1 be (z1 = a0)∨(z1 = a0 +1) . . .∨\n(z1 = a1), and let D2 be (z2 = b0) ∨(z2 = b0 + 1) . . . ∨(z2 = b1). Then there is a polynomial-size\n(in the size of D1, D2) R(lin) proof from D1, D2 of\n(z1 + z2 = a0 + b0) ∨(z1 + z2 = a0 + b0 + 1) ∨. . . ∨(z1 + z2 = a1 + b1) .\n(10)\nMoreover, if ⃗a and ⃗b consist of constant integers (which means that D1, D2 are R0(lin)-lines), then\nthere is a polynomial-size (in the size of D1, D2) R0(lin) proofs of (10) from D1, D2.\nLemma 8. Let ⃗a · ⃗x be a linear form with n variables, and let A := {⃗a · ⃗x | ⃗x ∈{0, 1}n} be the set\nof all possible values of ⃗a ·⃗x over Boolean assignments to ⃗x. Then there is a polynomial-size, in the\nsize of the linear form ⃗a · ⃗x,6 R(lin) proof of\n_\nα∈A\n(⃗a · ⃗x = α) .\n(11)\nMoreover, if the coefficients in ⃗a are constants, then there is a polynomial-size (in the size of ⃗a · ⃗x)\nR0(lin) proof of (11).\nProof: Without loss of generality, assume that all the coefficients in ⃗a are nonzero. Consider the\nBoolean axiom (x1 = 0) ∨(x1 = 1) and the (first) coefficient a1 from ⃗a. Assume that a1 ≥1. Add\n(x1 = 0) to itself a1 times, and arrive at (a1x1 = 0) ∨(x1 = 1). Then, in the resulted line, add\n(x1 = 1) to itself a1 times, until the following is reached:\n(a1x1 = 0) ∨(a1x1 = a1) .\nSimilarly, in case a1 ≤−1 we can subtract (|a1| + 1 many times) (x1 = 0) from itself in (x1 =\n0) ∨(x1 = 1), and then subtract (|a1| + 1 many times) (x1 = 1) from itself in the resulted line.\nIn the same manner, we can derive the disjunctions: (a2x2 = 0) ∨(a2x2 = a2), . . . , (anxn =\n0) ∨(anxn = an).\nConsider (a1x1 = 0) ∨(a1x1 = a1) and (a2x2 = 0) ∨(a2x2 = a2). From these two lines, by\nLemma 6, there is a polynomial-size in |a1| + |a2| derivation of:\n(a1x1 + a2x2 = 0) ∨(a1x1 + a2x2 = a1) ∨(a1x1 + a2x2 = a2) ∨(a1x1 + a2x2 = a1 + a2) .\n(12)\nIn a similar fashion, now consider (a3x3 = 0) ∨(a3x3 = a3) and apply again Lemma 6, to obtain\n_\nα∈A′\n(a1x1 + a2x2 + a3x3 = α) ,\n(13)\n6Recall that the size of ⃗a · ⃗x is Pn\ni=1 |ai|, that is, the size of the unary representation of ⃗a.\n14\n\nwhere A′ are all possible values to a1x1 + a2x2 + a3x3 over Boolean assignments to x1, x2, x3. The\nderivation of (13) is of size polynomial in |a1| + |a2| + |a3|.\nContinue to consider, successively, all other lines (a4x4 = 0) ∨(a4x4 = a4), . . . , (anxn = 0) ∨\n(anxn = an), and apply the same reasoning. Each step uses a derivation of size at most polynomial\nin Pn\ni=1 |ai|. And so overall we reach the desired line (11), with a derivation of size polynomial in\nthe size of ⃗a · ⃗x. This concludes the first part of the lemma.\nAssume that ⃗a consists of constant coefficients only. Then by inspecting the R(lin)-proof demon-\nstrated above (and by using the second part of Lemma 6), one can see that this proof is in fact\ncarried inside R0(lin).\nLemma 9. There is a polynomial-size (in n) R0(lin) proof from\n(x1 = 1) ∨· · · ∨(xn = 1)\n(14)\nof\n(x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n) .\n(15)\nProof: We show that for every i ∈[n], there is a polynomial-size (in n) R0(lin) proof from (xi = 1)\nof (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n). This concludes the proof since, by Lemma 5,\nwe then can derive from (14) (with a polynomial-size (in n) R0(lin) proof) the disjunction (14) in\nwhich each (xi = 1) (for all i ∈[n]) is replace by (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n),\nwhich is precisely the disjunction (15) (note that (15) is an R0(lin)-line).\nClaim 2. For every i ∈[n], there is a a polynomial-size (in n) R0(lin) proof from (xi = 1) of\n(x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n).\nProof of claim: By Lemma 8, for every i ∈[n] there is a polynomial-size (in n) R0(lin) proof\n(using only the Boolean axioms) of\n(x1 + . . . + xi−1 + xi+1 + . . . + xn = 0) ∨· · · ∨(x1 + . . . + xi−1 + xi+1 + . . . + xn = n −1) .\n(16)\nNow add successively (xi = 1) to every equation in (16) (note that this can be done in R0(lin)).\nWe obtain precisely (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n).\nLemma 10. There is a polynomial-size (in n) R0(lin) proof of (x1+. . .+xn = 0)∨(x1+. . .+xn = 1)\nfrom the collection of disjunctions consisting of (xi = 0) ∨(xj = 0), for all 1 ≤i < j ≤n.\nProof: We proceed by induction on n. The base case for n = 1 is immediate from the Boolean\naxiom (x1 = 0) ∨(x1 = 1). Assume we already have a polynomial-size proof of\n(x1 + . . . + xn = 0) ∨(x1 + . . . + xn = 1).\n(17)\nIf xn+1 = 0 we add xn+1 = 0 to both of the equations in (17), and reach:\n(x1 + . . . + xn+1 = 0) ∨(x1 + . . . + xn+1 = 1).\n(18)\nOtherwise, xn+1 = 1, and so we can cut-off(xn+1 = 0) in all the initial disjunctions (xi = 0) ∨\n(xn+1 = 0), for all 1 ≤i ≤n. We thus obtain (x1 = 0), . . . , (xn = 0). Adding together (x1 =\n0), . . . , (xn = 0) and (xn+1 = 1) we arrive at\n(x1 + . . . + xn+1 = 1) .\n(19)\nSo overall, either (18) holds or (19) holds; and so (using Lemma 5) we arrive at the disjunction of\n(19) and (18), which is precisely (18).\n15\n\n5. Implicational Completeness of R(lin) and its Subsystems\nIn this section we provide a proof of the implicational completeness of R(lin) and its subsystems.\nWe shall need this property in the sequel (see Section 6.2). The implicational completeness of a\nproof system is a stronger property than mere completeness. Essentially, a system is implicationally\ncomplete if whenever something is semantically implied by a set of initial premises, then it is\nalso derivable from the initial premises. In contrast to this, mere completeness means that any\ntautology (or in case of a refutation system, any unsatisfiable set of initial premises) has a proof in\nthe system (respectively, a refutation in the system). As a consequence, the proof of implicational\ncompleteness in this section establishes an alternative completeness proof to that obtained via\nsimulating resolution (see Proposition 1). Note that we are not concerned in this section with the\nsize of the proofs, but only with their existence.\nRecall the definition of the semantic implication relation |= from Section 3.1. Formally, we say\nthat R(lin) is implicationally complete if for every collection of disjunctions of linear equations\nD0, D1, . . . , Dm, it holds that D1, . . . , Dm |= D0 implies that there is an R(lin) proof of D0 from\nD1, . . . , Dm.\nTheorem 11. R(lin) is implicationally complete.\nProof: We proceed by induction on n, the number of variables x1, . . . , xn in D0, D1, . . . , Dm.\nThe base case n = 0. We need to show that D1, . . . , Dm |= D0 implies that there is an R(lin)\nproof of D0 from D1, . . . , Dm, where all Di’s (for 0 ≤i ≤m) have no variables but only constants.\nThis means that each Di is a disjunction of equations of the form (0 = a0) for some integer a0 (if\na linear equation have no variables, then the left hand side of this equation must be 0; see Section\n3.1).\nThere are two cases to consider. In the first case D0 is satisfiable. Since D0 has no variables,\nthis means precisely that D0 is the equation (0 = 0). Thus, D0 can be derived easily from any\naxiom in R(lin) (for instance, by subtracting each equation in (x1 = 0) ∨(x1 = 1) from itself, to\nreach (0 = 0) ∨(0 = 0), which is equal to (0 = 0), since we discard duplicate equations inside\ndisjunctions).\nIn the second case D0 is unsatisfiable. Thus, since D1, . . . , Dm |= D0, there is no assignment sat-\nisfying all D1, . . . , Dm. Hence, there must be at least one unsatisfiable disjunction Di in D1, . . . , Dm\n(as a disjunction with no variables is either tautological or unsatisfiable). Such an unsatisfiable Di\nis a disjunction of zero or more unsatisfiable equations of the form (0 = a0), for some integer a0 ̸= 0.\nWe can then use Simplification to cut-offall the unsatisfiable equations in Di to reach the empty\ndisjunction. By the Weakening rule, we can now derive D0 from the empty disjunction.\nThe induction step. Assume that the theorem holds for disjunctions with n variables. Let the\nunderlying variables of D0, D1, . . . , Dm be x1, . . . , xn+1, and assume that\nD1, . . . , Dm |= D0 .\n(20)\nWe write the disjunction D0 as:\nt_\nj=1\n n\nX\ni=1\na(j)\ni xi + a(j)\nn+1xn+1 = a(j)\n0\n!\n,\n(21)\nwhere the a(j)\ni ’s are integer coefficients. We need to show that there is an R(lin) proof of D0 from\nD1, . . . , Dm.\nLet D be a disjunction of linear equations, let xi be a variable and let b ∈{0, 1}. We shall\ndenote by D↾xi=b the disjunction D, where in every equation in D the variable xi is substituted by\nb, and the constant terms in the left hand sides of all resulting equations (after substituting b for\n16\n\nxi) switch sides (and change signs, obviously) to the right hand sides of the equations (we have to\nswitch sides of constant terms, as by definition linear equations in R(lin) proofs have all constant\nterms appearing only on the right hand sides of equations).\nWe now reason (slightly) informally inside R(lin) (as illustrated in Section 4.1). Fix some b ∈\n{0, 1}, and assume that xn+1 = b. Then, from D1, . . . , Dm we can derive (inside R(lin)):\nD1↾xn+1=b, . . . , Dm↾xn+1=b .\n(22)\nThe only variables occurring in (22) are x1, . . . , xn. From assumption (20) we clearly have D1↾xn+1=b\n, . . . , Dm↾xn+1=b |= D0↾xn+1=b. And so by the induction hypothesis there is an R(lin) derivation of\nD0↾xn+1=b from D1↾xn+1=b, . . . , Dm↾xn+1=b. So overall, assuming that xn+1 = b, there is an R(lin)\nderivation of D0↾xn+1=b from D1, . . . , Dm.\nWe now consider the two possible cases: xn+1 = 0 and xn+1 = 1.\nIn case xn+1 = 0, by the above discussion, we can derive D0↾xn+1=0 from D1, . . . , Dm. For every\nj ∈[t], add successively (a(j)\nn+1 times) the equation xn+1 = 0 to the jth equation in D0↾xn+1=0 (see\n(21)). We thus obtain precisely D0.\nIn case xn+1 = 1, again, by the above discussion, we can derive D0↾xn+1=1 from D1, . . . , Dm. For\nevery j ∈[t], add successively (a(j)\nn+1 times) the equation xn+1 = 1 to the jth equation in D0↾xn+1=1\n(recall that we switch sides of constant terms in every linear equation after the substitution of xn+1\nby 1 is performed in D0↾xn+1=1). Again, we obtain precisely D0.\nBy inspecting the proof of Theorem 11, it is possible to verify that if all the disjunctions\nD0, , . . . , Dm are R0(lin)-lines (see Definition 3.2), then the proof of D0 in R(lin) uses only R0(lin)-\nlines as well. Therefore, we have:\nCorollary 12. R0(lin) is implicationally complete.\nRemark 1. Corollary 12 states that any R0(lin)-line that is semantically implied by a set of initial\nR0(lin)-lines, is in fact derivable in R0(lin) from the initial R0(lin)-lines. On the other hand, it is\npossible that a certain proof of the same R0(lin)-line inside R(lin) will be significantly shorter than\nthe proof inside R0(lin). Indeed, we shall see in Section 8 that for certain CNF formulas R(lin) has\na super-polynomial speed-up over R0(lin).\n6. Short Proofs for Hard Tautologies\nIn this section we show that R0(lin) is already enough to admit small proofs for “hard” counting\nprinciples like the pigeonhole principle and the Tseitin graph formulas for constant degree graphs.\nOn the other hand, as we shall see in Section 8, R0(lin) inherits the same weakness that cutting\nplanes proofs have with respect to the clique-coloring tautologies. Nevertheless, we can efficiently\nprove the clique-coloring principle in (the stronger system) R(lin), but not by using R(lin) “ability\nto count”, rather by using its (straightforward) ability to simulate Res(2) proofs (that is, resolution\nproofs extended to operate with 2-DNF formulas, instead of clauses).\n6.1. The\nPigeonhole\nPrinciple\nTautologies\nin\nR0(lin). This\nsubsection\nillustrates\npolynomial-size R0(lin) proofs of the pigeonhole principle.\nThis will allow us to establish\npolynomial-size multilinear proofs operating with depth-3 multilinear formulas of the pigeonhole\nprinciple (in Section 9).\nThe m to n pigeonhole principle states that m pigeons cannot be mapped one-to-one into n < m\nholes. The negation of the pigeonhole principle, denoted ¬PHPm\nn , is formulated as an unsatisfiable\nCNF formula as follows (where clauses are translated to disjunctions of linear equations):\nDefinition 6.1. The ¬PHPm\nn is the following set of clauses:\n(1) Pigeons axioms:\n(xi,1 = 1) ∨· · · ∨(xi,n = 1), for all 1 ≤i ≤m;\n17\n\n(2) Holes axioms:\n(xi,k = 0) ∨(xj,k = 0),\nfor all 1 ≤i < j ≤m and for all 1 ≤k ≤n.\nThe intended meaning of each propositional variable xi,j is that the ith pigeon is mapped to the\njth hole.\nWe now describe a polynomial-size in n refutation of ¬PHPm\nn inside R0(lin). For this purpose it\nis sufficient to prove a polynomial-size refutation of the pigeonhole principle when the number of\npigeons m equals n + 1 (because the set of clauses pertaining to ¬PHPn+1\nn\nis already contained in\nthe set of clauses pertaining to ¬PHPm\nn , for any m > n). Thus, we fix m = n+1. In this subsection\nwe shall say a proof in R0(lin) is of polynomial-size, always intending polynomial-size in n (unless\notherwise stated).\nBy Lemma 9, for all i ∈[m] we can derive from the Pigeon axiom (for the ith pigeon):\n(xi,1 + . . . + xi,n = 1) ∨· · · ∨(xi,1 + . . . + xi,n = n)\n(23)\nwith a polynomial-size R0(lin) proof.\nBy Lemma 10, from the Hole axioms we can derive, with a polynomial-size R0(lin) proof\n(x1,j + . . . + xm,j = 0) ∨(x1,j + . . . + xm,j = 1),\n(24)\nfor all j ∈[n].\nLet S abbreviate the sum of all formal variables xi,j. In other words,\nS :=\nX\ni∈[m],j∈[n]\nxi,j .\nLemma 13. There is a polynomial-size R0(lin) proof from (23) (for all i ∈[m]) of\n(S = m) ∨(S = m + 1) · · · ∨(S = m · n).\nProof: For every i ∈[m] fix the abbreviation zi := xi,1 + . . . + xi,n.\nThus, by (23) we have\n(zi = 1) ∨· · · ∨(zi = n).\nConsider (z1 = 1) ∨· · · ∨(z1 = n) and (z2 = 1) ∨· · · ∨(z2 = n). By Corollary 7, we can derive\nfrom these two lines\n(z1 + z2 = 2) ∨(z1 + z2 = 3) ∨· · · ∨(z1 + z2 = 2n)\n(25)\nwith a polynomial-size R0(lin) proof.\nNow, consider (z3 = 1) ∨· · · ∨(z3 = n) and (25). By Corollary 7 again, from these two lines we\ncan derive with a polynomial-size R0(lin) proof:\n(z1 + z2 + z3 = 3) ∨(z1 + z2 + z3 = 4) ∨· · · ∨(z1 + z2 + z3 = 3n) .\n(26)\nContinuing in the same way, we eventually arrive at\n(z1 + . . . + zm = m) ∨(z1 + . . . + zm = m + 1) ∨· · · ∨(z1 + . . . + zm = m · n) ,\nwhich concludes the proof, since S equals z1 + . . . + zm.\nLemma 14. There is a polynomial-size R0(lin) proof from (24) of\n(S = 0) ∨· · · ∨(S = n).\nProof: For all j ∈[n], fix the abbreviation yj := x1,j + . . . + xm,j.\nThus, by (24) we have\n(yj = 0) ∨(yj = 1), for all j ∈[n]. Now the proof is similar to the proof of Lemma 8, except that\nhere single variables are abbreviations of linear forms.\nIf y1 = 0 then we can add y1 to the two sums in (y2 = 0) ∨(y2 = 1), and reach (y1 + y2 =\n0) ∨(y1 + y2 = 1) and if y1 = 1 we can do the same and reach (y1 + y2 = 1) ∨(y1 + y2 = 2). So, by\nLemma 5, we can derive with a polynomial-size R0(lin) proof\n(y1 + y2 = 0) ∨(y1 + y2 = 1) ∨(y1 + y2 = 2) .\n(27)\n18\n\nNow, we consider the three cases in (27): y1 +y2 = 0 or y1 +y2 = 1 or y1 +y2 = 2, and the clause\n(y3 = 0) ∨(y3 = 1). We arrive in a similar manner at (y1 + y2 + y3 = 0) ∨· · · ∨(y1 + y2 + y3 = 3).\nWe continue in the same way until we arrive at (S = 0) ∨· · · ∨(S = n).\nTheorem 15. There is a polynomial-size R0(lin) refutation of the m to n pigeonhole principle\n¬PHPm\nn .\nProof: By Lemmas 13 and 14 above, all we need is to show a polynomial-size refutation of (S =\nm) ∨· · · ∨(S = m · n) and (S = 0) ∨· · · ∨(S = n).\nSince n < m, for all 0 ≤k ≤n, if S = k then using the Resolution and Simplification rules we\ncan cut-offall the sums in (S = m) ∨· · · ∨(S = m · n) and arrive at the empty clause. Thus, by\nLemma 5, there is a polynomial-size R0(lin) proof of the empty clause from (S = 0) ∨· · · ∨(S = n)\nand (S = m) ∨· · · ∨(S = m · n).\n6.2. Tseitin mod p Tautologies in R0(lin). This subsection establishes polynomial-size R0(lin)\nproofs of Tseitin graph tautologies (for constant degree graphs). This will allow us (in Section 9)\nto extend the multilinear proofs of the Tseitin mod p tautologies to any field of characteristic 0\n(the proofs in [RT06] required working over a field containing a primitive pth root of unity when\nproving the Tseitin mod p tautologies; for more details see Section 9).\nTseitin mod p tautologies (introduced in [BGIP01]) are generalizations of the (original, mod 2)\nTseitin graph tautologies (introduced in [Tse68]). To build the intuition for the generalized version,\nwe start by describing the (original) Tseitin mod 2 principle.\nLet G = (V, E) be a connected\nundirected graph with an odd number of vertices n. The Tseitin mod 2 tautology states that there\nis no sub-graph G′ = (V, E′), where E′ ⊆E, so that for every vertex v ∈V , the number of edges\nfrom E′ incident to v is odd. This statement is valid, since otherwise, summing the degrees of all\nthe vertices in G′ would amount to an odd number (since n is odd), whereas this sum also counts\nevery edge in E′ twice, and so is even.\nAs mentioned above, the Tseitin mod 2 principle was generalized by Buss et al. [BGIP01] to\nobtain the Tseitin mod p principle. Let p ≥2 be some fixed integer and let G = (V, E) be a\nconnected undirected r-regular graph with n vertices and no double edges. Let G′ = (V, E′) be the\ncorresponding directed graph that results from G by replacing every (undirected) edge in G with\ntwo opposite directed edges. Assume that n ≡1 (mod p). Then, the Tseitin mod p principle states\nthat there is no way to assign to every edge in E′ a value from {0, . . . , p −1}, so that:\n(i): For every pair of opposite directed edges e, ̄e in E′, with assigned values a, b, respectively,\na + b ≡0 (mod p); and\n(ii): For every vertex v in V , the sum of the values assigned to the edges in E′ coming out of\nv is congruent to 1 (mod p).\nThe Tseitin mod p principle is valid, since if we sum the values assigned to all edges of E′ in\npairs we obtain 0 (mod p) (by (i)), where summing them by vertices we arrive at a total value of 1\n(mod p) (by (ii) and since n ≡1 (mod p)). We shall see in what follows, that this simple counting\nargument can be carried on in a natural (and efficient) way already inside R0(lin).\nAs an unsatisfiable propositional formula (in CNF form) the negation of the Tseitin mod p\nprinciple is formulated by assigning a variable xe,i for every edge e ∈E′ and every residue i modulo\np. The variable xe,i is an indicator variable for the fact that the edge e has an associated value i.\nThe following are the clauses of the Tseitin mod p CNF formula (as translated to disjunctions of\nlinear equations).\nDefinition 6.2 (Tseitin mod p formulas (¬TseitinG,p)). Let p ≥2 be some fixed integer and\nlet G = (V, E) be a connected undirected r-regular graph with n vertices and no double edges, and\n19\n\nassume that n ≡1 (mod p). Let G′ = (V, E′) be the corresponding directed graph that results\nfrom G by replacing every (undirected) edge in G with two opposite directed edges.\nGiven a vertex v ∈V , denote the edges in E′ coming out of v by e[v, 1], . . . , e[v, r] and define the\nfollowing set of (translation of) clauses:\nMODp,1(v):=\n( r_\nk=1\n(xe[v,k],ik = 0)\n i1, . . . , ir ∈{0, . . . , p −1} and\nr\nX\nk=1\nik ̸≡1 mod p\n)\n.\nThe Tseitin mod p formula, denoted ¬TseitinG,p, consists of the following (translation) of clauses:\n1.\np−1\nW\ni=0\n(xe,i = 1) , for all e ∈E′\n(expresses that every edge is assigned at least one value from 0, . . . , p −1);\n2. (xe,i = 0) ∨(xe,j = 0) , for all i ̸= j ∈{0, . . . , p −1} and all e ∈E′\n(expresses that every edge is assigned at most one value from 0, . . . , p −1);\n3. (xe,i = 1) ∨(x ̄e,p−i = 0) and (xe,i = 0) ∨(x ̄e,p−i = 1), 7\nfor all two opposite directed edges e, ̄e ∈E′ and all i ∈{0, . . . , p −1}\n(expresses condition (i) of the Tseitin mod p principle above);\n4. MODp,1(v) , for all v ∈V\n(expresses condition (ii) of the Tseitin mod p principle above).\nNote that for every edge e ∈E′, the polynomials of (1,2) in Definition 6.2, combined with the\nBoolean axioms of R0(lin), force any collection of edge-variables xe,0, . . . , xe,p−1 to contain exactly\none i ∈{0, . . . , p −1} so that xe,i = 1.\nAlso, it is easy to verify that, given a vertex v ∈V ,\nany assignment σ of 0, 1 values (to the relevant variables) satisfies both the disjunctions of (1,2)\nand the disjunctions of MODp,1(v) if and only if σ corresponds to an assignment of values from\n{0, . . . , p −1} to the edges coming out of v that sums up to 1 (mod p).\nUntil the rest of this subsection we fix an integer p ≥2 and a connected undirected r-regular\ngraph G = (V, E) with n vertices and no double edges, such that n ≡1 mod p and r is a constant.\nAs in Definition 6.2, we let G′ = (V, E′) be the corresponding directed graph that results from G\nby replacing every (undirected) edge in G with two opposite directed edges. We now proceed to\nrefute ¬TseitinG,p inside R0(lin) with a polynomial-size (in n) refutation.\nGiven a vertex v ∈V , and the edges in E′ coming out of v, denoted e[v, 1], . . . , e[v, r], define the\nfollowing abbreviation:\nαv :=\nr\nX\nj=1\np−1\nX\ni=0\ni · xe[v,j],i .\n(28)\nLemma 16. Let v ∈V be any vertex in G′. Then there is a constant-size R0(lin) proof from\n¬TseitinG,p of the following disjunction:\nr−1\n_\nl=0\n(αv = 1 + l· p) .\n(29)\nProof: Let Tv ⊆¬TseitinG,p be the set of all disjunctions of the form (1,2,4) from Definition 6.2\nthat contain only variables pertaining to vertex v (that is, all the variables xe,i, where e ∈E′ is an\nedge coming out of v, and i ∈{0, . . . , p −1}).\n7If i = 0 then x ̄e,p−i denotes x ̄e,0.\n20\n\nClaim 3. Tv semantically implies (29), that is:8\nTv |=\nr−1\n_\nl=0\n(αv = 1 + l· p) .\nProof of claim: Let σ be an assignment of 0, 1 values to the variables in Tv that satisfies both the\ndisjunctions of (1,2) and the disjunctions of MODp,1(v) in Definition 6.2. As mentioned above (the\ncomment after Definition 6.2), such a σ corresponds to an assignment of values from {0, . . . , p −1}\nto the edges coming out of v, that sums up to 1 mod p. This means precisely that αv = 1 mod p\nunder the assignment σ. Thus, there exists a nonnegative integer k, such that αv = 1 + kp under\nσ.\nIt remains to show that k ≤r −1 (and so the only possible values that αv can get under σ\nare 1, 1 + p, 1 + 2p, . . . , 1 + (r −1)p). Note that because σ gives the value 1 to only one variable\nfrom xe[v,j],0, . . . , xe[v,j],p−1 (for every j ∈[r]), then the maximal value that αv can have under σ is\nr(p −1). Thus, 1 + kp ≤rp −r and so k ≤r −1.\nFrom Claim 3 and from the implicational completeness of R0(lin) (Corollary 12), there exists an\nR0(lin) derivation of (29) from Tv. It remains to show that this derivation is of constant-size.\nSince the degree r of G′ and the modulus p are both constants, both Tv and (29) have constant\nnumber of variables and constant coefficients (including the free-terms). Thus, there is a constant-\nsize R0(lin) derivation of (29) from Tv.\nLemma 17. There is a polynomial-size (in n) R0(lin) derivation from ¬TseitinG,p of the following\ndisjunction:\n(r−1)·n\n_\nl=0\n X\nv∈V\nαv = n + l· p\n!\n.\nProof: Simply add successively all the equations pertaining to disjunctions (29), for all vertices\nv ∈V .\nFormally, we show that for every subset of vertices V ⊆V , with |V| = k, there is a\npolynomial-size (in n) R0(lin) derivation from ¬TseitinG,p of\n(r−1)·k\n_\nl=0\n X\nv∈V\nαv = k + l· p\n!\n,\n(30)\nand so putting V = V , will conclude the proof.\nWe proceed by induction on the size of V. The base case, |V| = 1, is immediate from Lemma 16.\nAssume that we already derived (30) with a polynomial-size (in n) R0(lin) proof, for some V ⊂V ,\nsuch that |V| = k < n. Let u ∈V \\ V. By Lemma 16, we can derive\nr−1\n_\nl=0\n(αu = 1 + l· p)\n(31)\nfrom ¬TseitinG,p with a constant-size proof. Now, by Lemma 6, each linear equation in (31) can\nbe added to each linear equation in (30), with a polynomial-size (in n) R0(lin) proof. This results\nin the following disjunction:\n(r−1)·(k+1)\n_\nl=0\n \n \nX\nv∈V∪{u}\nαv = k + 1 + l· p\n \n ,\n8Recall that we only consider assignments of 0, 1 values to variables when considering the semantic implication\nrelation |=.\n21\n\nwhich is precisely what we need to conclude the induction step.\nLemma 18. Let e, ̄e be any pair of opposite directed edges in G′ and let i ∈{0, . . . , p −1}. Let\nTe ⊆¬TseitinG,p be the set of all disjunctions of the form (1,2,3) from Definition 6.2 that contain\nonly variables pertaining to edges e, ̄e (that is, all the variables xe,j, x ̄e,j, for all j ∈{0, . . . , p −1}).\nThen, there is a constant-size R0(lin) proof from Te of the following disjunction:\n(i · xe,i + (p −i) · x ̄e,p−i = 0) ∨(i · xe,i + (p −i) · x ̄e,p−i = p) .\n(32)\nProof: First note that Te semantically implies\n(xe,i + x ̄e,p−i = 0) ∨(xe,i + x ̄e,p−i = 2) .\n(33)\nThe number of variables in Te and (33) is constant. Hence, there is a constant-size R0(lin)-proof\nof (32) from Te. Also note that\n(xe,i + x ̄e,p−i = 0) ∨(xe,i + x ̄e,p−i = 2) |=\n(i · xe,i + (p −i) · x ̄e,p−i = 0) ∨(i · xe,i + (p −i) · x ̄e,p−i = p) .\n(34)\nTherefore, there is also an R0(lin)-proof of constant-size from Te of the lower line in (34).\nWe are now ready to complete the polynomial-size R0(lin) refutation of ¬TseitinG,p. Using the\ntwo prior lemmas, the refutation idea is simple, as we now explain. Observe that\nX\nv∈V\nαv =\nX\n{e, ̄e}⊆E′\ni∈{0,...,p−1}\n(i · xe,i + (p −i) · x ̄e,p−i) ,\n(35)\nwhere by {e, ̄e} ⊆E′ we mean that e, ̄e is pair of opposite directed edges in G′.\nDerive by Lemma 17 the disjunction\n(r−1)·n\n_\nl=0\n X\nv∈V\nαv = n + l· p\n!\n.\n(36)\nThis disjunction expresses the fact that P\nv∈V αv = 1 mod p (since n = 1 mod p). On the other\nhand, using Lemma 18, we can “sum together” all the equations (32) (for all {e, ̄e} ⊆E′ and all\ni ∈{0, . . . , p −1}), to obtain a disjunction expressing the statement that\nX\n{e, ̄e}⊆E′\ni∈{0,...,p−1}\n(i · xe,i + (p −i) · x ̄e,p−i) = 0\nmod p .\nBy Equation (35), we then obtain the desired contradiction. This idea is formalized in the proof of\nthe following theorem:\nTheorem 19. Let G = (V, E) be an r-regular graph with n vertices, where r is a constant. Fix\nsome modulus p. Then, there are polynomial-size (in n) R0(lin) refutations of ¬TseitinG,p.\nProof: First, use Lemma 17 to derive\n(r−1)·n\n_\nl=0\n X\nv∈V\nαv = n + l· p\n!\n.\n(37)\nSecond, use Lemma 18 to derive\n(i · xe,i + (p −i) · x ̄e,p−i = p) ∨(i · xe,i + (p −i) · x ̄e,p−i = 0) ,\n(38)\nfor every pair of opposite directed edges in G′ = (V, E′) (as in Definition 6.2) and every residue\ni ∈{0, . . . , p −1}.\n22\n\nWe now reason inside R0(lin).\nPick a pair of opposite directed edges e, ̄e and a residue i ∈\n{0, . . . , p −1}. If i · xe,i + (p −i) · x ̄e,p−i = 0, then subtract this equation successively from every\nequation in (37). We thus obtain a new disjunction, similar to that of (37), but which does not\ncontain the xe,i and x ̄e,p−i variables, and with the same free-terms.\nOtherwise, i·xe,i+(p−i)·x ̄e,p−i = p, then subtract this equation successively from every equation\nin (37). Again, we obtain a new disjunction, similar to that of (37), but which does not contain\nthe xe,i and x ̄e,p−i variables, and such that p is subtracted from every free-term in every equation.\nSince, by assumption, n ≡1 mod p, the free-terms in every equation are (still) equal 1 mod p.\nSo overall, in both cases (i · xe,i + (p −i) · x ̄e,p−i = 0 and i · xe,i + (p −i) · x ̄e,p−i = p) we obtained\na new disjunction with all the free-terms in equations equal 1 mod p.\nWe now continue the same process for every pair e, ̄e of opposite directed edges in G′ and\nevery residue i. Eventually, we discard all the variables xe,i in the equations, for every e ∈E′\nand i ∈{0, . . . , p −1}, while all the free-terms in every equation remain to be equal 1 mod p.\nTherefore, we arrive at a disjunction of equations of the form (0 = γ) for some γ = 1 mod p.\nBy using the Simplification rule we can cut-offall such equations, and arrive finally at the empty\ndisjunction.\n6.3. The Clique-Coloring Principle in R(lin). In this section we observe that there are\npolynomial-size R(lin) proofs of the clique-coloring principle (for certain, weak, parameters). This\nimplies, in particular, that R(lin) does not possess the feasible monotone interpolation property\n(see more details on the interpolation method in Section 7).\nAtserias, Bonet & Esteban [ABE02] demonstrated polynomial-size Res(2) refutations of the\nclique-coloring formulas (for certain weak parameters; Theorem 20). Thus, it is sufficient to show\nthat R(lin) polynomially-simulates Res(2) proofs (Proposition 2). This can be shown in a straight-\nforward manner. As noted in the first paragraph of Section 6, because the proofs of the clique-\ncoloring formula we discuss here only follow the proofs inside Res(2), then in fact these proofs do\nnot take any advantage of the capacity “to count” inside R(lin) (this capacity is exemplified, for\ninstance, in Section 4.2).\nWe start with the clique-coloring formulas (these formulas will also be used in Section 8). These\nformulas express the clique-coloring principle that has been widely used in the proof complexity\nliterature (cf., [BPR97], [Pud97], [Kra97], [Kra98], [ABE02], [Kra07]). This principle is based on\nthe following basic combinatorial idea. Let G = (V, E) be an undirected graph with n vertices and\nlet k′ < k be two integers. Then, one of the following must hold:\n(i): The graph G does not contain a clique with k vertices;\n(ii): The graph G is not a complete k′-partite graph.\nIn other words, there is no way to\npartition G into k′ subgraphs G1, . . . , Gk′, such that every Gi is an independent set, and\nfor all i ̸= j ∈[k′], all the vertices in Gi are connected by edges (in E) to all the vertices in\nGj.\nObviously, if Item (ii) above is false (that is, if G is a complete k′-partite graph), then there\nexists a k′-coloring of the vertices of G; hence the name clique-coloring for the principle.\nThe propositional formulation of the (negation of the) clique-coloring principle is as follows.\nEach variable pi,j, for all i ̸= j ∈[n], is an indicator variable for the fact that there is an edge in\nG between vertex i and vertex j. Each variable ql,i, for all l∈[k] and all i ∈[n], is an indicator\nvariable for the fact that the vertex i in G is the lth vertex in the k-clique. Each variable rl,i, for\nall l∈[k′] and all i ∈[n], is an indicator variable for the fact that the vertex i in G pertains to the\nindependent set Gl.\nDefinition 6.3. The negation of the clique-coloring principle consists of the following unsatisfiable\ncollection of clauses (as translated to disjunctions of linear equations), denoted ¬cliquen\nk,k′:\n23\n\n(i) (ql,1 = 1) ∨· · · ∨(ql,n = 1), for all l∈[k]\n(expresses that there exists at least one vertex in G which constitutes the lth vertex of\nthe k-clique);\n(ii) (ql,i = 0) ∨(ql,j = 0), for all i ̸= j ∈[n], l∈[k]\n(expresses that there exists at most one vertex in G which constitutes the lth vertex of\nthe k-clique);\n(iii) (ql,i = 0) ∨(ql′,i = 0), for all i ∈[n], l̸= l′ ∈[k]\n(expresses that the ith vertex of G cannot be both the lth and the l′th vertex of the\nk-clique);\n(iv) (ql,i = 0) ∨(ql′,j = 0) ∨(pi,j = 1), for all l̸= l′ ∈[k], i ̸= j ∈[n]\n(expresses that if both the vertices i and j in G are in the k-clique, then there is an edge\nin G between i and j);\n(v) (r1,i = 1) ∨· · · ∨(rk′,i = 1), for all i ∈[n]\n(expresses that every vertex of G pertains to at least one independent set);\n(vi) (rl,i = 0) ∨(rl′,i = 0), for all l̸= l∈[k′], i ∈[n]\n(expresses that every vertex of G pertains to at most one independent set);\n(vii) (pi,j = 0) ∨(rt,i = 0) ∨(rt,j = 0), for all i ̸= j ∈[n], t ∈[k′]\n(expresses that if there is an edge between vertex i and j in G, then i and j cannot be\nin the same independent set);\nRemark 2. Our formulation of the clique-coloring formulas above is similar to the one used by\n[BPR97], except that we consider also the pi,j variables (we added the (iv) clauses and changed\naccordingly the (vii) clauses). This is done for the sake of clarity of the contradiction itself, and\nalso to make it clear that the formulas are in the appropriate form required by the interpolation\nmethod (see Section 7 for details on the interpolation method). By resolving over the pi,j variables\nin (iv) and (vii), one can obtain precisely the collection of clauses in [BPR97].\nAtserias, Bonet & Esteban [ABE02] demonstrated polynomial-size (in n) Res(2) refutations of\n¬cliquen\nk,k′, when k = √n and k′ = (log n)2/8 log log n. These are rather weak parameters, but\nthey suffice to establish the fact that Res(2) does not possess the feasible monotone interpolation\nproperty.\nThe Res(2) proof system (also called 2-DNF resolution), first considered in [Kra01], is resolution\nextended to operate with 2-DNF formulas, defined as follows.\nA 2-term is a conjunction of up to two literals. A 2-DNF is a disjunction of 2-terms. The size\nof a 2-term is the number of literals in it (that is, either 1 or 2). The size of a 2-DNF is the total\nsize of all the 2-terms in it.\nDefinition 6.4 (Res(2)). A Res(2) proof of a 2-DNF D from a collection K of 2-DNFs is a\nsequence of 2-DNFs D1, D2, . . . , Ds , such that Ds = D, and every Dj is either from K or was\nderived from previous line(s) in the sequence by the following inference rules:\nCut: Let A, B be two 2-DNFs.\nFrom A∨V2\ni=1 li and B∨W2\ni=1 ¬li derive A∨B, where the li’s are (not necessarily distinct)\nliterals (and ¬li is the negation of the literal li).\nAND-introduction: Let A, B be two 2-DNFs and l1, l2 two literals.\nFrom A ∨l1 and B ∨l2 derive A ∨B ∨V2\ni=1 li.\nWeakening: From a 2-DNF A derive A ∨V2\ni=1 li , where the li’s are (not necessarily dis-\ntinct) literals.\nA Res(2) refutation of a collection of 2-DNFs K is a Res(2) proof of the empty disjunction ✷from\nK (the empty disjunction stands for false). The size of a Res(2) proof is the total size of all the\n2-DNFs in it.\n24\n\nGiven a collection K of 2-DNFs we translate it into a collection of disjunctions of linear equations\nvia the following translation scheme. For a literal l, denote by bl the translation that maps a variable\nxi into xi, and ¬xi into 1−xi. A 2-term l1 ∧l2 is first transformed into the equation bl1 +bl2 = 2, and\nthen moving the free-terms in the left hand side of bl1 + bl2 = 2 (in case there are such free-terms)\nto the right hand side; So that the final translation of l1 ∧l2 has only a single free-term in the\nright hand side. A disjunction of 2-terms (that is, a 2-DNF) D = W\ni∈I(li,1 ∧li,2) is translated into\nthe disjunction of the translations of the 2-terms, denoted by bD. It is clear that every assignment\nsatisfies a 2-DNF D if and only if it satisfies bD.\nProposition 2. R(lin) polynomially simulates Res(2). In other words, if π is a Res(2) proof of D\nfrom a collection of 2-DNFs K1, . . . , Kt, then there is an R(lin) proof of bD from bK1, . . . , bKt whose\nsize is polynomial in the size of π.\nThe proof of Proposition 2 proceeds by induction on the length (that is, the number of proof-\nlines) in the Res(2) proof. This is pretty straightforward and similar to the simulation of resolution\nby R(lin), as illustrated in the proof of Proposition 1. We omit the details.\nTheorem 20 ([ABE02]). Let k = √n and k′ = (log n)2/8 log log n. Then ¬cliquen\nk,k′ has Res(2)\nrefutations of size polynomial in n.\nThus, Proposition 2 yields the following:\nCorollary 21. Let k, k′ be as in Theorem 20. Then ¬cliquen\nk,k′ has R(lin) refutations of size\npolynomial in n.\nThe following corollary is important (we refer the reader to Section A in the Appendix for the\nnecessary relevant definitions concerning the feasible monotone interpolation property and to Section\n7 for explanation and definitions concerning the general [non-monotone] interpolation method).\nCorollary 22. R(lin) does not possess the feasible monotone interpolation property.\nRemark 3. The proof of ¬cliquen\nk,k′ inside Res(2) demonstrated in [ABE02] (and hence, also\nthe corresponding proof inside R(lin)) proceeds along the following lines. First reduce ¬cliquen\nk,k′\nto the k to k′ pigeonhole principle.\nFor the appropriate values of the parameters k and k′ —\nand specifically, for the values in Theorem 20 — there is a short resolution proof of the k to k′\npigeonhole principle (this was shown by Buss & Pitassi [BP97]); (this resolution proof is polynomial\nin the number of pigeons k, but not in the number of holes k′, which is exponentially smaller than\nk).9 Therefore, in order to conclude the refutation of ¬cliquen\nk,k′ inside Res(2) (or inside R(lin)), it\nsuffices to simulate the short resolution refutation of the k to k′ pigeonhole principle. It is important\nto emphasize this point: After reducing, inside R(lin), ¬cliquen\nk,k′ to the pigeonhole principle, one\nsimulates the resolution refutation of the pigeonhole principle, and this has nothing to do with\nthe small-size R0(lin) refutations of the pigeonhole principle demonstrated in Section 6.1. This is\nbecause, the reduction (inside R(lin)) of ¬cliquen\nk,k′ to the k to k′ pigeonhole principle, results in\na substitution instance of the pigeonhole principle formulas; in other words, the reduction results\nin a collection of disjunctions that are similar to the pigeonhole principle disjunctions where each\noriginal pigeonhole principle variable is substituted by some big formula (and, in particular, these\ndisjunctions are not R0(lin)-lines at all). (Note that R0(lin) does not admit short proofs of the\nclique-coloring formulas as we show in Section 8.)\n9Whenever k ≥2k′ the k to k′ pigeonhole principle is referred to as the weak pigeonhole principle.\n25\n\n7. Interpolation Results for R0(lin)\nIn this section we study the applicability of the feasible (non-monotone) interpolation technique\nto R0(lin) refutations. In particular, we show that R0(lin) admits a polynomial (in terms of the\nR0(lin)-proofs) upper bound on the (non-monotone) circuit-size of interpolants. In the next section\nwe shall give a polynomial upper bound on the monotone circuit-size of interpolants, but only in the\ncase that the interpolant corresponds to the clique-coloring formulas (whereas, in this section we are\ninterested in the general case; that is, upper bounding circuit-size of interpolants corresponding to\nany formula [of the prescribed type; see below]). First, we shortly describe the feasible interpolation\nmethod and explain how this method can be applied to obtain (sometime, conditional) lower bounds\non proof size. Explicit usage of the interpolation method in proof complexity goes back to [Kra94].\nLet Ai(⃗p, ⃗q), i ∈I, and Bj(⃗p,⃗r), j ∈J, (I and J are sets of indices) be a collection of formulas (for\ninstance, a collection of disjunctions of linear equations) in the displayed variables only. Denote by\nA(⃗p, ⃗q) the conjunction of all Ai(⃗p, ⃗q), i ∈I, and by B(⃗p,⃗r), the conjunction of all Bj(⃗p,⃗r), j ∈J.\nAssume that ⃗p, ⃗q,⃗r are pairwise disjoint sets of distinct variables, and that there is no assignment\nthat satisfies both A(⃗p, ⃗q) and B(⃗p,⃗r). Fix an assignment ⃗α to the variables in ⃗p. The ⃗p variables\nare the only common variables of the Ai’s and the Bj’s. Therefore, either A(⃗α, ⃗q) is unsatisfiable\nor B(⃗α,⃗r) is unsatisfiable.\nThe interpolation technique transforms a refutation of A(⃗p, ⃗q) ∧B(⃗p,⃗r), in some proof system,\ninto a circuit (usually a Boolean circuit) separating those assignments ⃗α (for ⃗p) for which A(⃗α, ⃗q)\nis unsatisfiable, from those assignments ⃗α for which B(⃗α,⃗r) is unsatisfiable (the two cases are not\nnecessarily exclusive, so if both cases hold for an assignment, the circuit can output either that the\nfirst case holds or that the second case holds). In other words, given a refutation of A(⃗p, ⃗q)∧B(⃗p,⃗r),\nwe construct a circuit C(⃗p), called the interpolant, such that\nC(⃗α) = 1\n=⇒\nA(⃗α, ⃗q) is unsatisfiable, and\nC(⃗α) = 0\n=⇒\nB(⃗α,⃗r) is unsatisfiable.\n(39)\n(Note that if U denotes the set of those assignments ⃗α for which A(⃗α, ⃗q) is satisfiable, and V\ndenotes the set of those assignments ⃗α for which B(⃗α,⃗r) is satisfiable, then U and V are disjoint\n[since A(⃗p, ⃗q) ∧B(⃗p,⃗r) is unsatisfiable], and C(⃗p) separates U from V ; see Definition 7.2 below.)\nAssume that for a proof system P the transformation from refutations of A(⃗p, ⃗q), B(⃗p,⃗r) into\nthe corresponding interpolant circuit C(⃗p) results in a circuit whose size is polynomial in the size\nof the refutation. Then, an exponential lower bound on circuits for which (39) holds, implies an\nexponential lower bound on P-refutations of A(⃗p, ⃗q), B(⃗p,⃗r).\n7.1. Interpolation for Semantic Refutations. We now lay out the basic concepts needed to\nformally describe the feasible interpolation technique.\nWe use the general notion of semantic\nrefutations (which generalizes any standard propositional refutation system). We shall use a close\nterminology to that in [Kra97].\nDefinition 7.1 (Semantic refutation). Let N be a fixed natural number and let E1, . . . , Ek ⊆\n{0, 1}N, where Tk\ni=1 Ei = ∅. A semantic refutation from E1, . . . , Ek is a sequence D1, . . . , Dm ⊆\n{0, 1}N with Dm = ∅and such that for every i ∈[m], Di is either one of the Ej’s or is deduced\nfrom two previous Dj, Dl, 1 ≤j, l< i, by the following semantic inference rule:\n• From A, B ⊆{0, 1}N deduce any C, such that C ⊇(A ∩B).\nObserve that any standard propositional refutation (with inference rules that derive from at\nmost two proof-lines, a third line) can be regarded as a semantic refutation: just substitute each\nrefutation-line by the set of its satisfying assignments; and by the soundness of the inference rules\napplied in the refutation, it is clear that each refutation-line (considered as the set of assignments\nthat satisfy it) is deduced by the semantic inference rule from previous refutation-lines.\n26\n\nDefinition 7.2 (Separating circuit). Let U, V ⊆{0, 1}n, where U ∩V = ∅, be two disjoint sets. A\nBoolean circuit C with n input variables is said to separate U from V if C(⃗x) = 1 for every ⃗x ∈U,\nand C(⃗x) = 0 for every ⃗x ∈V. In this case we also say that U and V are separated by C.\nConvention: In what follows we sometime identify a Boolean formula with the set of its satisfying\nassignments.\nNotation: For two (or more) binary strings u, v ∈{0, 1}∗, we write (u, v) to denote the concate-\nnation of the u with v (where v comes to the right of u, obviously).\nLet N = n+s+t be fixed from now on. Let A1, . . . , Ak ⊆{0, 1}n+s and let B1, . . . , Bl⊆{0, 1}n+t.\nDefine the following two sets of assignments of length n (formally, 0, 1 strings of length n) that\ncan be extended to satisfying assignments of A1, . . . , Ak and B1, . . . , Bl, respectively (formally,\nthose 0, 1 string of length n + s and n + t, that are contained in all A1, . . . , Ak and B1, . . . , Bl,\nrespectively):\nUA :=\n(\nu ∈{0, 1}n\n ∃q ∈{0, 1}s , (u, q) ∈\nk\\\ni=1\nAi\n)\n,\nVB :=\n(\nv ∈{0, 1}n\n ∃r ∈{0, 1}t , (v, r) ∈\nl\\\ni=1\nBi\n)\n.\nDefinition 7.3 (polynomial upper bounds on interpolants). Let P be a propositional\nrefutation system.\nAssume that ⃗p, ⃗q,⃗r are pairwise disjoint sets of distinct variables, where\n⃗p has n variables, ⃗q has s variables and ⃗r has t variables.\nLet A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and\nB1(⃗p,⃗r), . . . , Bl(⃗p,⃗r) be two collections of formulas with the displayed variables only.\nAssume\nthat for any such A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r), if there exists a P-refutation of\nsize S for A1(⃗p, ⃗q) ∧· · · ∧Ak(⃗p, ⃗q) ∧B1(⃗p,⃗r) ∧. . . ∧Bl(⃗p,⃗r) then there exists a Boolean circuit\nseparating UA from VB of size polynomial in S.10 In this case we say that P has a polynomial upper\nbound on interpolant circuits.\n7.1.1. The Communication Game Technique. The feasible interpolation via communication game\ntechnique is based on transforming proofs into Boolean circuits, where the size of the resulting\ncircuit depends on the communication complexity of each proof-line. This technique goes back to\n[IPU94] and [Razb95] and was subsequently applied and extended in [BPR97] and [Kra97] ([IPU94]\nand [BPR97] did not use explicitly the notion of interpolation of tautologies or contradictions). We\nshall employ the interpolation theorem of Kraj ́ıˇcek in [Kra97], that demonstrates how to transform\na small semantic refutation with each proof-line having low communication complexity into a small\nBoolean circuit separating the corresponding sets.\nThe underlying idea of the interpolation via communication game technique is that a (semantic)\nrefutation, where each proof-line is of small (that is, logarithmic) communication complexity, can be\ntransformed into an efficient communication protocol for the Karchmer-Wigderson game (following\n[KW88]) for two players. In the Karchmer-Wigderson game the first player knows some binary\nstring u ∈U and the second player knows some different binary string v ∈V , where U and V are\ndisjoint sets of strings. The two players communicate by sending information bits to one another\n(following a protocol previously agreed on). The goal of the game is for the two players to decide\non an index i such that the ith bit of u is different from the ith bit of v. An efficient Karchmer-\nWigderson protocol (by which we mean a protocol that requires the players to exchange at most a\nlogarithmic number of bits in the worst-case) can then be transformed into a small circuit separating\n10Here UA and VB are defined as above, by identifying the Ai(⃗p, ⃗q)’s and the Bi(⃗p,⃗r)’s with the sets of assignments\nthat satisfy them.\n27\n\nU from V (see Definition 7.2). This efficient transformation from protocols for Karchmer-Wigderson\ngames (described in a certain way) into circuits, was demonstrated by Razborov in [Razb95]. So\noverall, given a semantic refutation with proof-lines of low communication complexity, one can\nobtain a small circuit for separating the corresponding sets.\nFirst, we need to define the concept of communication complexity in a suitable way for the\ninterpolation theorem.\nDefinition 7.4 (Communication complexity). Let N = n + s + t and A ⊆{0, 1}N. Let u, v ∈\n{0, 1}n, qu ∈{0, 1}s, rv ∈{0, 1}t. Denote by ui, vi the ith bit of u, v, respectively, and let (u, qu, rv)\nand (v, qu, rv) denote the concatenation of strings u, qu, rv and v, qu, rv, respectively. Consider the\nfollowing three tasks:\n(1) Decide whether (u, qu, rv) ∈A;\n(2) Decide whether (v, qu, rv) ∈A;\n(3) If one of the following holds:\n(i) (u, qu, rv) ∈A and (v, qu, rv) ̸∈A; or\n(ii) (u, qu, rv) ̸∈A and (v, qu, rv) ∈A,\nthen find an i ∈[n], such that ui ̸= vi;\nConsider a game between two players, Player I and Player II, where Player I knows u ∈{0, 1}n , qu ∈\n{0, 1}s and Player II knows v ∈{0, 1}n , rv ∈{0, 1}t. The two players communicate by exchanging\nbits of information between them (following a protocol previously agreed on). The communication\ncomplexity of A, denoted CC(A), is the minimal (over all protocols) number of bits that players I\nand II need to exchange in the worst-case in solving each of Tasks 1, 2 and 3 above.11\nFor A ⊆{0, 1}n+s define\n ̇A :=\n \n(a, b, c)\n (a, b) ∈A and c ∈{0, 1}t \n,\nwhere a and b range over {0, 1}n and {0, 1}s, respectively. Similarly, for B ⊆{0, 1}n+t define\n ̇B :=\n \n(a, b, c)\n (a, c) ∈B and b ∈{0, 1}t \n,\nwhere a and c range over {0, 1}n and {0, 1}t, respectively.\nTheorem 23 ([Kra97]). Let A1, . . . , Ak ⊆{0, 1}n+s and B1, . . . , Bl⊆{0, 1}n+t. Let D1, . . . , Dm\nbe a semantic refutation from ̇A1, . . . , ̇Ak and ̇B1, . . . , ̇Bl. Assume that CC(Di) ≤ζ, for all i ∈[m].\nThen, the sets UA and VB (as defined above) can be separated by a Boolean circuit of size (m +\nn)2O(ζ).\nIn light of Theorem 23, to demonstrate that a certain propositional refutation system P possesses\na polynomial upper bound on interpolant circuits (see Definition 7.3) it suffices to show that any\nproof-line of P induces a set of assignments with at most a logarithmic (in the number of variables)\ncommunication complexity (Definition 7.4).\n7.2. Polynomial Upper Bounds on Interpolants for R0(lin). Here we apply Theorem 23 to\nshow that R0(lin) has polynomial upper bounds on its interpolant circuits. Again, in what follows\nwe sometime identify a disjunction of linear equations with the set of its satisfying assignments.\nTheorem 24. R0(lin) has a polynomial upper bounds on interpolant circuits (Definition 7.3).\nAccording to the paragraph after Theorem 23, all we need in order to establish Theorem 24 is\nthe following lemma:\n11In other words, CC(A) is the minimal number ζ, for which there exists a protocol, such that for every input\n(u, qu to Player I and v, rv to Player II) and every task (from Tasks 1, 2 and 3), the players need to exchange at most\nζ bits in order to solve the task.\n28\n\nLemma 25. Let D be an R0(lin)-line with N variables and let eD be the set of assignments that\nsatisfy D.12Then, CC( eD) ≤O(log N).\nProof: Let N = n + s + t (and so eD ∈{0, 1}n+s+t). For the sake of convenience we shall assume\nthat the N variables in D are partitioned into (pairwise disjoint) three groups ⃗p := (p1 . . . , pn),\n⃗q := (q1, . . . , qs) and ⃗r := (r1, . . . , rt). Let u, v ∈{0, 1}n, qu ∈{0, 1}s, rv ∈{0, 1}t. Assume that\nPlayer I knows u, qu and Player II knows v, rv.\nBy the definition of an R0(lin)-line (see Definition 3.2) we can partition the disjunction D into\na constant number of disjuncts, where one disjunct is a (possibly empty, translation of a) clause in\nthe ⃗p, ⃗q,⃗r variables (see Section 3.1), and all other disjuncts have the following form:\n_\ni∈I\n \n⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r = li\n \n,\n(40)\nwhere I is (an unbounded) set of indices, li are integer numbers, for all i ∈I, and ⃗a,⃗b,⃗c denote\nvectors of n, s and t constant coefficients, respectively.\nLet us denote the (translation of the) clause from D in the ⃗p, ⃗q,⃗r variables by\nP ∨Q ∨R ,\nwhere P, Q and R denote the (translated) sub-clauses consisting of the ⃗p, ⃗q and ⃗r variables,\nrespectively.\nWe need to show that by exchanging O(log N) bits, the players can solve each of Tasks 1, 2 and\n3 from Definition 7.4, correctly.\nTask 1: The players need to decide whether (u, qu, rv) ∈eD. Player II, who knows rv, computes\nthe numbers ⃗c · rv, for every ⃗c pertaining to every disjunct of the form shown in Equation (40)\nabove. Then, Player II sends the (binary representation of) these numbers to Player I. Since there\nare only a constantly many such numbers and the coefficients in every ⃗c are also constants, this\namounts to O(log t) ≤O(log N) bits that Player II sends to Player I. Player II also computes the\ntruth value of the sub-clause R, and sends this (single-bit) value to Player I.\nNow, it is easy to see that Player I has sufficient data to compute by herself/himself whether\n(u, qu, rv) ∈eD (Player I can then send a single bit informing Player II whether (u, qu, rv) ∈eD).\nTask 2: This is analogous to Task 1.\nTask 3: Assume that (u, qu, rv) ∈eD and (v, qu, rv) ̸∈eD (the case (u, qu, rv) ̸∈eD and (v, qu, rv) ∈eD\nis analogous).\nThe first rounds of the protocol are completely similar to that described in Task 1 above: Player\nII, who knows rv, computes the numbers ⃗c · rv, for every ⃗c pertaining to every disjunct of the form\nshown in Equation (40) above. Then, Player II sends the (binary representation of) these numbers\nto Player I. Player II also computes the truth value of the sub-clause R, and sends this (single-bit)\nvalue to Player I. Again, this amounts to O(log N) bits that Player II sends to Player I.\nBy assumption (that (u, qu, rv) ∈eD and (v, qu, rv) ̸∈eD) the players need to deal only with the\nfollowing two cases:\nCase 1: The assignment (u, qu, rv) satisfies the clause P ∨Q∨R while (v, qu, rv) falsifies P ∨Q∨R.\nThus, it must be that ⃗u satisfies the sub-clause P while ⃗v falsifies P. This means that for any i ∈[n]\nsuch that ui sets to 1 a literal in P (there ought to exist at least one such i), it must be that ui ̸= vi.\nTherefore, all that Player I needs to do is to send the (binary representation of) index i to Player\nII. (This amounts to O(log N) bits that Player I sends to Player II.)\n12The notation eD has nothing to do with the same notation used in Section 3.\n29\n\nCase 2: There is some linear equation\n⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r = l\n(41)\nin D, such that ⃗a · u +⃗b · qu + ⃗c · rv = l. Note that (by assumption that (v, qu, rv) ̸∈eD) it must\nalso hold that: ⃗a · v +⃗b · qu + ⃗c · rv ̸= l(and so there is an i ∈[n], such that ui ̸= vi). Player I can\nfind linear equation (41), as he/she already received from Player II all the possible values of ⃗c · ⃗r\n(for all possible ⃗c ’s in D).\nRecall that the left hand side of a linear equation ⃗d · ⃗x = lis called the linear form of the\nequation. By the definition of an R0(lin)-line there are only constant many distinct linear forms\nin D. Since both players know these linear forms, we can assume that each linear form has some\nindex associated to it by both players. Player I sends to Player II the index of the linear form\n⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r from (41) in D. Since there are only constantly many such linear forms in D, it\ntakes only constant number of bits to send this index.\nNow both players need to apply a protocol for finding an i ∈[n] such that ui ̸= vi, where\n⃗a · ⃗u +⃗b · qu + ⃗c · rv = land ⃗a · ⃗v +⃗b · qu + ⃗c · rv ̸= l. Thus, it remains only to prove the following\nclaim:\nClaim 4. There is a communication protocol in which Player I and Player II need at most O(log N)\nbits of communication in order to find an i ∈[n] such that ui ̸= vi (under the above conditions).\nProof of claim: We invoke the well-known connection between Boolean circuit-depth and com-\nmunication complexity. Let f : {0, 1}N →{0, 1} be a Boolean function. Denote by dp(f) the\nminimal depth of a Boolean circuit computing f. Consider a game between two players: Player I\nknows some ⃗x ∈{0, 1}N and Player II knows some other ⃗y ∈{0, 1}N, such that f(⃗x) = 1 while\nf(⃗y) = 0. The goal of the game is to find an i ∈[N] such that xi ̸= yi. Denote by CC′(f) the\nminimal number of bits needed for the two players to communicate (in the worst case13) in order\nto solve this game.14 Then, for any function f it is known that dp(f) = CC′(f) (see [KW88]).\nTherefore, to conclude the proof of the claim it is enough to establish that the function f :\n{0, 1}N →{0, 1} that receives the input variables ⃗p, ⃗q,⃗r and computes the truth value of ⃗a · ⃗p +⃗b ·\n⃗q + ⃗c · ⃗r = lhas Boolean circuit of depth O(log N). In case all the coefficients in ⃗a,⃗b,⃗c are 1, it is\neasy to show15 that there is a Boolean circuit of depth O(log N) that computes the function f. In\nthe case that the coefficients in ⃗a,⃗b,⃗c are all constants, it is easy to show, by a reduction to the\ncase where all coefficients are 1,16 that there is a Boolean circuit of depth O(log N) that computes\nthe function f. We omit the details.\n8. Size Lower Bounds\nIn this section we establish an exponential-size lower bound on R0(lin) refutations of the clique-\ncoloring formulas. We shall employ the theorem of Bonet, Pitassi & Raz in [BPR97] that provides\nexponential-size lower bounds for any semantic refutation of the clique-coloring formulas, having\nlow communication complexity in each refutation-line.\n13Over all inputs ⃗x, ⃗y such that f(⃗x) = 1 and f(⃗y) = 0.\n14The measure CC′ is basically the same as CC defined earlier.\n15Using the known O(log N)-depth Boolean circuits for the threshold functions.\n16For instance, consider the simple case where we have only a single variable. That is, let c be a constant and\nassume that we wish to construct a circuit that computes c · x = l, for some integer l. Then, we take a circuit that\ncomputes the function f : {0, 1}c →{0, 1} that outputs the truth value of y1 + . . . + yc = l(thus, in f all coefficients\nare 1’s); and to compute c · x = lwe only have to substitute each yi in the circuit with the variable x.\n30\n\nFirst we recall the strong lower bound obtained by Alon & Boppana [AB87] (improving over\n[Razb85]; see also [And85]) for the (monotone) clique separator functions, defined as follows (a\nfunction f : {0, 1}n →{0, 1} is called monotone if for all α ∈{0, 1}n, α′ ≥α implies f(α′) ≥f(α)):\nDefinition 8.1 (Clique separator). A monotone boolean function Qn\nk,k′ is called a clique separator\nif it interprets its inputs as the edges of a graph on n vertices, and outputs 1 on every input\nrepresenting a k-clique, and 0 on every input representing a complete k′-partite graph (see Section\n6.3).\nRecall that a monotone Boolean circuit is a circuit that uses only monotone Boolean gates (for\ninstance, only the fan-in two gates ∧, ∨).\nTheorem 26 ([AB87]). Let k, k′ be integers such that 3 ≤k′ < k and k\n√\nk′ ≤n/(8 log n), then\nevery monotone Boolean circuit that computes a clique separator function Qn\nk,k′ requires size at least\n1\n8\n \nn\n4k\n√\nk′ log n\n (\n√\nk′+1)/2\n.\nFor the next theorem, we need a slightly different (and weaker) version of communication com-\nplexity, than that in Definition 7.4.\nDefinition 8.2 (Communication complexity (second definition)). Let X denote n Boolean variables\nx1, . . . , xn, and let S1, S2 be a partition of X into two disjoint sets of variables. The communication\ncomplexity of a Boolean function f : {0, 1}n →{0, 1} is the number of bits needed to be exchanged\nby two players, one knowing the values given to the S1 variables and the other knowing the values\ngiven to S2 variables, in the worst-case, over all possible partitions S1 and S2.\nTheorem 27 ([BPR97]). Every semantic refutation of ¬cliquen\nk,k′ (for k′ < k) with m refutation-\nlines and where each refutation-line (considered as a the characteristic function of the line) has\ncommunication complexity (as in Definition 8.2) ζ, can be transformed into a monotone circuit of\nsize m · 23ζ+1 that computes a separating function Qn\nk,k′.\nIn light of Theorem 26, in order to be able to apply Theorem 27 to R0(lin), and arrive at an\nexponential-size lower bound for R0(lin) refutations of the clique-coloring formulas, it suffices to\nshow that R0(lin) proof-lines have logarithmic communication complexity:\nLemma 28. Let D be an R0(lin)-line with N variables. Then, the communication complexity (as\nin Definition 8.2) of D is at most O(log N) (where D is identified here with the characteristic\nfunction of D).\nProof: The proof is similar to the proof of Lemma 25 for solving Task 1 (and the analogous Task\n2) in Definition 7.4.\nBy direct calculations we obtain the following lower bound from Theorems 26, 27 and Lemma\n28:\nCorollary 29. Let k be an integer such that 3 ≤k′ = k −1 and assume that 1\n2 · n/(8 log n) ≤\nk\n√\nk ≤n/(8 log n). Then, for all ε < 1/3, every R0(lin) refutation of ¬cliquen\nk,k′ is of size at least\n2Ω(nε).\nWhen considering the parameters of Theorem 20, we obtain a super-polynomial separation be-\ntween R0(lin) refutations and R(lin) refutations, as described below.\nFrom Theorems 26,27 and Lemma 28 we have (by direct calculations):\n31\n\nCorollary 30. Let k = √n and k′ = (log n)2/8 log log n.\nThen, every R0(lin) refutation of\n¬cliquen\nk,k′ has size at least nΩ\n“\nlog n\n√log log n\n”\n.\nBy Corollary 21, R(lin) admits polynomial-size in n refutations of ¬cliquen\nk,k′ under the param-\neters in Corollary 30. Thus we obtain the following separation result:\nCorollary 31. R(lin) is super-polynomially stronger than R0(lin).\nComment 1. Note that we do not need to assume that the coefficients in R0(lin)-lines are constants\nfor the lower bound argument. If the coefficients in R0(lin)-lines are only polynomially bounded\n(in the number of variables) then the same lower bound as in Corollary 30 also applies. This is\nbecause R0(lin)-lines in which coefficients are polynomially bounded integers, still have low (that\nis, logarithmic) communication complexity (as in Definition 8.2).\n9. Applications to Multilinear Proofs\nIn this section we arrive at one of the main benefits of the work we have done so far; Namely,\napplying results on resolution over linear equations in order to obtain new results for multilinear\nproof systems. Subsection 9.1 that follows, contains definitions, sufficient for the current paper,\nconcerning the notion of multilinear proofs introduced in [RT06].\n9.1. Background on Algebraic and Multilinear Proofs.\n9.1.1. Arithmetic and Multilinear Formulas.\nDefinition 9.1 (Arithmetic formula). Fix a field F. An arithmetic formula is a tree, with edges\ndirected from the leaves to the root, and with unbounded (finite) fan-in. Every leaf of the tree\n(namely, a node of fan-in 0) is labeled with either an input variable or a field element. A field\nelement can also label an edge of the tree. Every other node of the tree is labeled with either + or\n× (in the first case the node is a plus gate and in the second case a product gate). We assume that\nthere is only one node of out-degree zero, called the root. The size of an arithmetic formula F is\nthe total number of nodes in its graph and is denoted by |F|. An arithmetic formula computes a\npolynomial in the ring of polynomials F[x1, . . . , xn] in the following way. A leaf just computes the\ninput variable or field element that labels it. A field element that labels an edge means that the\npolynomial computed at its tail (namely, the node where the edge is directed from) is multiplied\nby this field element. A plus gate computes the sum of polynomials computed by the tails of all\nincoming edges. A product gate computes the product of the polynomials computed by the tails of\nall incoming edges. (Subtraction is obtained using the constant −1.) The output of the formula is\nthe polynomial computed by the root. The depth of a formula F is the maximal number of edges\nin a path from a leaf to the root of F.\nWe say that an arithmetic formula has a plus (resp., product) gate at the root if the root of the\nformula is labeled with a plus (resp., product) gate.\nA polynomial is multilinear if in each of its monomials the power of every input variable is at\nmost one.\nDefinition 9.2 (Multilinear formula). An arithmetic formula is a multilinear formula (or equiva-\nlently, multilinear arithmetic formula) if the polynomial computed by each gate of the formula is\nmultilinear (as a formal polynomial, that is, as an element of F[x1, . . . , xn]).\nAn additional definition we shall need is the following linear operator, called the multilinearization\noperator:\n32\n\nDefinition 9.3 (Multilinearization operator). Given a field F and a polynomial q ∈F[x1, . . . , xn],\nwe denote by M[q] the unique multilinear polynomial equal to q modulo the ideal generated by all\nthe polynomials x2\ni −xi, for all variables xi.\nFor example, if q = x2\n1x2 + ax3\n4 (for some a ∈F) then M[q] = x1x2 + ax4 .\nThe simulation of R0(lin) by multilinear proofs will rely heavily on the fact that multilinear\nsymmetric polynomials have small depth-3 multilinear formulas over fields of characteristic 0 (see\n[SW01] for a proof of this fact). To this end we define precisely the concept of symmetric polyno-\nmials.\nA renaming of the variables x1, . . . , xn is a permutation σ ∈Sn (the symmetric group on [n])\nsuch that xi is mapped to xσ(i) for every 1 ≤i ≤n.\nDefinition 9.4 (Symmetric polynomial). Given a set of variables X = {x1, . . . , xn}, a symmetric\npolynomial f over X is a polynomial in (all the variables of) X such that renaming of variables\ndoes not change the polynomial (as a formal polynomial).\n9.1.2. Polynomial Calculus with Resolution. Here we define the PCR proof system, introduced by\nAlekhnovich et al. in [ABSRW02].\nDefinition 9.5 (Polynomial Calculus with Resolution (PCR)). Let F be some fixed field and\nlet Q := {Q1, . . . , Qm} be a collection of multivariate polynomials from the ring of polynomials\nF[x1, . . . , xn, ̄x1, . . . , ̄xn]. The variables ̄x1, . . . , ̄xn are treated as new formal variables. Call the\nset of polynomials x2 −x, for x ∈{x1, . . . , xn, ̄x1, . . . , ̄xn}, plus the polynomials xi + ̄xi −1, for\nall 1 ≤i ≤n, the set of Boolean axioms of PCR. A PCR proof from Q of a polynomial g is\na finite sequence π = (p1, ..., pl) of multivariate polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] (each\npolynomial pi is interpreted as the polynomial equation pi = 0), where pl= g and for each i ∈[l],\neither pi = Qj for some j ∈[m], or pi is a Boolean axiom, or pi was deduced from pj, pk , where\nj, k < i, by one of the following inference rules:\nProduct: From p deduce xi · p , for some variable xi ;\nFrom p deduce ̄xi · p , for some variable ̄xi ;\nAddition: From p and q deduce α · p + β · q, for some α, β ∈F.\nA PCR refutation of Q is a proof of 1 (which is interpreted as 1 = 0) from Q. The number of\nsteps in a PCR proof is the number of proof-lines in it (that is, lin the case of π above).\nNote that the Boolean axioms of PCR have only 0, 1 solutions, where ̄xi = 0 if xi = 1 and ̄xi = 1\nif xi = 0.\n9.1.3. Multilinear Proof Systems. In [RT06] the authors introduced a natural (semantic) algebraic\nproof system that operates with multilinear arithmetic formulas denoted fMC (which stands for\nformula multilinear calculus), defined as follows:\nDefinition 9.6 (Formula Multilinear Calculus (fMC)). Fix a field F and let Q := {Q1, . . . , Qm} be\na collection of multilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] (the variables ̄x1, . . . , ̄xn are\ntreated as formal variables). Call the set of polynomials consisting of xi + ̄xi −1 and xi · ̄xi for\n1 ≤i ≤n , the Boolean axioms of fMC. An fMC proof from Q of a polynomial g is a finite sequence\nπ = (p1, ..., pl) of multilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] , such that pl= g and for\neach i ∈[l], either pi = Qj for some j ∈[m], or pi is a Boolean axiom of fMC, or pi was deduced\nby one of the following inference rules using pj, pk for j, k < i:\nProduct: from p deduce q · p , for some polynomial q ∈F[x1, . . . , xn, ̄x1, . . . , ̄xn] such that\np · q is multilinear;\nAddition: from p, q deduce α · p + β · q, for some α, β ∈F.\n33\n\nAll the polynomials in an fMC proof are represented as multilinear formulas. (A polynomial pi in\nan fMC proof is interpreted as the polynomial equation pi = 0.) An fMC refutation of Q is a proof\nof 1 (which is interpreted as 1 = 0) from Q. The size of an fMC proof π is defined as the total\nsum of all the formula sizes in π and is denoted by |π|.\nNote that the Boolean axioms have only 0, 1 solutions, where ̄xi = 0 if xi = 1 and ̄xi = 1 if\nxi = 0, for each 1 ≤i ≤n .\nDefinition 9.7 (Depth-k Formula Multilinear Calculus (depth-k fMC)). For a natural number k,\ndepth-k fMC denotes a restriction of the fMC proof system, in which proofs consist of multilinear\npolynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] represented as multilinear formulas of depth at most k.\n9.2. From R(lin) Proofs to PCR Proofs. We now demonstrate a general and straightforward\ntranslation from R(lin) proofs into PCR proofs over fields of characteristic 0. We use the term\n“translation” in order to distinguish it from a simulation; since here we are not interested in the\nsize of PCR proofs. In fact we have not defined the size of PCR proofs at all. We shall be interested\nonly in the number of steps in PCR proofs.\nFrom now on, all polynomials and arithmetic formulas are considered over some fix field F of\ncharacteristic 0.\nRecall that any field of characteristic 0 contains (an isomorphic copy of) the\ninteger numbers, and so we can use integer coefficients in the field.\nDefinition 9.8 (Polynomial translation of R(lin) proof-lines). Let D be a disjunction of linear\nequations:\n \na(1)\n1 x1 + . . . + a(1)\nn xn = a(1)\n0\n \n∨· · · ∨\n \na(t)\n1 x1 + . . . + a(t)\nn xn = a(t)\n0\n \n.\n(42)\nWe denote by bD its translation into the following polynomial:17\n \na(1)\n1 x1 + . . . + a(1)\nn xn −a(1)\n0\n \n· · ·\n \na(t)\n1 x1 + . . . + a(t)\nn xn −a(t)\n0\n \n.\n(43)\nIf D is the empty disjunction, we define bD to be the polynomial 1.\nIt is clear that every 0, 1 assignment to the variables in D, satisfies D, if and only if bD evaluates\nto 0 under the assignment.\nProposition 3. Let π = (D1, . . . , Dl) be an R(lin) proof sequence of Dl, from some collection\nof initial disjunctions of linear equations Q1, . . . , Qm. Then, there exists a PCR proof of bDlfrom\nbQ1, . . . , bQm with at most a polynomial in |π| number of steps.\nProof: We proceed by induction on the number of lines in π.\nThe base case is the translation of the axioms of R(lin) via the translation scheme in Definition\n9.8. An R(lin) Boolean axiom (xi = 0) ∨(xi = 1) is translated into xi · (xi −1) which is already a\nBoolean axiom of PCR.\nFor the induction step, we translate every R(lin) inference rule application into a polynomial-size\nPCR proof sequence as follows. We use the following simple claim:\nClaim 5. Let p and q be two polynomials and let s be the minimal size of an arithmetic formula\ncomputing q. Then one can derive in PCR, with only a polynomial in s number of steps, from p\nthe product q · p.18\nProof of claim: By induction on s.\n17This notation should not be confused with the same notation in Section 6.3.\n18Again, note that we only require that the number of steps in the proof is polynomial. We do not consider here\nthe size of the PCR proof.\n34\n\nAssume that Di = Dj ∨L was derived from Dj using the Weakening inference rule of R(lin),\nwhere j < i ≤land L is some linear equation. Then, by Claim 5, bDi = bDj · bL can be derived from\nbDj with a derivation of at most polynomial in |Dj ∨L| many steps.\nAssume that Di was derived from Dj where Dj is Di ∨(0 = k), using the Simplification inference\nrule of R(lin), where j < i ≤land k is a non-zero integer.\nThen, bDi can be derived from\nbDj = bDi · −k by multiplying with −k−1 (via the Addition rule of PCR).\nThus, it remains to simulate the resolution rule application of R(lin). Let A, B be two disjunc-\ntions of linear equations and assume that A ∨B ∨((⃗a + ⃗b) · ⃗x = a0 + b0) was derived in π from\nA ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0) (the case where A ∨B ∨((⃗a −⃗b) · ⃗x = a0 −b0) was derived from\nA ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0), is similar).\nWe need to derive bA · bB · ((⃗a +⃗b) · ⃗x −a0 −b0) from bA · (⃗a · ⃗x −a0) and bB · (⃗b · ⃗x −b0). This is\ndone by multiplying bA · (⃗a · ⃗x −a0) with bB and multiplying bB · (⃗b · ⃗x −b0) with bA (using Claim 5),\nand then adding the resulted polynomials together.\nRemark 4. When translating R(lin) proofs into PCR proofs we actually do not make any use of\nthe “negative” variables ̄x1, . . . , ̄xn. Nevertheless, the multilinear proof systems make use of these\nvariables in order to polynomially simulate PCR proofs (see Theorem 33 and its proof in [RT06]).\nWe shall need the following corollary in the sequel:\nCorollary 32. Let π = D1, . . . , Dlbe an R0(lin) proof of Dl, and let s be the maximal size of an\nR0(lin)-line in π. Then there is a PCR proof π′ of bDlwith polynomial-size in |π| number of steps\nand such that every line of π′ is a translation (via Definition 9.8) of an R0(lin)-line (Definition\n3.2), where the size of the R0(lin)-line is polynomial in s.\nProof: The simulation of R(lin) by PCR shown above, can be thought of as, first, considering\nbD1, . . . , bDlas the “skeleton” of a PCR proof of bDl. And second, for each Di that was deduced by\none of R(lin)’s inference rules from previous lines, one inserts the corresponding PCR proof sequence\nthat simulates the appropriate inference rule application (as described in the proof of Proposition\n3). By definition, those PCR proof-lines that correspond to lines in the skeleton bD1, . . . , bDlare\ntranslations of R0(lin)-lines (with size at most polynomial in s). Thus, to conclude the proof of\nthe corollary, one needs only to check that for any R0(lin)-line Di that was deduced by one of\nR(lin)’s inference rules from previous R0(lin)-lines (as demonstrated in the proof of Proposition 3),\nthe inserted corresponding PCR proof sequence uses only translations of R0(lin)-lines (with size\npolynomial in s). This can be verified by a straightforward inspection.\n9.3. From PCR Proofs to Multilinear Proofs. We now recall the general simulation result\nproved in [RT06] stating the following: Let π be a PCR refutation of some initial collection of\nmultilinear polynomials Q over some fixed field. Assume that π has polynomially many steps (that\nis, the number of proof lines in the PCR proof sequence is polynomial). If the ‘multilinearization’\n(namely, the result of applying the M[·] operator – see Definition 9.3) of each of the polynomials\nin π has a polynomial-size depth d multilinear formula (with a plus gate at the root), then there is\na polynomial-size depth-d fMC refutation of Q. More formally, we have:\nTheorem 33 ([RT06]). Fix a field F (not necessarily of characteristic 0) and let Q be a set of\nmultilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn]. Let π = (p1, . . . , pm) be a PCR refutation\nof Q. For each pi ∈π, let Φi be a multilinear formula for the polynomial M[pi]. Let s be the total\nsize of all formulas Φi, that is, s = Σm\ni=1|Φi|, and let d ≥2 be the maximal depth of all formulas\nΦi. Assume that the depth of all the formulas Φi that have a product gate at the root is at most\nd −1. Then there is a depth-d fMC refutation of Q of size polynomial in s.\n35\n\n9.3.1. Depth-3 Multilinear Proofs. Here we show that multilinear proofs operating with depth-3\nmultilinear formulas (that is, depth-3 fMC) over fields of characteristic 0 polynomially simulate\nR0(lin) proofs. In light of Proposition 32 and Theorem 33, to this end it suffices to show that any\nR0(lin)-line D translates into a corresponding polynomial p (via the translation in Definition 9.8)\nsuch that M[p] has a multilinear formula of size polynomial (in the number of variables) and depth\nat most 3 (with a plus gate at the root) over fields of characteristic 0.\nWe need the following proposition from [RT06]:\nProposition 4 ([RT06]). Let F be a field of characteristic 0. For a constant c, let X1, . . . , Xc\nbe c finite sets of variables (not necessarily disjoint), where Σc\ni=1|Xi| = n . Let f1, . . . , fc be c\nsymmetric polynomials over X1, . . . , Xc (over the field F), respectively. Then, there is a depth-3\nmultilinear formula for M[f1 · · · fc] of size polynomial (in n), with a plus gate at the root.\nThe following is the key lemma of the simulation:\nLemma 34. Let D be an R0(lin)-line with n variables and let p = bD (see Definition 9.8). Then,\nM[p] has a depth-3 multilinear formula over fields of characteristic 0, with a plus gate at the root\nand size at most polynomial in the size of D.\nProof: Assume that the underlying variables of D are ⃗x = x1 . . . , xn. By the definition of an\nR0(lin)-line (see Definition 3.2) we can partition the disjunction D into a constant number of\ndisjuncts, where one disjunct is a (possibly empty, translation of a) clause C,19 and all other\ndisjuncts have the following form:\nm\n_\ni=1\n(⃗a · ⃗x = li) ,\n(44)\nwhere the li’s are integers, m is not necessarily bounded and ⃗a denotes a vector of n constant\ninteger coefficients.\nLet us denote by q the polynomial representing the clause C.20\nConsider a disjunct as shown in (44). Since the coefficients ⃗a are constants, ⃗a · ⃗x can be written\nas a sum of constant number of linear forms, each with the same constant coefficient. In other\nwords, ⃗a · ⃗x can be written as z1 + . . . + zd, for some constant d, where for all i ∈[d]:\nzi := b ·\nX\nj∈J\nxj ,\n(45)\nfor some J ⊆[n] and some constant integer b. We shall assume without loss of generality that d is\nthe same constant for every disjunct of the form (44) inside D (otherwise, take d to be the maximal\nsuch d).\nThus, (44) is translated (via the translation scheme in Definition 9.8) into:\nm\nY\ni=1\n(z1 + ... + zd −li) .\n(46)\nBy fully expanding the product in (46), we arrive at:\nX\nr1+...+rd+1=m\n \nαrd+1 ·\nd\nY\nk=1\nzrk\nk\n!\n,\n(47)\n19If there is more than one clause in D, we simply combine all the clauses into a single clause.\n20C is a translation of a clause (that is, disjunction of literals) into a disjunction of linear equations, as defined\nin Section 3.1. The polynomial q is then the polynomial translation of this disjunction of linear equations, as in\nDefinition 9.8.\n36\n\nwhere the ri’s are non-negative integers, and where the αr’s, for every 0 ≤r ≤m are just integer\ncoefficients, formally defined as follows (this definition is not essential; we present it only for the\nsake of concreteness):\nαr :=\nX\nU⊆[m]\n| U|=r\nY\nj∈U\n(−lj) .\n(48)\nClaim 6. The polynomial bD (the polynomial translation of D) is a linear combination (over F) of\npolynomially (in |D|) many terms, such that each term can be written as\nq ·\nY\nk∈K\nzrk\nk ,\nwhere K is a collection of a constant number of indices, rk’s are non-negative integers, and the zk’s\nand q are as above (that is, the zk’s are linear forms, where each zk has a single coefficient for all\nvariables in it, as in (45), and q is a polynomial translation of a clause).\nProof of claim: Denote the total number of disjuncts of the form (44) in D by h. By definition\n(of R0(lin)-line), h is a constant. Consider the polynomial (47) above. In bD, we actually need to\nmultiply h many polynomials of the form shown in (47) and the polynomial q.\nFor every j ∈[h] we write the (single) linear form in the jth disjunct as a sum of constantly\nmany linear forms zj,1 + . . . + zj,d, where each linear form zj,k has the same coefficient for every\nvariable in it. Thus, bD can be written as:\nq ·\nh\nY\nj=1\n \n \n \n \n \n \n \nX\nr1+...+rd+1=mj\n \nα(j)\nrd+1 ·\nd\nY\nk=1\nzrk\nj,k\n!\n|\n{z\n}\n(⋆)\n \n \n \n \n \n \n \n,\n(49)\n(where the mj’s are not bounded, and the coefficients α(j)\nrd+1 are as defined in (48) except that here\nwe add the index (j) to denote that they depend on the jth disjunct in D). Denote the maximal\nmj, for all j ∈[h], by m0. The size of D, denoted |D|, is at least m0. Note that since d is a\nconstant, the number of summands in each (middle) sum in (49) is polynomial in m0, which is at\nmost polynomial in |D|. Thus, by expanding the outermost product in (49), we arrive at a sum of\npolynomially in |D| many summands. Each summand in this sum is a product of h terms of the\nform (⋆) multiplied by q.\nIt remains to apply the multilinearization operator (Definition 9.3) on bD, and verify that the re-\nsulting polynomial has a depth-3 multilinear formula with a plus gate at the root and of polynomial-\nsize (in |D|). Since M[·] is a linear operator, it suffices to show that when applying M[·] on each\nsummand in bD, as described in Claim 6, one obtains a (multilinear) polynomial that has a depth-3\nmultilinear formula with a plus gate at the root, and of polynomial-size in the number of variables\nn (note that clearly n ≤|D|). This is established in the following claim:\nClaim 7. The polynomial M\n \nq · Q\nk∈K zrk\nk\n \nhas a depth-3 multilinear formula of polynomial-size\nin n (the overall number of variables) and with a plus gate at the root (over fields of characteristic\n0), under the same notation as in Claim 6.\nProof of claim: Recall that a power of a symmetric polynomial is a symmetric polynomial in itself.\nSince each zk (for all k ∈K) is a symmetric polynomial, then its power zrk\nk is also symmetric. The\npolynomial q is a translation of a clause, hence it is a product of two symmetric polynomials: the\n37\n\nsymmetric polynomial that is the translation of the disjunction of literals with positive signs, and\nthe symmetric polynomial that is the translation of the disjunction of literals with negative signs.\nTherefore, q ·Q\nk∈K zrk\nk is a product of constant number of symmetric polynomials. By Proposition\n4, M\n \nq · Q\nk∈K zrk\nk\n \n(where here the M[·] operator operates on the ⃗x variables in the zk’s and q) is\na polynomial for which there is a polynomial-size (in n) depth-3 multilinear formula with a plus\ngate at the root (over fields of characteristic 0).\nWe now come to the main corollary of this section.\nCorollary 35. Multilinear proofs operating with depth-3 multilinear formulas (that is, depth-3 fMC\nproofs) polynomially-simulate R0(lin) proofs.\nProof: Immediate from Corollary 32, Theorem 33 and Proposition 34.\nFor the sake of clarity we repeat the chain of transformations needed to prove the simulation.\nGiven an R0(lin) proof π, we first use Corollary 32 to transform π into a PCR proof π′, with number\nof steps that is at most polynomial in |π|, and where each line in π′ is a polynomial translation of\nsome R0(lin)-line with size at most polynomial in the maximal line in π (which is clearly at most\npolynomial in |π|). Thus, by Proposition 34 each polynomial in π′ has a corresponding multilinear\npolynomial with a polynomial-size in |π| depth-3 multilinear formula (and a plus gate at the root).\nTherefore, by Theorem 33, we can transform π′ into a depth-3 fMC proof with only a polynomial\n(in |π|) increase in size.\n9.4. Small Depth-3 Multilinear Proofs. Since R0(lin) admits polynomial-size (in n) refutations\nof the m to n pigeonhole principle (for any m > n) (as defined in 6.1), Corollary 35 and Theorem\n15 yield:\nTheorem 36. For any m > n there are polynomial-size (in n) depth-3 fMC refutations of the m\nto n pigeonhole principle PHPm\nn (over fields of characteristic 0).\nThis improves over the result in [RT06] that demonstrated a polynomial-size (in n) depth-3 fMC\nrefutations of a weaker principle, namely the m to n functional pigeonhole principle.\nFurthermore, corollary 35 and Theorem 19 yield:\nTheorem 37. Let G be an r-regular graph with n vertices, where r is a constant, and fix some\nmodulus p. Then there are polynomial-size (in n) depth-3 fMC refutations of Tseitin mod p formulas\n¬TseitinG,p (over fields of characteristic 0).\nThe polynomial-size refutations of Tseitin graph tautologies here are different than those demon-\nstrated in [RT06]. Theorem 37 establishes polynomial-size refutations over any field of characteristic\n0 of Tseitin mod p formulas, whereas [RT06] required the field to contain a primitive pth root of\nunity. On the other hand, the refutations in [RT06] of Tseitin mod p formulas do not make any use\nof the semantic nature of the fMC proof system, in the sense that they do not utilize the fact that\nthe base field is of characteristic 0 (which in turn enables one to efficiently represent any symmetric\n[multilinear] polynomial by a depth-3 multilinear formula).\n10. Relations with Extensions of Cutting Planes\nIn this section we tie some loose ends by showing that, in full generality, R(lin) polynomially\nsimulates R(CP) with polynomially bounded coefficients, denoted R(CP*). First we define the\nR(CP*) proof system – introduced in [Kra98] – which is a common extension of resolution and\nCP* (the latter is cutting planes with polynomially bounded coefficients). The system R(CP*),\nthus, is essentially resolution operating with disjunctions of linear inequalities (with polynomially\nbounded integral coefficients) augmented with the cutting planes inference rules.\n38\n\nA linear inequality is written as\n⃗a · ⃗x ≥a0 ,\n(50)\nwhere ⃗a is a vector of integral coefficients a1, . . . , an, ⃗x is a vector of variables x1, . . . , xn, and a0\nis an integer. The size of the linear inequality (50) is the sum of all a0, . . . , an written in unary\nnotation (this is similar to the size of linear equations in R(lin)). A disjunction of linear inequalities\nis just a disjunction of inequalities of the form in (50). The semantics of a disjunction of inequalities\nis the natural one, that is, a disjunction is true under an assignment of integral values to ⃗x if and\nonly if at least one of the inequalities is true under the assignment. The size of a disjunction of\nlinear inequalities is the total size of all linear inequalities in it. We can also add in the obvious\nway linear inequalities, that is, if L1 is the linear inequality ⃗a·⃗x ≥a0 and L2 is the linear inequality\n⃗b · ⃗x ≥b0, then L1 + L2 is the linear inequality (⃗a +⃗b) · ⃗x ≥a0 + b0.\nThe proof system R(CP*) operates with disjunctions of linear inequalities with integral coeffi-\ncients (written in unary representation), and is defined as follows (our formulation is similar to\nthat in [Koj07]):21\nDefinition 10.1 (R(CP*)). Let K := {K1, . . . , Km} be a collection of disjunctions of linear in-\nequalities (whose coefficients are written in unary representation). An R(CP*)-proof from K of a\ndisjunction of linear inequalities D is a finite sequence π = (D1, ..., Dl) of disjunctions of linear\ninequalities, such that Dl= D and for each i ∈[l]: either Di = Kj for some j ∈[m]; or Di is one\nof the following R(CP*)-axioms:\n(1) xi ≥0, for any variable xi;\n(2) −xi ≥−1, for any variable xi;\n(3) (⃗a · ⃗x ≥a0) ∨(−⃗a · ⃗x ≥1 −a0), where all coefficients (including a0) are integers;\nor Di was deduced from previous lines by one of the following R(CP*)-inference rules:\n(1) Let A, B be two disjunctions of linear inequalities and let L1, L2 be two linear inequalities.22\nFrom A ∨L1 and B ∨L2 derive A ∨B ∨(L1 + L2).\n(2) Let L be some linear equation.\nFrom a disjunction of linear equations A derive A ∨L.\n(3) Let A be a disjunction of linear equations\nFrom A ∨(0 ≥1) derive A.\n(4) Let c be a non-negative integer.\nFrom (⃗a · ⃗x ≥a0) ∨A derive (c⃗a · ⃗x ≥ca0) ∨A.\n(5) Let A be a disjunction of linear inequalities, and let c ≥1 be an integer.\nFrom (c⃗a · ⃗x ≥a0) ∨A derive (a · ⃗x ≥⌈a0/c⌉) ∨A.\nAn R(CP*) refutation of a collection of disjunctions of linear inequalities K is a proof of the empty\ndisjunction from K. The size of a proof π in R(CP*) is the total size of all the disjunctions of\nlinear inequalities in π, denoted |π|.\nIn order for R(lin) to simulate R(CP*) proofs, we need to fix the following translation scheme.\nEvery linear inequality L of the form ⃗a·⃗x ≥a0 is translated into the following disjunction, denoted\nbL:\n(⃗a · ⃗x = a0) ∨(⃗a · ⃗x = a0 + 1) ∨· · · ∨(⃗a · ⃗x = a0 + k) ,\n(51)\nwhere k is such that a0 + k equals the sum of all positive coefficients in ⃗a, that is, a0 + k =\nmax\n⃗x∈{0,1}n (⃗a · ⃗x) (in case the sum of all positive coefficients in ⃗a is less than a0, then we put k = 0).\nAn inequality with no variables of the form 0 ≥a0 is translated into 0 = a0 in case it is false (that\n21When we allow coefficients to be written in binary representation, instead of unary representation, the resulting\nproof system is denoted R(CP).\n22In all R(CP*)-inference rules, A, B are possibly the empty disjunctions.\n39\n\nis, in case 0 < a0), and into 0 = 0 in case it is true (that is, in case 0 ≥a0). Note that since the\ncoefficients of linear inequalities (and linear equations) are written in unary representation, any\nlinear inequality of size s translates into a disjunction of linear equations of size O(s2). Clearly,\nevery 0, 1 assignment to the variables ⃗x satisfies L if and only if it satisfies its translation bL. A\ndisjunction of linear inequalities D is translated into the disjunction of the translations of all the\nlinear inequalities in it, denoted bD.\nA collection K := {K1, . . . , Km} of disjunctions of linear\ninequalities, is translated into the collection\nn\nbK1, . . . , bKm\no\n.\nTheorem 38. R(lin) polynomially-simulates R(CP*). In other words, if π is an R(CP*) proof of\na linear inequality D from a collection of disjunctions of linear inequalities K1, . . . , Kt, then there\nis an R(lin) proof of bD from bK1, . . . , bKt whose size is polynomial in |π|.\nProof: By induction on the number of proof-lines in π.\nBase case: Here we only need to show that the axioms of R(CP*) translates into axioms of\nR(lin), or can be derived with polynomial-size (in the size of the original R(CP*) axiom) R(lin)\nderivations (from R(lin)’s axioms).\nR(CP*) axiom number (1): xi ≥0 translates into the R(lin) axiom (xi = 0) ∨(xi = 1).\nR(CP*) axiom number (2): −xi ≥−1, translates into (−xi = −1) ∨(−xi = 0).\nFrom the\nBoolean axiom (xi = 1) ∨(xi = 0) of R(lin), one can derive with a constant-size R(lin) proof the\nline (−xi = −1)∨(−xi = 0) (for instance, by subtracting twice each equation in (xi = 1)∨(xi = 0)\nfrom itself).\nR(CP*) axiom number (3): (⃗a ·⃗x ≥a0) ∨(−⃗a ·⃗x ≥1−a0). The inequality (⃗a ·⃗x ≥a0) translates\ninto\nh_\nb=a0\n(⃗a · ⃗x = b) ,\nwhere h is the maximal value of ⃗a · ⃗x over 0, 1 assignments to ⃗x (that is, h is just the sum of all\npositive coefficients in ⃗a). The inequality (−⃗a · ⃗x ≥1 −a0) translates into\nf_\nb=1−a0\n(−⃗a · ⃗x = b) ,\nwhere f is the maximal value of −⃗a · ⃗x over 0, 1 assignments to ⃗x (that is, f is just the sum of\nall negative coefficients in ⃗a). Note that one can always flip the sign of any equation ⃗a · ⃗x = b in\nR(lin). This is done, for instance, by subtracting twice ⃗a · ⃗x = b from itself. So overall R(CP*)\naxiom number (3) translates into\nh_\nb=a0\n(⃗a · ⃗x = b) ∨\nf_\nb=1−a0\n(−⃗a · ⃗x = b) ,\nthat can be converted inside R(lin) into\na0−1\n_\nb=−f\n(⃗a · ⃗x = b) ∨\nh_\nb=a0\n(⃗a · ⃗x = b) .\n(52)\nLet A′ := {−f, −f + 1, . . . , a0 −1, a0, a0 + 1, . . . , h} and let A be the set of all possible values that\n⃗a · ⃗x can get over all possible Boolean assignments to ⃗x. Notice that A ⊆A′. By Lemma 8, for any\n⃗a · ⃗x, there is a polynomial-size (in the size of the linear form ⃗a · ⃗x) derivation of W\nα∈A(⃗a · ⃗x = α).\nBy using the R(lin) Weakening rule we can then derive W\nα∈A′(⃗a · ⃗x = α) which is equal to (52).\nInduction step: Here we simply need to show how to polynomially simulate inside R(lin) every\ninference rule application of R(CP*).\n40\n\nRule (1): Let A, B be two disjunctions of linear inequalities and let L1, L2 be two linear inequalities.\nAssume we already have a R(lin) proofs of bA ∨bL1 and bB ∨bL2. We need to derive bA ∨bB ∨\\\nL1 + L2.\nCorollary 7 shows that there is a polynomial-size (in the size of bL1 and bL2; which is polynomial\nin the size of L1 and L2) derivation of\n\\\nL1 + L2 from bL1 and bL2, from which the desired derivation\nimmediately follows.\nRule (2): The simulation of this rule in R(lin) is done using the R(lin) Weakening rule.\nRule (3): The simulation of this rule in R(lin) is done using the R(lin) Simplification rule (remem-\nber that 0 ≥1 translates into 0 = 1 under our translation scheme).\nRule (4): Let c be a non-negative integer. We need to derive\n\\\n(c⃗a · ⃗x ≥ca0)∨bA from\n\\\n(⃗a · ⃗x ≥a0)∨bA\nin R(lin).\nThis amounts only to “adding together” c times the disjunction\n\\\n(⃗a · ⃗x ≥a0) in\n\\\n(⃗a · ⃗x ≥a0) ∨bA. This can be achieved by c many applications of Corollary 7. We omit the details.\nRule (5): We need to derive\n\\\n(⃗a · ⃗x ≥⌈a0/c⌉) ∨bA, from\n\\\n(c⃗a · ⃗x ≥a0) ∨bA. Consider the disjunction\nof linear equations\n\\\n(c⃗a · ⃗x ≥a0), which can be written as:\n(c⃗a · ⃗x = a0) ∨(c⃗a · ⃗x = a0 + 1) ∨. . . ∨(c⃗a · ⃗x = a0 + r) ,\n(53)\nwhere a0 + r is the maximal value c⃗a · ⃗x can get over 0, 1 assignments to ⃗x. By Lemma 8 there is a\npolynomial-size (in the size of ⃗a · ⃗x) R(lin) proof of\n_\nα∈A\n(⃗a · ⃗x = α) ,\n(54)\nwhere A is the set of all possible values of ⃗a · ⃗x over 0, 1 assignments to ⃗x.\nWe now use (53) to cut-offfrom (54) all equations (⃗a · ⃗x = β) for all β < ⌈a0/c⌉(this will give\nus the desired disjunction of linear equations). Consider the equation (⃗a · ⃗x = β) in (54) for some\nfixed β < ⌈a0/c⌉. Use the resolution rule of R(lin) to add this equation to itself c times inside (54).\nWe thus obtain\n(c⃗a · ⃗x = cβ) ∨\n_\nα∈A\\{β}\n(⃗a · ⃗x = α) .\n(55)\nSince β is an integer and β < ⌈a0/c⌉, we have cβ < a0. Thus, the equation (c⃗a · ⃗x = cβ) does\nnot appear in (53). We can then successively resolve (c⃗a · ⃗x = cβ) in (55) with each equation\n(c⃗a · ⃗x = a0), . . . , (c⃗a · ⃗x = a0 + r) in (53). Hence, we arrive at W\nα∈A\\{β} (⃗a · ⃗x = α). Overall, we\ncan cut-offall equations (⃗a · ⃗x = β), for β < ⌈a0/c⌉, from (54). We then get the disjunction\n_\nα∈A′\n(⃗a · ⃗x = α) ,\nwhere A′ is the set of all elements of A greater or equal to ⌈a0/c⌉(in other words, all values greater\nor equal to ⌈a0/c⌉that ⃗a·⃗x can get over 0, 1 assignments to ⃗x). Using the Weakening rule of R(lin)\n(if necessary) we can arrive finally at the desired disjunction\n\\\n(⃗a · ⃗x ≥⌈a0/c⌉), which concludes the\nR(lin) simulation of R(CP*)’s inference Rule (5).\nAppendix A. Feasible Monotone Interpolation\nHere we formally define the feasible monotone interpolation property. The definition is taken\nmainly from [Kra97].\nRecall that for two binary strings of length n (or equivalently, Boolean assignments for n propo-\nsitional variables) α, α′, we denote by α′ ≥α that α′ is bitwise greater than α, that is, that for all\ni ∈[n], α′\ni ≥αi (where α′\ni and αi are the ith bits of α′ and α, respectively). Let A(⃗p, ⃗q), B(⃗p,⃗r)\nbe two collections of formulas in the displayed variables only, where ⃗p, ⃗q,⃗r are pairwise disjoint\n41\n\nsequences of distinct variables (similar to the notation at the beginning of Section 7). Assume\nthat there is no assignment that satisfies both A(⃗p, ⃗q) and B(⃗p,⃗r). We say that A(⃗p, ⃗q), B(⃗p,⃗r) are\nmonotone if one of the following conditions hold:\n(1) If ⃗α is an assignment to ⃗p and ⃗β is an assignment to ⃗q such that A(⃗α, ⃗β) = 1, then for any\nassignment ⃗α′ ≥⃗α it holds that A(⃗α′, ⃗β) = 1.\n(2) If ⃗α is an assignment to ⃗p and ⃗β is an assignment to ⃗r such that B(⃗α, ⃗β) = 1, then for any\nassignment ⃗α′ ≤⃗α it holds that B(⃗α′, ⃗β) = 1.\nFix a certain proof system P. Recall the definition of the interpolant function (corresponding\nto a given unsatisfiable A(⃗p, ⃗q) ∧B(⃗p,⃗r); that is, functions for which (39) in Section 7 hold).\nAssume that for every monotone A(⃗p, ⃗q), B(⃗p,⃗r) there is a transformation from every P-refutation\nof A(⃗p, ⃗q)∧B(⃗p,⃗r) into the corresponding interpolant monotone Boolean circuit C(⃗p) (that is, C(⃗p)\nuses only monotone gates23) and whose size is polynomial in the size of the refutation (note that\nfor every monotone A(⃗p, ⃗q), B(⃗p,⃗r) the corresponding interpolant circuit must compute a monotone\nfunction;24 the interpolant circuit itself, however, might not be monotone, namely, it may use non-\nmonotone gates). In such a case, we say that P has the feasible monotone interpolation property.\nThis means that, if a proof system P has the feasible monotone interpolation property, then an\nexponential lower bound on monotone circuits that compute the interpolant function corresponding\nto A(⃗p, ⃗q) ∧B(⃗p,⃗r) implies an exponential-size lower bound on P-refutations of A(⃗p, ⃗q) ∧B(⃗p,⃗r).\nDefinition A.1 (Feasible monotone interpolation property). Let P be a propositional refu-\ntation system. Let A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r) be two collections of formulas\nwith the displayed variables only (where ⃗p has n variables, ⃗q has s variables and ⃗r has t variables),\nsuch that either (the set of satisfying assignments of) A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) meet condition 1 above\nor (the set of satisfying assignments of) B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r) meet condition 2 above. Assume\nthat for any such A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r), if there exists a P-refutation for\nA1(⃗p, ⃗q) ∧· · · ∧Ak(⃗p, ⃗q) ∧B1(⃗p,⃗r) ∧. . . ∧Bl(⃗p,⃗r) of size S then there exists a monotone Boolean\ncircuit separating UA from VB (as defined in Section 7.1) of size polynomial in S. In this case we\nsay that P possesses the feasible monotone interpolation property.\nAcknowledgments\nWe wish to thank Arist Kojevnikov for useful correspondence on his paper.\nThis work was\ncarried out in partial fulfillment of the requirements for the Ph.D. degree of the second author.\nReferences\n[AB87]\nNoga Alon and Ravi B. Boppana. The monotone circuit complexity of boolean functions. Combinatorica,\n7(1):1–22, 1987. 8, 26\n[ABE02]\nAlbert Atserias, Maria L. Bonet, and Juan L. Esteban. Lower bounds for the weak pigeonhole principle\nand random formulas beyond resolution. Information and Computation, 176:152–136, August 2002. 1.2,\n3, 6.3, 6.3, 20, 3\n[ABSRW02] Michael Alekhnovich, Eli Ben-Sasson, Alexander A. Razborov, and Avi Wigderson. Space complexity\nin propositional calculus. SIAM J. Comput., 31(4):1184–1211 (electronic), 2002. 9.1.2\n[And85]\nA. E. Andreev. On a method for obtaining lower bounds for the complexity of individual monotone\nfunctions. Dokl. Akad. Nauk SSSR (in Russian), 282(5):1033–1037, 1985. [Engl. Transl. Soviet Math.\nDokl., vol. 31 (1985), pp. 530-534]. 8\n23For instance, a monotone Boolean circuit is a circuit that uses only ∧, ∨gates of fan-in two (see also Section\n8). In certain cases, the monotone interpolation technique is also applicable for a larger class of circuits, that is,\ncircuits that compute with real numbers and that can use any nondecreasing real functions as gates (this was proved\nby Pudl ́ak in [Pud97]).\n24That is, if α′ ≥α then C(α′) ≥C(α).\n42\n\n[BGIP01]\nSamuel Buss, Dima Grigoriev, Russell Impagliazzo, and Toniann Pitassi. Linear gaps between degrees for\nthe polynomial calculus modulo distinct primes. J. Comput. System Sci., 62(2):267–289, 2001. Special\nissue on the 14th Annual IEEE Conference on Computational Complexity (Atlanta, GA, 1999). 6.2\n[BP97]\nSamuel Buss and Toniann Pitassi. Resolution and the weak pigeonhole principle. In Computer science\nlogic (Aarhus, 1997), volume 1414 of Lecture Notes in Comput. Sci., pages 149–156. Springer, Berlin,\n1997. 3\n[BPR97]\nMaria Bonet, Toniann Pitassi, and Ran Raz. Lower bounds for cutting planes proofs with small coeffi-\ncients. The Journal of Symbolic Logic, 62(3):708–728, 1997. 1.1, 1.2, 1.2, 6.3, 2, 7.1.1, 8, 27\n[CR79]\nStephen A. Cook and Robert A. Reckhow. The relative efficiency of propositional proof systems. The\nJournal of Symbolic Logic, 44(1):36–50, 1979. 2\n[Hak85]\nArmin Haken. The intractability of resolution. Theoret. Comput. Sci., 39(2-3):297–308, 1985. 1\n[HK06]\nEdward Hirsch and Arist Kojevnikov. Several notes on the power of Gomory-Chv ́atal cuts. Annals of\nPure and Applied Logic, 141:429–436, 2006. 1.1\n[IPU94]\nRussel Impagliazzo, Toniann Pitassi, and Alasdair Urquhart. Upper and lower bounds for tree-like\ncutting planes proofs. In Ninth Annual Symposium on Logic in Computer Science, pages 220–228. IEEE\nComput. Soc. Press, 1994. 7.1.1\n[Koj07]\nArist Kojevnikov. Improved lower bounds for tree-like resolution over linear inequalities. In In Proceed-\nings of the 10th International Conference on Theory and Applications of Satisfiability Testing (SAT),\n2007. Preliminary version in Electronic Colloquium on Computational Complexity, ECCC, January 2007.\nReport No. TR07-010. 1.1, 10\n[Kra94]\nJan Kraj ́ıˇcek. Lower bounds to the size of constant-depth propositional proofs. The Journal of Symbolic\nLogic, 59(1):73–86, 1994. 7\n[Kra97]\nJan Kraj ́ıˇcek. Interpolation theorems, lower bounds for proof systems, and independence results for\nbounded arithmetic. The Journal of Symbolic Logic, 62(2):457–486, 1997. 1.1, 1.2, 1.2, 6.3, 7.1, 7.1.1,\n23, A\n[Kra98]\nJan Kraj ́ıˇcek. Discretely ordered modules as a first-order extension of the cutting planes proof system.\nThe Journal of Symbolic Logic, 63(4):1582–1596, 1998. 1.1, 1.2, 6.3, 10\n[Kra01]\nJan Kraj ́ıˇcek. On the weak pigeonhole principle. Fund. Math., 170(1-2):123–140, 2001. Dedicated to the\nmemory of Jerzy Lo ́s. 6.3\n[Kra07]\nJan Kraj ́ıˇcek. An exponential lower bound for a constraint propagation proof system based on ordered\nbinary decision diagrams. To appear in The Journal of Symbolic Logic. Preliminary version available in\nElectronic Colloquium on Computational Complexity, ECCC, January 2007. Report No. TR07-007. 6.3\n[KW88]\nMauricio Karchmer and Avi Wigderson. Monotone circuits for connectivity require super-logarithmic\ndepth. In Proceedings of the 20th Annual ACM Symposium on Theory of Computing, pages 539–550.\nACM, 1988. 7.1.1, 7.2\n[Pud97]\nPavel Pudl ́ak. Lower bounds for resolution and cutting plane proofs and monotone computations. The\nJournal of Symbolic Logic, 62(3):981–998, Sept. 1997. 6.3, 23\n[Razb85]\nAlexander A. Razborov. Lower bounds on the monotone complexity of some Boolean functions. Dokl.\nAkad. Nauk SSSR (in Russian), 281(4):798–801, 1985. [English translation in Sov. Math. Dokl., vol . 31\n(1985), pp. 354-357.]. 8\n[Razb95]\nAlexander A. Razborov. Unprovability of lower bounds on circuit size in certain fragments of bounded\narithmetic. Izv. Ross. Akad. Nauk Ser. Mat., 59(1):201–224, 1995. 7.1.1\n[Razb02]\nAlexander A. Razborov. Proof complexity of pigeonhole principles. In Developments in language theory\n(Vienna, 2001), volume 2295 of Lecture Notes in Comput. Sci., pages 110–116. Springer, Berlin, 2002. 1\n[Raz04]\nRan Raz. Multi-linear formulas for permanent and determinant are of super-polynomial size. In Pro-\nceedings of the 36th Annual ACM Symposium on the Theory of Computing, pages 633–641, Chicago, IL,\n2004. ACM. 1\n[Raz06]\nRan Raz. Separation of multilinear circuit and formula size. Theory of Computing, Vol. 2, article 6,\n2006. 1\n[RT06]\nRan Raz and Iddo Tzameret. The strength of multilinear proofs. Comput. Complexity (to appear). Pre-\nliminary version in Electronic Colloquium on Computational Complexity, ECCC, January 2006. Report\nNo. TR06-001. (document), 1, 1.1, 1.2, 1.2, 1, 2, 6.2, 9, 9.1.3, 4, 9.3, 33, 9.3.1, 4, 9.4, 9.4\n[SW01]\nAmir Shpilka and Avi Wigderson. Depth-3 arithmetic circuits over fields of characteristic zero. Comput.\nComplexity, 10:1–27, 2001. 9.1.1\n[Tse68]\nG. C. Tseitin. On the complexity of derivations in propositional calculus. Studies in constructive math-\nematics and mathematical logic Part II. Consultants Bureau, New-York-London, 1968. 6.2\n43\n\nDepartment of Applied Mathematics and Computer Science, Weizmann Institute, Rehovot 76100,\nIsrael\nE-mail address: ranraz@wisdom.weizmann.ac.il\nSchool of Computer Science, Tel Aviv University, Tel Aviv 69978, Israel\nE-mail address: tzameret@tau.ac.il\n44","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0708.1529v1 [cs.CC] 10 Aug 2007\nRESOLUTION OVER LINEAR EQUATIONS\nAND MULTILINEAR PROOFS\nRAN RAZ AND IDDO TZAMERET\nAbstract. We develop and study the complexity of propositional proof systems of varying strength\nextending resolution by allowing it to operate with disjunctions of linear equations instead of\nclauses. We demonstrate polynomial-size refutations for hard tautologies like the pigeonhole prin-\nciple, Tseitin graph tautologies and the clique-coloring tautologies in these proof systems. Using\nthe (monotone) interpolation by a communication game technique we establish an exponential-size\nlower bound on refutations in a certain, considerably strong, fragment of resolution over linear\nequations, as well as a general polynomial upper bound on (non-monotone) interpolants in this\nfragment.\nWe then apply these results to extend and improve previous results on multilinear proofs (over\nfields of characteristic 0), as studied in [RT06]. Specifically, we show the following:\n• Proofs operating with depth-3 multilinear formulas polynomially simulate a certain, consid-\nerably strong, fragment of resolution over linear equations.\n• Proofs operating with depth-3 multilinear formulas admit polynomial-size refutations of the\npigeonhole principle and Tseitin graph tautologies. The former improve over a previous result\nthat established small multilinear proofs only for the functional pigeonhole principle. The\nlatter are different than previous proofs, and apply to multilinear proofs of Tseitin mod p\ngraph tautologies over any field of characteristic 0.\nWe conclude by connecting resolution over linear equations with extensions of the cutting planes\nproof system.\nContents\n1.\nIntroduction\n2\n1.1.\nComparison to Earlier Work\n4\n1.2.\nSummary of Results\n5\n2.\nNotation and Background on Propositional Proof Systems\n7\n3.\nResolution over Linear Equations and its Subsystems\n8\n3.1.\nDisjunctions of Linear Equations\n8\n3.2.\nResolution over Linear Equations – R(lin)\n9\n3.3.\nFragment of Resolution over Linear Equations – R0(lin)\n10\n4.\nReasoning and Counting inside R(lin) and its Subsystems\n11\n4.1.\nBasic Reasoning inside R(lin) and its Subsystems\n12\n4.2.\nBasic Counting inside R(lin) and R0(lin)\n13\n5.\nImplicational Completeness of R(lin) and its Subsystems\n16\n6.\nShort Proofs for Hard Tautologies\n17\n6.1.\nThe Pigeonhole Principle Tautologies in R0(lin)\n17\n6.2.\nTseitin mod p Tautologies in R0(lin)\n19\n2000 Mathematics Subject Classification.\n03F20, 68Q17, 68Q15.\nKey words and phrases. proof complexity, resolution, algebraic proof systems, multilinear proofs, cutting planes,\nfeasible monotone interpolation.\nThe first author was supported by The Israel Science Foundation and The Minerva Foundation.\nThe second author was supported by The Israel Science Foundation (grant no. 250/05).\n1"},{"paragraph_id":"p2","order":2,"text":"6.3.\nThe Clique-Coloring Principle in R(lin)\n23\n7.\nInterpolation Results for R0(lin)\n26\n7.1.\nInterpolation for Semantic Refutations\n26\n7.2.\nPolynomial Upper Bounds on Interpolants for R0(lin)\n28\n8.\nSize Lower Bounds\n30\n9.\nApplications to Multilinear Proofs\n32\n9.1.\nBackground on Algebraic and Multilinear Proofs\n32\n9.2.\nFrom R(lin) Proofs to PCR Proofs\n34\n9.3.\nFrom PCR Proofs to Multilinear Proofs\n35\n9.4.\nSmall Depth-3 Multilinear Proofs\n38\n10.\nRelations with Extensions of Cutting Planes\n38\nAppendix A.\nFeasible Monotone Interpolation\n41\nAcknowledgments\n42\nReferences\n42\n1. Introduction\nThis paper considers two kinds of proof systems. The first kind are extensions of resolution\nthat operate with disjunctions of linear equations with integral coefficients instead of clauses. The\nsecond kind are algebraic proof systems operating with multilinear arithmetic formulas. Proofs in\nboth kinds of systems establish the unsatisfiability of formulas in conjunctive normal form (CNF).\nWe are primarily concerned with connections between these two families of proof systems and with\nextending and improving previous results on multilinear proofs.\nThe resolution system is a popular propositional proof system that establishes the unsatisfiability\nof CNF formulas (or equivalently, the truth of tautologies in disjunctive normal form) by operating\nwith clauses (a clause is a disjunction of propositional variables and their negations). It is well\nknown that resolution cannot provide small (that is, polynomial-size) proofs for many basic count-\ning arguments. The most notable example of this are the strong exponential lower bounds on the\nresolution refutation size of the pigeonhole principle and its different variants (Haken [Hak85] was\nthe first to establish such a lower bound; see also [Razb02] for a survey on the proof complexity of\nthe pigeonhole principle). Due to the popularity of resolution both in practice, as the core of many\nautomated theorem provers, and as a theoretical case-study in propositional proof complexity, it\nis natural to consider weak extensions of resolution that can overcome its inefficiency in provid-\ning proofs of counting arguments. The proof systems we present in this paper are extensions of\nresolution, of various strength, that are suited for this purpose.\nPropositional proof systems of a different nature that also attracted much attention in proof\ncomplexity theory are algebraic proof systems, which are proof systems operating with (multivariate)\npolynomials over a field. In this paper, we are particularly interested in algebraic proof systems\nthat operate with multilinear polynomials represented as multilinear arithmetic formulas, called by\nthe generic name multilinear proofs (a polynomial is multilinear if the power of each variable in its\nmonomials is at most one). The investigation into such proof systems was initiated in [RT06], and\nhere we continue this line of research. This research is motivated on the one hand by the apparent\nconsiderable strength of such systems; and on the other hand, by the known super-polynomial\nsize lower bounds on multilinear formulas computing certain important functions [Raz04, Raz06],\ncombined with the general working assumption that establishing lower bounds on the size of objects\na proof system manipulates (in this case, multilinear formulas) is close to establishing lower bounds\non the size of the proofs themselves.\n2"},{"paragraph_id":"p3","order":3,"text":"The basic proof system we shall study is denoted R(lin). The proof-lines1 in R(lin) proofs are\ndisjunctions of linear equations with integral coefficients over the variables ⃗x = x1, . . . , xn.\nIt\nturns out that (already proper subsystems of) R(lin) can handle very elegantly basic counting\narguments.\nThe following defines the R(lin) proof system. Given an initial CNF, we translate\nevery clause W\ni∈I xi ∨W\nj∈J ¬xj (where I are the indices of variables with positive polarities and\nJ are the indices of variables with negative polarities) pertaining to the CNF, into the disjunction\nW\ni∈I(xi = 1)∨W\nj∈J(xj = 0). Let A and B be two disjunctions of linear equations, and let ⃗a·⃗x = a0\nand ⃗b · ⃗x = b0 be two linear equations (where ⃗a,⃗b are two vectors of n integral coefficients, and\n⃗a · ⃗x is the scalar product Pn\ni=1 aixi; and similarly for ⃗b · ⃗x). The rules of inference belonging to\nR(lin) allow to derive A ∨B ∨((⃗a +⃗b) · ⃗x = a0 + b0) from A ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0) (or\nsimilarly, to derive A ∨B ∨((⃗a −⃗b) ·⃗x = a0 −b0) from A ∨(⃗a ·⃗x = a0) and B ∨(⃗b ·⃗x = b0)). We can\nalso simplify disjunctions by discarding (unsatisfiable) equations of the form (0 = k), for k ̸= 0.\nIn addition, for every variable xi, we shall add an axiom (xi = 0) ∨(xi = 1), which forces xi to\ntake on only Boolean values. A derivation of the empty disjunction (which stands for false) from\nthe (translated) clauses of a CNF is called an R(lin) refutation of the given CNF. This way, every\nunsatisfiable CNF has an R(lin) refutation (this can be proved by a straightforward simulation of\nresolution by R(lin)).\nThe basic idea connecting resolution operating with disjunctions of linear equations and multilin-\near proofs is this: Whenever a disjunction of linear equations is simple enough — and specifically,\nwhen it is close to a symmetric function, in a manner made precise — then it can be represented\nby a small size and small depth multilinear arithmetic formula over fields of characteristic 0. This\nidea was already used (somewhat implicitly) in [RT06] to obtain polynomial-size multilinear proofs\noperating with depth-3 multilinear formulas of the functional pigeonhole principle (this principle\nis weaker than the pigeonhole principle). In the current paper we generalize previous results on\nmultilinear proofs by fully using this idea: We show how to polynomially simulate with multilinear\nproofs, operating with small depth multilinear formulas, certain short proofs carried inside resolu-\ntion over linear equations. This enables us to provide new polynomial-size multilinear proofs for\ncertain hard tautologies, improving results from [RT06].\nMore specifically, we introduce a certain fragment of R(lin), which can be polynomially simu-\nlated by depth-3 multilinear proofs (that is, multilinear proofs operating with depth-3 multilinear\nformulas). On the one hand this fragment of resolution over linear equations already is sufficient\nto formalize in a transparent way basic counting arguments, and so it admits small proofs of the\npigeonhole principle and the Tseitin mod p formulas (which yields some new upper bounds on\nmultilinear proofs); and on the other hand we can use the (monotone) interpolation technique to\nestablish an exponential-size lower bound on refutations in this fragment as well as demonstrating a\ngeneral (non-monotone) polynomial upper bound on interpolants for this fragment. The possibility\nthat multilinear proofs (possibly, operating with depth-3 multilinear formulas) possess the feasible\nmonotone interpolation property (and hence, admit exponential-size lower bounds) remains open.\nAnother family of propositional proof systems we discuss in relation to the systems mentioned\nabove are the cutting planes system and its extensions. The cutting planes proof system operates\nwith linear inequalities with integral coefficients, and this system is very close to the extensions\nof resolution we present in this paper. In particular, the following simple observation can be used\nto polynomially simulate cutting planes proofs with polynomially bounded coefficients (and some\nof its extensions) inside resolution over linear equations: The truth value of a linear inequality\n⃗a · ⃗x ≥a0 (where ⃗a is a vector of n integral coefficients and ⃗x is a vector of n Boolean variables) is\n1Each element (usually a formula) of a proof-sequence is referred to as a proof-line.\n3"},{"paragraph_id":"p4","order":4,"text":"equivalent to the truth value of the following disjunction of linear equalities:\n(⃗a · ⃗x = a0) ∨(⃗a · ⃗x = a0 + 1) ∨· · · ∨(⃗a · ⃗x = a0 + k) ,\nwhere a0 + k equals the sum of all positive coefficients in ⃗a (that is, a0 + k =\nmax\n⃗x∈{0,1}n (⃗a · ⃗x)).\nNote on terminology. All the proof systems considered in this paper intend to prove the unsatis-\nfiability over 0, 1 values of collections of clauses (possibly, of translation of the clauses to disjunctions\nof linear equations). In other words, proofs in such proof systems intend to refute the collections\nof clauses, which is to validate their negation. Therefore, throughout this paper we shall sometime\nspeak about refutations and proofs interchangeably, always intending refutations, unless otherwise\nstated.\n1.1. Comparison to Earlier Work. To the best of our knowledge this paper is the first that\nconsiders resolution proofs operating with disjunctions of linear equations. Previous works consid-\nered extensions of resolution over linear inequalities augmented with the cutting planes inference\nrules (the resulting proof system denoted R(CP)). In full generality, we show that resolution over\nlinear equations can polynomially simulate R(CP) when the coefficients in all the inequalities are\npolynomially bounded (however, the converse is not known to hold). On the other hand, we shall\nconsider a certain fragment of resolution over linear equations, in which we do not even know how to\npolynomially simulate cutting planes proofs with polynomially bounded coefficients in inequalities\n(let alone R(CP) with polynomially bounded coefficients in inequalities). We now shortly discuss\nthe previous work on R(CP) and related proof systems.\nExtensions of resolution to disjunctions of linear inequalities were first considered by Kraj ́ıˇcek\n[Kra98] who developed the proof systems LK(CP) and R(CP). The LK(CP) system is a first-order\n(Gentzen-style) sequent calculus that operates with linear inequalities instead of atomic formulas\nand augments the standard first-order sequent calculus inference rules with the cutting planes\ninference rules. The R(CP) proof system is essentially resolution over linear inequalities, that is,\nresolution that operates with disjunctions of linear inequalities instead of clauses.\nThe main motivation of [Kra98] is to extend the feasible interpolation technique and consequently\nthe lower bounds results, from cutting planes and resolution to stronger proof systems. That paper\nestablishes an exponential-size lower bound on a restricted version of R(CP) proofs, namely, when\nthe number of inequalities in each proof-line is O(nε), where n is the number of variables of the\ninitial formulas, ε is a small enough constant and the coefficients in the cutting planes inequalities\nare polynomially bounded.\nOther papers considering extensions of resolution over linear inequalities are the more recent\npapers by Hirsch & Kojevnikov [HK06] and Kojevnikov [Koj07]. The first paper [HK06] considers\na combination of resolution with LP (an incomplete subsystem of cutting planes based on simple\nlinear programming reasoning), with the ‘lift and project’ proof system (L&P), and with the cutting\nplanes proof system. The second paper [Koj07] deals with improving the parameters of the tree-like\nR(CP) lower-bounds obtained in [Kra98].\nWhereas previous results concerned primarily with extending the cutting planes proof system,\nour foremost motivation is to extend and improve previous results on algebraic proof systems\noperating with multilinear formulas obtained in [RT06]. In that paper the concept of multilinear\nproofs was introduced and several basic results concerning multilinear proofs were proved.\nIn\nparticular, polynomial-size proofs of two important combinatorial principles were demonstrated:\nthe functional pigeonhole principle and the Tseitin (mod p) graph tautologies. In the current paper\nwe improve both these results.\nAs mentioned above, motivated by relations with multilinear proofs operating with depth-3 mul-\ntilinear formulas, we shall consider a certain subsystem of resolution over linear equations. For\nthis subsystem we apply twice the interpolation by a communication game technique. The first\n4"},{"paragraph_id":"p5","order":5,"text":"application is of the non-monotone version of the technique, and the second application is of the\nmonotone version. Namely, the first application provides a general (non-monotone) interpolation\ntheorem that demonstrates a polynomial (in the size of refutations) upper bound on interpolants;\nThe proof uses the general method of transforming a refutation into a Karchmer-Wigderson com-\nmunication game for two players, from which a Boolean circuit is then attainable. In particular,\nwe shall apply the interpolation theorem of Kraj ́ıˇcek from [Kra97]. The second application of the\n(monotone) interpolation by a communication game technique is implicit and proceeds by using\nthe lower bound criterion of Bonet, Pitassi & Raz in [BPR97]. This criterion states that (semantic)\nproof systems (of a certain natural and standard kind) whose proof-lines (considered as Boolean\nfunctions) have low communication complexity cannot prove efficiently a certain tautology (namely,\nthe clique-coloring tautologies).\n1.2. Summary of Results. This paper introduces and connects several new concepts and ideas\nwith some known ones. It identifies new extensions of resolution operating with linear equations,\nand relates (a certain) such extension to multilinear proofs. The upper bounds for the pigeonhole\nprinciple and Tseitin mod p formulas in fragments of resolution over linear equations are new. By\ngeneralizing the machinery developed in [RT06], these upper bounds yield new and improved re-\nsults concerning multilinear proofs. The lower bound for the clique-coloring formulas in a fragment\nof resolution over linear equations employs the standard monotone interpolation by a communica-\ntion game technique, and specifically utilizes the theorem of Bonet, Pitassi & Raz from [BPR97].\nThe general (non-monotone) interpolation result for a fragment of resolution over linear equations\nemploys the theorem of Kraj ́ıˇcek from [Kra97]. The upper bound in (the stronger variant of –\nas described in the introduction) resolution over linear equations of the clique-coloring formulas\nfollows that of Atserias, Bonet & Esteban [ABE02]. We now give a detailed outline of the results\nin this paper.\nThe proof systems. In Section 3 we formally define two extensions of resolution of decreasing\nstrength allowing resolution to operate with disjunctions of linear equations. The size of a linear\nequation a1x1 + . . . + anxn = a0 is the sum of all a0, . . . , an written in unary notation. The size of\na disjunction of linear equations is the total size of all linear equations in the disjunction. The size\nof a proof operating with disjunctions of linear equations is the total size of all the disjunctions in\nit.\nR(lin): This is the stronger proof system (described in the introduction) that operates with\ndisjunctions of linear equations with integer coefficients.\nR0(lin): This is a (provably proper) fragment of R(lin). It operates with disjunctions of (arbi-\ntrarily many) linear equations whose variables have constant coefficients, under the restriction that\nevery disjunction can be partitioned into a constant number of sub-disjunctions, where each sub-\ndisjunction either consists of linear equations that differ only in their free-terms or is a (translation\nof a) clause.\nNote that any single linear inequality with Boolean variables can be represented by a disjunction\nof linear equations that differ only in their free-terms (see the example in the introduction section).\nSo the R0(lin) proof system is close to a proof system operating with disjunctions of constant\nnumber of linear inequalities (with constant integral coefficients). In fact, disjunctions of linear\nequations varying only in their free-terms, have more (expressive) strength than a single inequality.\nFor instance, the parity function can be easily represented by a disjunction of linear equations,\nwhile it cannot be represented by a single linear inequality (or even by a disjunction of linear\ninequalities).\nAs already mentioned, the motivation to consider the restricted proof system R0(lin) comes from\nits relation to multilinear proofs operating with depth-3 multilinear formulas (in short, depth-3\n5"},{"paragraph_id":"p6","order":6,"text":"multilinear proofs): R0(lin) corresponds roughly to the subsystem of R(lin) that we know how\nto simulate by depth-3 multilinear proofs via the technique in [RT06] (the technique is based on\nconverting disjunctions of linear forms into symmetric polynomials, which are known to have small\ndepth-3 multilinear formulas). This simulation is then applied in order to improve over known\nupper bounds for depth-3 multilinear proofs, as R0(lin) is already sufficient to efficiently prove\ncertain “hard tautologies”.\nMoreover, we are able to establish an exponential lower bound on\nR0(lin) refutations size (see below for both upper and lower bounds on R0(lin) proofs). We also\nestablish a super-polynomial separation of R(lin) from R0(lin) (via the clique-coloring principle, for\na certain choice of parameters; see below).\nShort refutations. We demonstrate the following short refutations in R0(lin) and R(lin):\n(1) Polynomial-size refutations of the pigeonhole principle in R0(lin);\n(2) Polynomial-size refutations of Tseitin mod p graph formulas in R0(lin);\n(3) Polynomial-size refutations of the clique-coloring formulas in R(lin) (for certain parameters).\nThe refutations here follow by direct simulation of the Res(2) refutations of clique-coloring\nformulas from [ABE02].\nAll the three families of formulas above are prominent “hard tautologies” in proof complexity\nliterature, which means that strong size lower bounds on proofs in various proof systems are known\nfor them (for the exact formulation of these families of formulas see Section 6).\nInterpolation results. We provide a polynomial upper-bound on (non-monotone) interpolants\ncorresponding to R0(lin) refutations; Namely, we show that any R0(lin)-refutation of a given formula\ncan be transformed into a (non-monotone) Boolean circuit computing the corresponding interpolant\nfunction of the formula (if there exists such a function), with at most a polynomial increase in size.\nWe employ the general interpolation theorem of Kraj ́ıˇcek [Kra97] for semantic proof systems.\nLower bounds. We provide the following exponential lower bound:\nTheorem 1. R0(lin) does not have sub-exponential refutations for the clique-coloring formulas.\nThis result is proved by applying a result of Bonet, Pitassi & Raz [BPR97], that (implicitly) use\nthe monotone interpolation by a communication game technique for establishing an exponential-\nsize lower bound on refutations of general semantic proof systems operating with proof-lines of low\ncommunication complexity.\nApplications to multilinear proofs. Multilinear proof systems are (semantic) refutation sys-\ntems operating with multilinear polynomials over a fixed field, where every multilinear polynomial\nis represented by a multilinear arithmetic formula.\nIn this paper we shall consider multilinear\nformulas over fields of characteristic 0 only. The size of a multilinear proof (that is, a proof in\na multilinear proof system) is the total size of all multilinear formulas in the proof (for formal\ndefinitions concerning multilinear proofs see Section 9).\nWe shall first connect multilinear proofs with resolution over linear equations by the following\nresult:\nTheorem 2. Multilinear proofs operating with depth-3 multilinear formulas over characteristic 0\npolynomially-simulate R0(lin).\nAn immediate corollary of this theorem and the upper bounds in R0(lin) described above are\npolynomial-size multilinear proofs for the pigeonhole principle and the Tseitin mod p formulas.\n(1) Polynomial-size depth-3 multilinear refutations for the pigeonhole principle over fields of\ncharacteristic 0. This improves over [RT06] that shows a similar upper bound for a weaker\nprinciple, namely, the functional pigeonhole principle.\n(2) Polynomial-size depth-3 multilinear refutations for the Tseitin mod p graph formulas over\nfields of characteristic 0. These refutations are different than those demonstrated in [RT06],\n6"},{"paragraph_id":"p7","order":7,"text":"and further they establish short multilinear refutations of the Tseitin mod p graph formulas\nover any field of characteristic 0 (the proof in [RT06] showed how to refute the Tseitin mod\np formulas by multilinear refutations only over fields that contain a primitive pth root of\nunity).\nRelations with cutting planes proofs. As mentioned in the introduction, a proof system com-\nbining resolution with cutting planes was presented by Kraj ́ıˇcek in [Kra98]. The resulting system\nis denoted R(CP) (see Section 10 for a definition). When the coefficients in the linear inequalities\ninside R(CP) proofs are polynomially bounded, the resulting proof system is denoted R(CP*). We\nestablish the following simulation result:\nTheorem 3. R(lin) polynomially simulates resolution over cutting planes inequalities with polyno-\nmially bounded coefficients R(CP*).\nWe do not know if the converse also holds.\n2. Notation and Background on Propositional Proof Systems\nFor a natural number n, we use [n] to denote {1, . . . , n}. For a vector of n (integral) coefficients\n⃗a and a vector of n variables ⃗x, we denote by ⃗a · ⃗x the scalar product Pn\ni=1 aixi. If ⃗b is another\nvector (of length n), then ⃗a +⃗b denotes the addition of ⃗a and ⃗b as vectors, and c⃗a (for an integer\nc) denotes the product of the scalar c with ⃗a (where, −⃗a denotes −1⃗a). For two linear equations\nL1 : ⃗a · ⃗x = a0 and L2 : ⃗b · ⃗x = b0, their addition (⃗a +⃗b) · ⃗x = a0 + b0 is denoted L1 + L2 (and their\nsubtraction (⃗a −⃗b) · ⃗x = a0 −b0 is denoted L1 −L2). For two Boolean assignments (identified as\n0, 1 strings) α, α′ ∈{0, 1}n we write α′ ≥α if α′\ni ≥αi, for all i ∈[n] (where αi, α′\ni are the ith bits\nof α and α′, respectively).\nWe now recall some basic concepts on propositional proof systems. For background on algebraic\nproof systems (and specifically multilinear proofs) see Section 9.\nResolution. In order to put our work in context, we need to define the resolution refutation system.\nA CNF formula over the variables x1, . . . , xn is defined as follows. A literal is a variable xi or\nits negation ¬xi. A clause is a disjunction of literals. A CNF formula is a conjunction of clauses.\nThe size of a clause is the number of literals in it.\nResolution is a complete and sound proof system for unsatisfiable CNF formulas. Let C and D\nbe two clauses containing neither xi nor ¬xi, the resolution rule allows one to derive C ∨D from\nC ∨xi and D ∨¬xi. The clause C ∨D is called the resolvent of the clauses C ∨xi and D ∨¬xi on\nthe variable xi, and we also say that C ∨xi and D ∨¬xi were resolved over xi. The weakening rule\nallows to derive the clause C ∨D from the clause C, for any two clauses C, D.\nDefinition 2.1 (Resolution). A resolution proof of the clause D from a CNF formula K is a\nsequence of clauses D1, D2, . . . , Dl, such that: (1) each clause Dj is either a clause of K or a\nresolvent of two previous clauses in the sequence or derived by the weakening rule from a previous\nclause in the sequence; (2) the last clause Dl= D. The size of a resolution proof is the sum of all\nthe sizes of the clauses in it. A resolution refutation of a CNF formula K is a resolution proof of\nthe empty clause ✷from K (the empty clause stands for false; that is, the empty clause has no\nsatisfying assignments).\nA proof in resolution (or any of its extensions) is called also a derivation or a proof-sequence.\nEach sequence-element in a proof-sequence is called also a proof-line. A proof-sequence containing\nthe proof-lines D1, . . . , Dlis also said to be a derivation of D1, . . . , Dl.\n7"},{"paragraph_id":"p8","order":8,"text":"Cook-Reckhow proof systems. Following [CR79], a Cook-Reckhow proof system is a polynomial-\ntime algorithm A that receives a Boolean formula F (for instance, a CNF) and a string π over some\nfinite alphabet (“the (proposed) refutation” of F), such that there exists a π with A(F, π) = 1 if\nand only if F is unsatisfiable. The completeness of a (Cook-Reckhow) proof system (with respect\nto the set of all unsatisfiable Boolean formulas; or for a subset of it, e.g. the set of unsatisfiable\nCNF formulas) stands for the fact that every unsatisfiable formula F has a string π (“the refutation\nof F”) so that A(F, π) = 1. The soundness of a (Cook-Reckhow) proof system stands for the fact\nthat every formula F so that A(F, π) = 1 for some string π is unsatisfiable (in other words, no\nsatisfiable formula has a refutation).\nFor instance, resolution is a Cook-Reckhow proof system, since it is complete and sound for the\nset of unsatisfiable CNF formulas, and given a CNF formula F and a string π it is easy to check in\npolynomial-time (in both F and π) whether π constitutes a resolution refutation of F.\nWe shall also consider proof systems that are not necessarily (that is, not known to be) Cook-\nReckhow proof systems. Specifically, multilinear proof systems (over large enough fields) meet the\nrequirements in the definition of Cook-Reckhow proof systems, except that the condition on A\nabove is relaxed: we allow A to be in probabilistic polynomial-time BPP (which is not known to\nbe equal to deterministic polynomial-time).\nPolynomial simulations of proof systems. When comparing the strength of different proof\nsystems we shall confine ourselves to CNF formulas only. That is, we consider propositional proof\nsystems as proof systems for the set of unsatisfiable CNF formulas. For that purpose, if a proof\nsystem does not operate with clauses directly, then we fix a (direct) translation from clauses to\nthe objects operated by the proof system. This is done for both resolution over linear equations\n(which operate with disjunctions of linear equations) and its fragments, and also for multilinear\nproofs (which operate with multilinear polynomials, represented as multilinear formulas); see for\nexample Subsection 3.1 for such a direct translation.\nDefinition 2.2. Let P1, P2 be two proof systems for the set of unsatisfiable CNF formulas (we\nidentify a CNF formula with its corresponding translation, as discussed above). We say that P2\npolynomially simulates P1 if given a P1 refutation π of a CNF formula F, then there exists a\nrefutation of F in P2 of size polynomial in the size of π. In case P2 polynomially simulates P1 while\nP1 does not polynomially simulates P2 we say that P2 is strictly stronger than P1.\n3. Resolution over Linear Equations and its Subsystems\nThe proof systems we consider in this section are extensions of resolution. Proof-lines in res-\nolution are clauses.\nInstead of this, the extensions of resolution we consider here operate with\ndisjunctions of linear equations with integral coefficients. For this section we use the convention\nthat all the formal variables in the propositional proof systems considered are taken from the set\nX := {x1, . . . , xn}.\n3.1. Disjunctions of Linear Equations. For L a linear equation a1x1 + . . . + anxn = a0, the\nright hand side a0 is called the free-term of L and the left hand side a1x1 + . . . + anxn is called the\nlinear form of L (the linear form can be 0). A disjunction of linear equations is of the following\ngeneral form:"},{"paragraph_id":"p9","order":9,"text":"a(1)\n1 x1 + . . . + a(1)\nn xn = a(1)\n0"},{"paragraph_id":"p10","order":10,"text":"∨· · · ∨"},{"paragraph_id":"p11","order":11,"text":"a(t)\n1 x1 + . . . + a(t)\nn xn = a(t)\n0"},{"paragraph_id":"p12","order":12,"text":",\n(1)\nwhere t ≥0 and the coefficients a(j)\ni\nare integers (for all 0 ≤i ≤n, 1 ≤j ≤t). We discard duplicate\nlinear equations from a disjunction of linear equations. The semantics of such a disjunction is the\nnatural one: We say that an assignment of integral values to the variables x1, ..., xn satisfies (1)\n8"},{"paragraph_id":"p13","order":13,"text":"if and only if there exists j ∈[t] so that the equation a(j)\n1 x1 + . . . + a(j)\nn xn = a(j)\n0\nholds under the\ngiven assignment.\nThe symbol |= denotes the semantic implication relation, that is, for every collection D1, . . . , Dm\nof disjunctions of linear equations,\nD1, . . . , Dm |= D0\nmeans that every assignment of 0, 1 values that satisfies all D1, . . . , Dm also satisfies D0.2 In this\ncase we also say that D1, . . . , Dm semantically imply D0.\nThe size of a linear equation a1x1 + . . . + anxn = a0 is Pn\ni=0 |ai|, i.e., the sum of the bit sizes\nof all ai written in unary notation. Accordingly, the size of the linear form a1x1 + . . . + anxn is\nPn\ni=1 |ai|. The size of a disjunction of linear equations is the total size of all linear equations in it.\nSince all linear equations considered in this paper are of integral coefficients, we shall speak\nof linear equations when we actually mean linear equations with integral coefficients. Similar to\nresolution, the empty disjunction is unsatisfiable and stands for the truth value false.\nTranslation of clauses. As described in the introduction, we can translate any CNF formula to\na collection of disjunctions of linear equations in a direct manner: Every clause W\ni∈I xi ∨W\nj∈J ¬xj\n(where I and J are sets of indices of variables) pertaining to the CNF is translated into the\ndisjunction W\ni∈I(xi = 1) ∨W\nj∈J(xj = 0). For a clause D we denote by eD its translation into a\ndisjunction of linear equations. It is easy to verify that any Boolean assignment to the variables\nx1, . . . , xn satisfies a clause D if and only if it satisfies eD (where true is treated as 1 and false as\n0).\n3.2. Resolution over Linear Equations – R(lin). Defined below is our basic proof system\nR(lin) that enables resolution to reason with disjunctions of linear equations. As we wish to reason\nabout Boolean variables we augment the system with the axioms (xi = 0)∨(xi = 1), for all i ∈[n],\ncalled the Boolean axioms.\nDefinition 3.1 (R(lin)). Let K := {K1, . . . , Km} be a collection of disjunctions of linear equations.\nAn R(lin)-proof from K of a disjunction of linear equations D is a finite sequence π = (D1, ..., Dl)\nof disjunctions of linear equations, such that Dl= D and for every i ∈[l], either Di = Kj for\nsome j ∈[m], or Di is a Boolean axiom (xh = 0) ∨(xh = 1) for some h ∈[n], or Di was deduced\nby one of the following R(lin)-inference rules, using Dj, Dk for some j, k < i:\nResolution: Let A, B be two disjunctions3of linear equations and let L1, L2 be two linear\nequations.\nFrom A ∨L1 and B ∨L2 derive A ∨B ∨(L1 + L2).\nSimilarly, from A ∨L1 and B ∨L2 derive A ∨B ∨(L1 −L2).\nWeakening: From a disjunction of linear equations A derive A∨L , where L is an arbitrary\nlinear equation over X.\nSimplification: From A ∨(0 = k) derive A, where A is a disjunction of linear equations\nand k ̸= 0.\nAn R(lin) refutation of a collection of disjunctions of linear equations K is a proof of the empty\ndisjunction from K. The size of an R(lin)-proof π is the total size of all the disjunctions of linear\nequations in π, denoted |π|.\nSimilar to resolution, in case A ∨B ∨(L1 + L2) is derived from A ∨L1 and B ∨L2 by the\nresolution rule, we say that A ∨L1 and B ∨L2 were resolved over L1 and L2, respectively, and we\n2Alternatively, we can consider assignments of any integral values (instead of only Boolean values) to the variables\nin D1, . . . , Dm, stipulating that the collection D1, . . . , Dm contains all disjunctions of the form (xj = 0) ∨(xj = 1)\nfor all the variables xj ∈X (these formulas force any satisfying assignment to give only 0, 1 values to the variables).\n3Possibly the empty disjunction. This remark also applies to the inference rules below.\n9"},{"paragraph_id":"p14","order":14,"text":"call A ∨B ∨(L1 + L2) the resolvent of A ∨L1 and B ∨L2 (and similarly, when A ∨B ∨(L1 −L2)\nis derived from A ∨L1 and B ∨L2 by the resolution rule; we use the same terminology for both\naddition and subtraction, and it should be clear from the context which operation is actually\napplied). We also describe such an application of the resolution rule by saying that L1 was added\n(resp., subtracted) to (resp. from) L2 in A ∨L1 and B ∨L2.\nIn light of the direct translation between CNF formulas and collections of disjunctions of linear\nequations (described in the previous subsection), we can consider R(lin) to be a proof system for\nthe set of unsatisfiable CNF formulas:\nProposition 1. The R(lin) refutation system is a sound and complete Cook-Reckhow (see Sec-\ntion 2) refutation system for unsatisfiable CNF formulas (translated into unsatisfiable collection of\ndisjunctions of linear equations).\nProof: Completeness of R(lin) (for the set of unsatisfiable CNF formulas) stems from a straight-\nforward simulation of resolution, as we now show.\nClaim 1. R(lin) polynomially simulates resolution.\nProof of claim: Proceed by induction on the length of the resolution refutation to show that any\nresolution derivation of a clause A can be translated with only a linear increase in size into an R(lin)\nderivation of the corresponding disjunction of linear equations eA (see the previous subsection for\nthe definition of eA).\nThe base case: An initial clause A is translated into its corresponding disjunction of linear\nequations eA.\nThe induction step: If a resolution clause A ∨B was derived by the resolution rule from A ∨xi\nand B ∨¬xi, then in R(lin) we subtract (xi = 0) from (xi = 1) in eB ∨(xi = 0) and eA ∨(xi = 1),\nrespectively, to obtain eA ∨eB ∨(0 = 1). Then, using the Simplification rule, we can cut-off(0 = 1)\nfrom eA ∨eB ∨(0 = 1), and arrive at eA ∨eB.\nIf a clause A ∨B was derived in resolution from A by the Weakening rule, then we derive eA ∨eB\nfrom eA by the Weakening rule in R(lin).\nSoundness of R(lin) stems from the soundness of the inference rules (which means that: If D\nwas derived from C, B by the R(lin) resolution rule then any assignment that satisfies both C and\nB also satisfies D; and if D was derived from C by either the Weakening rule or the Simplification\nrule, then any assignment that satisfies C also satisfies D).\nThe R(lin) proof system is a Cook-Reckhow proof system, as it is easy to verify in polynomial-\ntime whether an R(lin) proof-line is inferred, by an application of one of R(lin)’s inference rules,\nfrom a previous proof-line (or proof-lines). Thus, any sequence of disjunctions of linear equations,\ncan be checked in polynomial-time (in the size of the sequence) to decide whether or not it is a\nlegitimate R(lin) proof-sequence.\nIn Section 5 we shall see that a stronger notion of completeness (that is, implicational complete-\nness) holds for R(lin) and its subsystems.\n3.3. Fragment of Resolution over Linear Equations – R0(lin). Here we consider a restriction\nof R(lin), denoted R0(lin). As discussed in the introduction section, R0(lin) is roughly the fragment\nof R(lin) we know how to polynomially simulate with depth-3 multilinear proofs.\nBy results established in the sequel (Sections 6.3 and 8) R(lin) is strictly stronger than R0(lin),\nwhich means that R(lin) polynomially simulates R0(lin), while the converse does not hold.\nR0(lin) operates with disjunctions of (arbitrarily many) linear equations with constant coefficients\n(excluding the free terms), under the following restriction: Every disjunction can be partitioned\n10"},{"paragraph_id":"p15","order":15,"text":"into a constant number of sub-disjunctions, where each sub-disjunction either consists of linear\nequations that differ only in their free-terms or is a (translation of a) clause.\nAs mentioned in the introduction, every linear inequality with Boolean variables can be rep-\nresented by a disjunction of linear equations that differ only in their free-terms. So the R0(lin)\nproof system resembles, to some extent, a proof system operating with disjunctions of constant\nnumber of linear inequalities with constant integral coefficients (on the other hand, it is probable\nthat R0(lin) is stronger than such a proof system, as a disjunction of linear equations that differ\nonly in their free terms is [expressively] stronger than a linear inequality [or even a disjunction of\nlinear inequalities]: the former can define the parity function while the latter cannot).\nExample of an R0(lin)-line:\n(x1 + . . . + xl= 1) ∨· · · ∨(x1 + . . . + xl= l) ∨(xl+1 = 1) ∨· · · ∨(xn = 1),\nfor some 1 ≤l≤n. The next section contains other concrete (and natural) examples of R0(lin)-\nlines.\nLet us define formally what it means to be an R0(lin) proof-line, that is, a proof-line inside an\nR0(lin) proof, called R0(lin)-line:\nDefinition 3.2 (R0(lin)-line). Let D be a disjunction of linear equations whose variables have\nconstant integer coefficients (the free-terms are unbounded). Assume D can be partitioned into a\nconstant number k of sub-disjunctions D1, . . . , Dk, where each Di either consists of (an unbounded)\ndisjunction of linear equations that differ only in their free-terms, or is a translation of a clause (as\ndefined in Subsection 3.1). Then the disjunction D is called an R0(lin)-line.\nThus, any R0(lin)-line is of the following general form:\n_\ni∈I1"},{"paragraph_id":"p16","order":16,"text":"⃗a(1) · ⃗x = l(1)\ni"},{"paragraph_id":"p17","order":17,"text":"∨· · · ∨\n_\ni∈Ik"},{"paragraph_id":"p18","order":18,"text":"⃗a(k) · ⃗x = l(k)\ni"},{"paragraph_id":"p19","order":19,"text":"∨\n_\nj∈J\n(xj = bj) ,\n(2)\nwhere k and all at\nr (for r ∈[n] and t ∈[k]) are integer constants and bj ∈{0, 1} (for all j ∈J) (and\nI1, . . . , Ik, J are unbounded sets of indices). Note that a disjunction of clauses can be combined\ninto a single clause. Hence, without loss of generality we can assume that in any R0(lin)-line only\na single (translation of a) clause occurs. This is depicted in (2) (where in addition we have ignored\nin (2) the possibility that the single clause obtained by combining several clauses contains xj ∨¬xj,\nfor some j ∈[n]).\nDefinition 3.3 (R0(lin)). The R0(lin) proof system is a restriction of the R(lin) proof system in\nwhich each proof-line is an R0(lin)-line (as in Definition 3.2).\nFor a completeness proof of R0(lin) see Section 5.4\n4. Reasoning and Counting inside R(lin) and its Subsystems\nIn this section we illustrate a simple way to reason by case-analysis inside R(lin) and its subsys-\ntems. This kind of reasoning will simplify the presentation of proofs inside R(lin) (and R0(lin)) in\nthe sequel (essentially, a similar – though weaker – kind of reasoning is applicable already in reso-\nlution). We will then demonstrate efficient and transparent proofs for simple counting arguments\nthat will also facilitate us in the sequel.\n4The simulation of resolution inside R(lin) (in the proof of Proposition 1) is carried on with each R(lin) proof-line\nbeing in fact a translation of a clause, and hence, an R0(lin)-line (notice that the Boolean axioms of R(lin) are\nR0(lin)-lines). This already implies that R0(lin) is a complete refutation system for the set of unsatisfiable CNF\nformulas. In section 5 we give a proof of a stronger notion of completeness for R0(lin).\n11"},{"paragraph_id":"p20","order":20,"text":"4.1. Basic Reasoning inside R(lin) and its Subsystems. Given K a collection of disjunc-\ntions of linear equations {K1, . . . , Km} and C a disjunction of linear equations, denote by K ∨C\nthe collection {K1 ∨C, . . . , Km ∨C}.\nRecall that the formal variables in our proof system are\nx1, . . . , xn.\nLemma 4. Let K be a collection of disjunctions of linear equations, and let z abbreviate some linear\nform with integer coefficients. Let E1, . . . , Elbe ldisjunctions of linear equations. Assume that for\nall i ∈[l] there is an R(lin) derivation of Ei from z = ai and K with size at most s where a1, . . . , al\nare distinct integers. Then, there is an R(lin) proof of Wl\ni=1 Ei from K and (z = a1)∨· · ·∨(z = al),\nwith size polynomial in s and l.\nProof: Denote by D the disjunction (z = a1) ∨· · · ∨(z = al) and by πi the R(lin) proof of Ei from\nK and z = ai (with size at most s), for all i ∈[l]. It is easy to verify that for all i ∈[l] the sequence\nπi ∨W\nj∈[l]\\{i}(z = aj) is an R(lin) proof of Ei ∨W\nj∈[l]\\{i}(z = aj) from K and D. So overall, given\nD and K as premises, there is an R(lin) derivation of size polynomial in s and lof the following\ncollection of disjunctions of linear equations:\nE1 ∨\n_\nj∈[l]\\{1}\n(z = aj), . . . , El∨\n_\nj∈[l]\\{l}\n(z = aj) .\n(3)\nWe now use the Resolution rule to cut-offall the equations (z = ai) inside all the disjunctions\nin (3). Formally, we prove that for every 1 ≤k ≤lthere is a polynomial-size (in s and l) R(lin)\nderivation from (3) of\nE1 ∨· · · ∨Ek ∨\n_\nj∈[l]\\[k]\n(z = aj) ,\n(4)\nand so putting k = l, will conclude the proof of the lemma.\nWe proceed by induction on k.\nThe base case for k = 1 is immediate (from (3)).\nFor the\ninduction case, assume that for some 1 ≤k < lwe already have an R(lin) proof of (4), with size\npolynomial in s and l.\nConsider the line\nEk+1 ∨\n_\nj∈[l]\\{k+1}\n(z = aj) .\n(5)\nWe can now cut-offthe disjunctions W\nj∈[l]\\[k](z = aj) and W\nj∈[l]\\{k+1}(z = aj) from (4) and (5),\nrespectively, using the Resolution rule (since the aj’s in (4) and in (5) are disjoint).\nWe will\ndemonstrate this derivation in some detail now, in order to exemplify a proof carried inside R(lin).\nWe shall be less formal sometime in the sequel.\nResolve (4) with (5) over (z = ak+1) and (z = a1), respectively, to obtain\n(0 = a1 −ak+1) ∨E1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,k+1}\n(z = aj) .\n(6)\nSince a1 ̸= ak+1, we can use the Simplification rule to cut-off(0 = a1 −ak+1) from (6), and we\narrive at\nE1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,k+1}\n(z = aj) .\n(7)\nNow, similarly, resolve (4) with (7) over (z = ak+1) and (z = a2), respectively, and use Simplification\nto obtain\nE1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,2,k+1}\n(z = aj) .\n12"},{"paragraph_id":"p21","order":21,"text":"Continue in a similar manner until you arrive at\nE1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,2,...,k,k+1}\n(z = aj) ,\nwhich is precisely what we need.\nUnder the appropriate conditions, Lemma 4 also holds for R0(lin) proofs. This is stated in the\nfollowing lemma.\nLemma 5. Let K be a collection of disjunctions of linear equations, and let z abbreviate a linear\nform with integer coefficients. Let E1, . . . , Elbe ldisjunctions of linear equations. Assume that for\nall i ∈[l] there is an R0(lin) derivation of Ei from z = ai and K with size at most s, where the\nai’s are distinct integers. Then, assuming Wl\ni=1 Ei is an R0(lin)-line, there is an R0(lin) proof of\nWl\ni=1 Ei from K and (z = a1) ∨· · · ∨(z = al), with size polynomial in s and l.\nProof: It can be verified by simple inspection that, under the conditions spelled out in the state-\nment of the lemma, each proof-line in the R(lin) derivations in the proof of Lemma 4 is actually\nan R0(lin)-line.5\nAbbreviations. Lemmas 4 and 5 will sometime facilitate us to proceed inside R(lin) and R0(lin)\nwith a slightly less formal manner. For example, the situation in Lemma 4 above can be depicted\nby saying that “if z = ai implies Ei (with a polynomial-size proof) for all i ∈[l], then Wl\ni=1(z = ai)\nimplies Wl\ni=1 Ei (with a polynomial-size proof)”.\nIn case Wl\ni=1(z = ai) above is just the Boolean axiom (xi = 0) ∨(xi = 1), for some i ∈[n], and\nxi = 0 implies E0 and xi = 1 implies E1 (both with polynomial-size proofs), then to simplify the\nwriting we shall sometime not mention the Boolean axiom at all. For example, the latter situation\ncan be depicted by saying that “if xi = 0 implies E0 with a polynomial-size proof and xi = 1 implies\nE1 with a polynomial-size proof, then we can derive E0 ∨E1 with a polynomial-size proof”.\n4.2. Basic Counting inside R(lin) and R0(lin). In this subsection we illustrate how to effi-\nciently prove several basic counting arguments inside R(lin) and R0(lin). This will facilitate us in\nshowing short proofs for hard tautologies in the sequel. In accordance with the last paragraph in\nthe previous subsection, we shall carry the proofs inside R(lin) and R0(lin) with a slightly less rigor.\nLemma 6. Let z1 abbreviate ⃗a · ⃗x and z2 abbreviate ⃗b · ⃗x. Let D1 be W\nα∈A(z1 = α) and let D2 be\nW\nβ∈B (z2 = β), where A, B are two (finite) sets of integers. Then there is a polynomial-size (in the\nsize of D1, D2) R(lin) proof from D1, D2 of:\n_\nα∈A,β∈B\n(z1 + z2 = α + β) .\n(8)\nMoreover, if ⃗a and ⃗b consist of constant integers (which means that D1, D2 are R0(lin)-lines), then\nthere is a polynomial-size (in the size of D1, D2) R0(lin) proof of (8) from D1, D2.\nProof: Denote the elements of A by α1, . . . , αk. In case z1 = αi, for some i ∈[k] then we can add\nz1 = αi to every equation in W\nβ∈B (z2 = β) to get W\nβ∈B(z1 + z2 = αi + β). Therefore, there exist\nk R(lin) proofs, each with polynomial-size (in |D1| and |D2|), of\n_\nβ∈B\n(z1 + z2 = α1 + β) ,\n_\nβ∈B\n(z1 + z2 = α2 + β) ,\n. . .\n,\n_\nβ∈B\n(z1 + z2 = αk + β)\n5Note that when the proofs of Ei from z = ai, for all i ∈[l], are all done inside R0(lin), then the linear form z\nought to have constant coefficients.\n13"},{"paragraph_id":"p22","order":22,"text":"from z1 = α1, z1 = α2 ,. . . ,z1 = αk, respectively.\nThus, by Lemma 4, we can derive\n_\nα∈A,β∈B\n(z1 + z2 = α + β)\n(9)\nfrom D1 and D2 in a polynomial-size (in |D1| and |D2|) R(lin)-proof. This concludes the first part\nof the lemma.\nAssume that ⃗a and ⃗b consist of constant coefficients only. Then by inspecting the R(lin)-proof\nof (9) from D1 and D2 demonstrated above (and by using Lemma 5 instead of Lemma 4), one can\nverify that this proof is in fact carried inside R0(lin).\nAn immediate corollary of Lemma 6 is the efficient formalization in R(lin) of the following obvious\ncounting argument: If a linear form equals some value in the interval (of integer numbers) [a0, a1]\nand another linear form equals some value in [b0, b1] (for some a0 ≤a1 and b0 ≤b1), then their\naddition equals some value in [a0 + b0, a1 + b1]. More formally:\nCorollary 7. Let z1 abbreviate ⃗a·⃗x and z2 abbreviate ⃗b·⃗x. Let D1 be (z1 = a0)∨(z1 = a0 +1) . . .∨\n(z1 = a1), and let D2 be (z2 = b0) ∨(z2 = b0 + 1) . . . ∨(z2 = b1). Then there is a polynomial-size\n(in the size of D1, D2) R(lin) proof from D1, D2 of\n(z1 + z2 = a0 + b0) ∨(z1 + z2 = a0 + b0 + 1) ∨. . . ∨(z1 + z2 = a1 + b1) .\n(10)\nMoreover, if ⃗a and ⃗b consist of constant integers (which means that D1, D2 are R0(lin)-lines), then\nthere is a polynomial-size (in the size of D1, D2) R0(lin) proofs of (10) from D1, D2.\nLemma 8. Let ⃗a · ⃗x be a linear form with n variables, and let A := {⃗a · ⃗x | ⃗x ∈{0, 1}n} be the set\nof all possible values of ⃗a ·⃗x over Boolean assignments to ⃗x. Then there is a polynomial-size, in the\nsize of the linear form ⃗a · ⃗x,6 R(lin) proof of\n_\nα∈A\n(⃗a · ⃗x = α) .\n(11)\nMoreover, if the coefficients in ⃗a are constants, then there is a polynomial-size (in the size of ⃗a · ⃗x)\nR0(lin) proof of (11).\nProof: Without loss of generality, assume that all the coefficients in ⃗a are nonzero. Consider the\nBoolean axiom (x1 = 0) ∨(x1 = 1) and the (first) coefficient a1 from ⃗a. Assume that a1 ≥1. Add\n(x1 = 0) to itself a1 times, and arrive at (a1x1 = 0) ∨(x1 = 1). Then, in the resulted line, add\n(x1 = 1) to itself a1 times, until the following is reached:\n(a1x1 = 0) ∨(a1x1 = a1) .\nSimilarly, in case a1 ≤−1 we can subtract (|a1| + 1 many times) (x1 = 0) from itself in (x1 =\n0) ∨(x1 = 1), and then subtract (|a1| + 1 many times) (x1 = 1) from itself in the resulted line.\nIn the same manner, we can derive the disjunctions: (a2x2 = 0) ∨(a2x2 = a2), . . . , (anxn =\n0) ∨(anxn = an).\nConsider (a1x1 = 0) ∨(a1x1 = a1) and (a2x2 = 0) ∨(a2x2 = a2). From these two lines, by\nLemma 6, there is a polynomial-size in |a1| + |a2| derivation of:\n(a1x1 + a2x2 = 0) ∨(a1x1 + a2x2 = a1) ∨(a1x1 + a2x2 = a2) ∨(a1x1 + a2x2 = a1 + a2) .\n(12)\nIn a similar fashion, now consider (a3x3 = 0) ∨(a3x3 = a3) and apply again Lemma 6, to obtain\n_\nα∈A′\n(a1x1 + a2x2 + a3x3 = α) ,\n(13)\n6Recall that the size of ⃗a · ⃗x is Pn\ni=1 |ai|, that is, the size of the unary representation of ⃗a.\n14"},{"paragraph_id":"p23","order":23,"text":"where A′ are all possible values to a1x1 + a2x2 + a3x3 over Boolean assignments to x1, x2, x3. The\nderivation of (13) is of size polynomial in |a1| + |a2| + |a3|.\nContinue to consider, successively, all other lines (a4x4 = 0) ∨(a4x4 = a4), . . . , (anxn = 0) ∨\n(anxn = an), and apply the same reasoning. Each step uses a derivation of size at most polynomial\nin Pn\ni=1 |ai|. And so overall we reach the desired line (11), with a derivation of size polynomial in\nthe size of ⃗a · ⃗x. This concludes the first part of the lemma.\nAssume that ⃗a consists of constant coefficients only. Then by inspecting the R(lin)-proof demon-\nstrated above (and by using the second part of Lemma 6), one can see that this proof is in fact\ncarried inside R0(lin).\nLemma 9. There is a polynomial-size (in n) R0(lin) proof from\n(x1 = 1) ∨· · · ∨(xn = 1)\n(14)\nof\n(x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n) .\n(15)\nProof: We show that for every i ∈[n], there is a polynomial-size (in n) R0(lin) proof from (xi = 1)\nof (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n). This concludes the proof since, by Lemma 5,\nwe then can derive from (14) (with a polynomial-size (in n) R0(lin) proof) the disjunction (14) in\nwhich each (xi = 1) (for all i ∈[n]) is replace by (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n),\nwhich is precisely the disjunction (15) (note that (15) is an R0(lin)-line).\nClaim 2. For every i ∈[n], there is a a polynomial-size (in n) R0(lin) proof from (xi = 1) of\n(x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n).\nProof of claim: By Lemma 8, for every i ∈[n] there is a polynomial-size (in n) R0(lin) proof\n(using only the Boolean axioms) of\n(x1 + . . . + xi−1 + xi+1 + . . . + xn = 0) ∨· · · ∨(x1 + . . . + xi−1 + xi+1 + . . . + xn = n −1) .\n(16)\nNow add successively (xi = 1) to every equation in (16) (note that this can be done in R0(lin)).\nWe obtain precisely (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n).\nLemma 10. There is a polynomial-size (in n) R0(lin) proof of (x1+. . .+xn = 0)∨(x1+. . .+xn = 1)\nfrom the collection of disjunctions consisting of (xi = 0) ∨(xj = 0), for all 1 ≤i < j ≤n.\nProof: We proceed by induction on n. The base case for n = 1 is immediate from the Boolean\naxiom (x1 = 0) ∨(x1 = 1). Assume we already have a polynomial-size proof of\n(x1 + . . . + xn = 0) ∨(x1 + . . . + xn = 1).\n(17)\nIf xn+1 = 0 we add xn+1 = 0 to both of the equations in (17), and reach:\n(x1 + . . . + xn+1 = 0) ∨(x1 + . . . + xn+1 = 1).\n(18)\nOtherwise, xn+1 = 1, and so we can cut-off(xn+1 = 0) in all the initial disjunctions (xi = 0) ∨\n(xn+1 = 0), for all 1 ≤i ≤n. We thus obtain (x1 = 0), . . . , (xn = 0). Adding together (x1 =\n0), . . . , (xn = 0) and (xn+1 = 1) we arrive at\n(x1 + . . . + xn+1 = 1) .\n(19)\nSo overall, either (18) holds or (19) holds; and so (using Lemma 5) we arrive at the disjunction of\n(19) and (18), which is precisely (18).\n15"},{"paragraph_id":"p24","order":24,"text":"5. Implicational Completeness of R(lin) and its Subsystems\nIn this section we provide a proof of the implicational completeness of R(lin) and its subsystems.\nWe shall need this property in the sequel (see Section 6.2). The implicational completeness of a\nproof system is a stronger property than mere completeness. Essentially, a system is implicationally\ncomplete if whenever something is semantically implied by a set of initial premises, then it is\nalso derivable from the initial premises. In contrast to this, mere completeness means that any\ntautology (or in case of a refutation system, any unsatisfiable set of initial premises) has a proof in\nthe system (respectively, a refutation in the system). As a consequence, the proof of implicational\ncompleteness in this section establishes an alternative completeness proof to that obtained via\nsimulating resolution (see Proposition 1). Note that we are not concerned in this section with the\nsize of the proofs, but only with their existence.\nRecall the definition of the semantic implication relation |= from Section 3.1. Formally, we say\nthat R(lin) is implicationally complete if for every collection of disjunctions of linear equations\nD0, D1, . . . , Dm, it holds that D1, . . . , Dm |= D0 implies that there is an R(lin) proof of D0 from\nD1, . . . , Dm.\nTheorem 11. R(lin) is implicationally complete.\nProof: We proceed by induction on n, the number of variables x1, . . . , xn in D0, D1, . . . , Dm.\nThe base case n = 0. We need to show that D1, . . . , Dm |= D0 implies that there is an R(lin)\nproof of D0 from D1, . . . , Dm, where all Di’s (for 0 ≤i ≤m) have no variables but only constants.\nThis means that each Di is a disjunction of equations of the form (0 = a0) for some integer a0 (if\na linear equation have no variables, then the left hand side of this equation must be 0; see Section\n3.1).\nThere are two cases to consider. In the first case D0 is satisfiable. Since D0 has no variables,\nthis means precisely that D0 is the equation (0 = 0). Thus, D0 can be derived easily from any\naxiom in R(lin) (for instance, by subtracting each equation in (x1 = 0) ∨(x1 = 1) from itself, to\nreach (0 = 0) ∨(0 = 0), which is equal to (0 = 0), since we discard duplicate equations inside\ndisjunctions).\nIn the second case D0 is unsatisfiable. Thus, since D1, . . . , Dm |= D0, there is no assignment sat-\nisfying all D1, . . . , Dm. Hence, there must be at least one unsatisfiable disjunction Di in D1, . . . , Dm\n(as a disjunction with no variables is either tautological or unsatisfiable). Such an unsatisfiable Di\nis a disjunction of zero or more unsatisfiable equations of the form (0 = a0), for some integer a0 ̸= 0.\nWe can then use Simplification to cut-offall the unsatisfiable equations in Di to reach the empty\ndisjunction. By the Weakening rule, we can now derive D0 from the empty disjunction.\nThe induction step. Assume that the theorem holds for disjunctions with n variables. Let the\nunderlying variables of D0, D1, . . . , Dm be x1, . . . , xn+1, and assume that\nD1, . . . , Dm |= D0 .\n(20)\nWe write the disjunction D0 as:\nt_\nj=1\n n\nX\ni=1\na(j)\ni xi + a(j)\nn+1xn+1 = a(j)\n0\n!\n,\n(21)\nwhere the a(j)\ni ’s are integer coefficients. We need to show that there is an R(lin) proof of D0 from\nD1, . . . , Dm.\nLet D be a disjunction of linear equations, let xi be a variable and let b ∈{0, 1}. We shall\ndenote by D↾xi=b the disjunction D, where in every equation in D the variable xi is substituted by\nb, and the constant terms in the left hand sides of all resulting equations (after substituting b for\n16"},{"paragraph_id":"p25","order":25,"text":"xi) switch sides (and change signs, obviously) to the right hand sides of the equations (we have to\nswitch sides of constant terms, as by definition linear equations in R(lin) proofs have all constant\nterms appearing only on the right hand sides of equations).\nWe now reason (slightly) informally inside R(lin) (as illustrated in Section 4.1). Fix some b ∈\n{0, 1}, and assume that xn+1 = b. Then, from D1, . . . , Dm we can derive (inside R(lin)):\nD1↾xn+1=b, . . . , Dm↾xn+1=b .\n(22)\nThe only variables occurring in (22) are x1, . . . , xn. From assumption (20) we clearly have D1↾xn+1=b\n, . . . , Dm↾xn+1=b |= D0↾xn+1=b. And so by the induction hypothesis there is an R(lin) derivation of\nD0↾xn+1=b from D1↾xn+1=b, . . . , Dm↾xn+1=b. So overall, assuming that xn+1 = b, there is an R(lin)\nderivation of D0↾xn+1=b from D1, . . . , Dm.\nWe now consider the two possible cases: xn+1 = 0 and xn+1 = 1.\nIn case xn+1 = 0, by the above discussion, we can derive D0↾xn+1=0 from D1, . . . , Dm. For every\nj ∈[t], add successively (a(j)\nn+1 times) the equation xn+1 = 0 to the jth equation in D0↾xn+1=0 (see\n(21)). We thus obtain precisely D0.\nIn case xn+1 = 1, again, by the above discussion, we can derive D0↾xn+1=1 from D1, . . . , Dm. For\nevery j ∈[t], add successively (a(j)\nn+1 times) the equation xn+1 = 1 to the jth equation in D0↾xn+1=1\n(recall that we switch sides of constant terms in every linear equation after the substitution of xn+1\nby 1 is performed in D0↾xn+1=1). Again, we obtain precisely D0.\nBy inspecting the proof of Theorem 11, it is possible to verify that if all the disjunctions\nD0, , . . . , Dm are R0(lin)-lines (see Definition 3.2), then the proof of D0 in R(lin) uses only R0(lin)-\nlines as well. Therefore, we have:\nCorollary 12. R0(lin) is implicationally complete.\nRemark 1. Corollary 12 states that any R0(lin)-line that is semantically implied by a set of initial\nR0(lin)-lines, is in fact derivable in R0(lin) from the initial R0(lin)-lines. On the other hand, it is\npossible that a certain proof of the same R0(lin)-line inside R(lin) will be significantly shorter than\nthe proof inside R0(lin). Indeed, we shall see in Section 8 that for certain CNF formulas R(lin) has\na super-polynomial speed-up over R0(lin).\n6. Short Proofs for Hard Tautologies\nIn this section we show that R0(lin) is already enough to admit small proofs for “hard” counting\nprinciples like the pigeonhole principle and the Tseitin graph formulas for constant degree graphs.\nOn the other hand, as we shall see in Section 8, R0(lin) inherits the same weakness that cutting\nplanes proofs have with respect to the clique-coloring tautologies. Nevertheless, we can efficiently\nprove the clique-coloring principle in (the stronger system) R(lin), but not by using R(lin) “ability\nto count”, rather by using its (straightforward) ability to simulate Res(2) proofs (that is, resolution\nproofs extended to operate with 2-DNF formulas, instead of clauses).\n6.1. The\nPigeonhole\nPrinciple\nTautologies\nin\nR0(lin). This\nsubsection\nillustrates\npolynomial-size R0(lin) proofs of the pigeonhole principle.\nThis will allow us to establish\npolynomial-size multilinear proofs operating with depth-3 multilinear formulas of the pigeonhole\nprinciple (in Section 9).\nThe m to n pigeonhole principle states that m pigeons cannot be mapped one-to-one into n < m\nholes. The negation of the pigeonhole principle, denoted ¬PHPm\nn , is formulated as an unsatisfiable\nCNF formula as follows (where clauses are translated to disjunctions of linear equations):\nDefinition 6.1. The ¬PHPm\nn is the following set of clauses:\n(1) Pigeons axioms:\n(xi,1 = 1) ∨· · · ∨(xi,n = 1), for all 1 ≤i ≤m;\n17"},{"paragraph_id":"p26","order":26,"text":"(2) Holes axioms:\n(xi,k = 0) ∨(xj,k = 0),\nfor all 1 ≤i < j ≤m and for all 1 ≤k ≤n.\nThe intended meaning of each propositional variable xi,j is that the ith pigeon is mapped to the\njth hole.\nWe now describe a polynomial-size in n refutation of ¬PHPm\nn inside R0(lin). For this purpose it\nis sufficient to prove a polynomial-size refutation of the pigeonhole principle when the number of\npigeons m equals n + 1 (because the set of clauses pertaining to ¬PHPn+1\nn\nis already contained in\nthe set of clauses pertaining to ¬PHPm\nn , for any m > n). Thus, we fix m = n+1. In this subsection\nwe shall say a proof in R0(lin) is of polynomial-size, always intending polynomial-size in n (unless\notherwise stated).\nBy Lemma 9, for all i ∈[m] we can derive from the Pigeon axiom (for the ith pigeon):\n(xi,1 + . . . + xi,n = 1) ∨· · · ∨(xi,1 + . . . + xi,n = n)\n(23)\nwith a polynomial-size R0(lin) proof.\nBy Lemma 10, from the Hole axioms we can derive, with a polynomial-size R0(lin) proof\n(x1,j + . . . + xm,j = 0) ∨(x1,j + . . . + xm,j = 1),\n(24)\nfor all j ∈[n].\nLet S abbreviate the sum of all formal variables xi,j. In other words,\nS :=\nX\ni∈[m],j∈[n]\nxi,j .\nLemma 13. There is a polynomial-size R0(lin) proof from (23) (for all i ∈[m]) of\n(S = m) ∨(S = m + 1) · · · ∨(S = m · n).\nProof: For every i ∈[m] fix the abbreviation zi := xi,1 + . . . + xi,n.\nThus, by (23) we have\n(zi = 1) ∨· · · ∨(zi = n).\nConsider (z1 = 1) ∨· · · ∨(z1 = n) and (z2 = 1) ∨· · · ∨(z2 = n). By Corollary 7, we can derive\nfrom these two lines\n(z1 + z2 = 2) ∨(z1 + z2 = 3) ∨· · · ∨(z1 + z2 = 2n)\n(25)\nwith a polynomial-size R0(lin) proof.\nNow, consider (z3 = 1) ∨· · · ∨(z3 = n) and (25). By Corollary 7 again, from these two lines we\ncan derive with a polynomial-size R0(lin) proof:\n(z1 + z2 + z3 = 3) ∨(z1 + z2 + z3 = 4) ∨· · · ∨(z1 + z2 + z3 = 3n) .\n(26)\nContinuing in the same way, we eventually arrive at\n(z1 + . . . + zm = m) ∨(z1 + . . . + zm = m + 1) ∨· · · ∨(z1 + . . . + zm = m · n) ,\nwhich concludes the proof, since S equals z1 + . . . + zm.\nLemma 14. There is a polynomial-size R0(lin) proof from (24) of\n(S = 0) ∨· · · ∨(S = n).\nProof: For all j ∈[n], fix the abbreviation yj := x1,j + . . . + xm,j.\nThus, by (24) we have\n(yj = 0) ∨(yj = 1), for all j ∈[n]. Now the proof is similar to the proof of Lemma 8, except that\nhere single variables are abbreviations of linear forms.\nIf y1 = 0 then we can add y1 to the two sums in (y2 = 0) ∨(y2 = 1), and reach (y1 + y2 =\n0) ∨(y1 + y2 = 1) and if y1 = 1 we can do the same and reach (y1 + y2 = 1) ∨(y1 + y2 = 2). So, by\nLemma 5, we can derive with a polynomial-size R0(lin) proof\n(y1 + y2 = 0) ∨(y1 + y2 = 1) ∨(y1 + y2 = 2) .\n(27)\n18"},{"paragraph_id":"p27","order":27,"text":"Now, we consider the three cases in (27): y1 +y2 = 0 or y1 +y2 = 1 or y1 +y2 = 2, and the clause\n(y3 = 0) ∨(y3 = 1). We arrive in a similar manner at (y1 + y2 + y3 = 0) ∨· · · ∨(y1 + y2 + y3 = 3).\nWe continue in the same way until we arrive at (S = 0) ∨· · · ∨(S = n).\nTheorem 15. There is a polynomial-size R0(lin) refutation of the m to n pigeonhole principle\n¬PHPm\nn .\nProof: By Lemmas 13 and 14 above, all we need is to show a polynomial-size refutation of (S =\nm) ∨· · · ∨(S = m · n) and (S = 0) ∨· · · ∨(S = n).\nSince n < m, for all 0 ≤k ≤n, if S = k then using the Resolution and Simplification rules we\ncan cut-offall the sums in (S = m) ∨· · · ∨(S = m · n) and arrive at the empty clause. Thus, by\nLemma 5, there is a polynomial-size R0(lin) proof of the empty clause from (S = 0) ∨· · · ∨(S = n)\nand (S = m) ∨· · · ∨(S = m · n).\n6.2. Tseitin mod p Tautologies in R0(lin). This subsection establishes polynomial-size R0(lin)\nproofs of Tseitin graph tautologies (for constant degree graphs). This will allow us (in Section 9)\nto extend the multilinear proofs of the Tseitin mod p tautologies to any field of characteristic 0\n(the proofs in [RT06] required working over a field containing a primitive pth root of unity when\nproving the Tseitin mod p tautologies; for more details see Section 9).\nTseitin mod p tautologies (introduced in [BGIP01]) are generalizations of the (original, mod 2)\nTseitin graph tautologies (introduced in [Tse68]). To build the intuition for the generalized version,\nwe start by describing the (original) Tseitin mod 2 principle.\nLet G = (V, E) be a connected\nundirected graph with an odd number of vertices n. The Tseitin mod 2 tautology states that there\nis no sub-graph G′ = (V, E′), where E′ ⊆E, so that for every vertex v ∈V , the number of edges\nfrom E′ incident to v is odd. This statement is valid, since otherwise, summing the degrees of all\nthe vertices in G′ would amount to an odd number (since n is odd), whereas this sum also counts\nevery edge in E′ twice, and so is even.\nAs mentioned above, the Tseitin mod 2 principle was generalized by Buss et al. [BGIP01] to\nobtain the Tseitin mod p principle. Let p ≥2 be some fixed integer and let G = (V, E) be a\nconnected undirected r-regular graph with n vertices and no double edges. Let G′ = (V, E′) be the\ncorresponding directed graph that results from G by replacing every (undirected) edge in G with\ntwo opposite directed edges. Assume that n ≡1 (mod p). Then, the Tseitin mod p principle states\nthat there is no way to assign to every edge in E′ a value from {0, . . . , p −1}, so that:\n(i): For every pair of opposite directed edges e, ̄e in E′, with assigned values a, b, respectively,\na + b ≡0 (mod p); and\n(ii): For every vertex v in V , the sum of the values assigned to the edges in E′ coming out of\nv is congruent to 1 (mod p).\nThe Tseitin mod p principle is valid, since if we sum the values assigned to all edges of E′ in\npairs we obtain 0 (mod p) (by (i)), where summing them by vertices we arrive at a total value of 1\n(mod p) (by (ii) and since n ≡1 (mod p)). We shall see in what follows, that this simple counting\nargument can be carried on in a natural (and efficient) way already inside R0(lin).\nAs an unsatisfiable propositional formula (in CNF form) the negation of the Tseitin mod p\nprinciple is formulated by assigning a variable xe,i for every edge e ∈E′ and every residue i modulo\np. The variable xe,i is an indicator variable for the fact that the edge e has an associated value i.\nThe following are the clauses of the Tseitin mod p CNF formula (as translated to disjunctions of\nlinear equations).\nDefinition 6.2 (Tseitin mod p formulas (¬TseitinG,p)). Let p ≥2 be some fixed integer and\nlet G = (V, E) be a connected undirected r-regular graph with n vertices and no double edges, and\n19"},{"paragraph_id":"p28","order":28,"text":"assume that n ≡1 (mod p). Let G′ = (V, E′) be the corresponding directed graph that results\nfrom G by replacing every (undirected) edge in G with two opposite directed edges.\nGiven a vertex v ∈V , denote the edges in E′ coming out of v by e[v, 1], . . . , e[v, r] and define the\nfollowing set of (translation of) clauses:\nMODp,1(v):=\n( r_\nk=1\n(xe[v,k],ik = 0)\n i1, . . . , ir ∈{0, . . . , p −1} and\nr\nX\nk=1\nik ̸≡1 mod p\n)\n.\nThe Tseitin mod p formula, denoted ¬TseitinG,p, consists of the following (translation) of clauses:\n1.\np−1\nW\ni=0\n(xe,i = 1) , for all e ∈E′\n(expresses that every edge is assigned at least one value from 0, . . . , p −1);\n2. (xe,i = 0) ∨(xe,j = 0) , for all i ̸= j ∈{0, . . . , p −1} and all e ∈E′\n(expresses that every edge is assigned at most one value from 0, . . . , p −1);\n3. (xe,i = 1) ∨(x ̄e,p−i = 0) and (xe,i = 0) ∨(x ̄e,p−i = 1), 7\nfor all two opposite directed edges e, ̄e ∈E′ and all i ∈{0, . . . , p −1}\n(expresses condition (i) of the Tseitin mod p principle above);\n4. MODp,1(v) , for all v ∈V\n(expresses condition (ii) of the Tseitin mod p principle above).\nNote that for every edge e ∈E′, the polynomials of (1,2) in Definition 6.2, combined with the\nBoolean axioms of R0(lin), force any collection of edge-variables xe,0, . . . , xe,p−1 to contain exactly\none i ∈{0, . . . , p −1} so that xe,i = 1.\nAlso, it is easy to verify that, given a vertex v ∈V ,\nany assignment σ of 0, 1 values (to the relevant variables) satisfies both the disjunctions of (1,2)\nand the disjunctions of MODp,1(v) if and only if σ corresponds to an assignment of values from\n{0, . . . , p −1} to the edges coming out of v that sums up to 1 (mod p).\nUntil the rest of this subsection we fix an integer p ≥2 and a connected undirected r-regular\ngraph G = (V, E) with n vertices and no double edges, such that n ≡1 mod p and r is a constant.\nAs in Definition 6.2, we let G′ = (V, E′) be the corresponding directed graph that results from G\nby replacing every (undirected) edge in G with two opposite directed edges. We now proceed to\nrefute ¬TseitinG,p inside R0(lin) with a polynomial-size (in n) refutation.\nGiven a vertex v ∈V , and the edges in E′ coming out of v, denoted e[v, 1], . . . , e[v, r], define the\nfollowing abbreviation:\nαv :=\nr\nX\nj=1\np−1\nX\ni=0\ni · xe[v,j],i .\n(28)\nLemma 16. Let v ∈V be any vertex in G′. Then there is a constant-size R0(lin) proof from\n¬TseitinG,p of the following disjunction:\nr−1\n_\nl=0\n(αv = 1 + l· p) .\n(29)\nProof: Let Tv ⊆¬TseitinG,p be the set of all disjunctions of the form (1,2,4) from Definition 6.2\nthat contain only variables pertaining to vertex v (that is, all the variables xe,i, where e ∈E′ is an\nedge coming out of v, and i ∈{0, . . . , p −1}).\n7If i = 0 then x ̄e,p−i denotes x ̄e,0.\n20"},{"paragraph_id":"p29","order":29,"text":"Claim 3. Tv semantically implies (29), that is:8\nTv |=\nr−1\n_\nl=0\n(αv = 1 + l· p) .\nProof of claim: Let σ be an assignment of 0, 1 values to the variables in Tv that satisfies both the\ndisjunctions of (1,2) and the disjunctions of MODp,1(v) in Definition 6.2. As mentioned above (the\ncomment after Definition 6.2), such a σ corresponds to an assignment of values from {0, . . . , p −1}\nto the edges coming out of v, that sums up to 1 mod p. This means precisely that αv = 1 mod p\nunder the assignment σ. Thus, there exists a nonnegative integer k, such that αv = 1 + kp under\nσ.\nIt remains to show that k ≤r −1 (and so the only possible values that αv can get under σ\nare 1, 1 + p, 1 + 2p, . . . , 1 + (r −1)p). Note that because σ gives the value 1 to only one variable\nfrom xe[v,j],0, . . . , xe[v,j],p−1 (for every j ∈[r]), then the maximal value that αv can have under σ is\nr(p −1). Thus, 1 + kp ≤rp −r and so k ≤r −1.\nFrom Claim 3 and from the implicational completeness of R0(lin) (Corollary 12), there exists an\nR0(lin) derivation of (29) from Tv. It remains to show that this derivation is of constant-size.\nSince the degree r of G′ and the modulus p are both constants, both Tv and (29) have constant\nnumber of variables and constant coefficients (including the free-terms). Thus, there is a constant-\nsize R0(lin) derivation of (29) from Tv.\nLemma 17. There is a polynomial-size (in n) R0(lin) derivation from ¬TseitinG,p of the following\ndisjunction:\n(r−1)·n\n_\nl=0\n X\nv∈V\nαv = n + l· p\n!\n.\nProof: Simply add successively all the equations pertaining to disjunctions (29), for all vertices\nv ∈V .\nFormally, we show that for every subset of vertices V ⊆V , with |V| = k, there is a\npolynomial-size (in n) R0(lin) derivation from ¬TseitinG,p of\n(r−1)·k\n_\nl=0\n X\nv∈V\nαv = k + l· p\n!\n,\n(30)\nand so putting V = V , will conclude the proof.\nWe proceed by induction on the size of V. The base case, |V| = 1, is immediate from Lemma 16.\nAssume that we already derived (30) with a polynomial-size (in n) R0(lin) proof, for some V ⊂V ,\nsuch that |V| = k < n. Let u ∈V \\ V. By Lemma 16, we can derive\nr−1\n_\nl=0\n(αu = 1 + l· p)\n(31)\nfrom ¬TseitinG,p with a constant-size proof. Now, by Lemma 6, each linear equation in (31) can\nbe added to each linear equation in (30), with a polynomial-size (in n) R0(lin) proof. This results\nin the following disjunction:\n(r−1)·(k+1)\n_\nl=0"},{"paragraph_id":"p30","order":30,"text":"X\nv∈V∪{u}\nαv = k + 1 + l· p"},{"paragraph_id":"p31","order":31,"text":",\n8Recall that we only consider assignments of 0, 1 values to variables when considering the semantic implication\nrelation |=.\n21"},{"paragraph_id":"p32","order":32,"text":"which is precisely what we need to conclude the induction step.\nLemma 18. Let e, ̄e be any pair of opposite directed edges in G′ and let i ∈{0, . . . , p −1}. Let\nTe ⊆¬TseitinG,p be the set of all disjunctions of the form (1,2,3) from Definition 6.2 that contain\nonly variables pertaining to edges e, ̄e (that is, all the variables xe,j, x ̄e,j, for all j ∈{0, . . . , p −1}).\nThen, there is a constant-size R0(lin) proof from Te of the following disjunction:\n(i · xe,i + (p −i) · x ̄e,p−i = 0) ∨(i · xe,i + (p −i) · x ̄e,p−i = p) .\n(32)\nProof: First note that Te semantically implies\n(xe,i + x ̄e,p−i = 0) ∨(xe,i + x ̄e,p−i = 2) .\n(33)\nThe number of variables in Te and (33) is constant. Hence, there is a constant-size R0(lin)-proof\nof (32) from Te. Also note that\n(xe,i + x ̄e,p−i = 0) ∨(xe,i + x ̄e,p−i = 2) |=\n(i · xe,i + (p −i) · x ̄e,p−i = 0) ∨(i · xe,i + (p −i) · x ̄e,p−i = p) .\n(34)\nTherefore, there is also an R0(lin)-proof of constant-size from Te of the lower line in (34).\nWe are now ready to complete the polynomial-size R0(lin) refutation of ¬TseitinG,p. Using the\ntwo prior lemmas, the refutation idea is simple, as we now explain. Observe that\nX\nv∈V\nαv =\nX\n{e, ̄e}⊆E′\ni∈{0,...,p−1}\n(i · xe,i + (p −i) · x ̄e,p−i) ,\n(35)\nwhere by {e, ̄e} ⊆E′ we mean that e, ̄e is pair of opposite directed edges in G′.\nDerive by Lemma 17 the disjunction\n(r−1)·n\n_\nl=0\n X\nv∈V\nαv = n + l· p\n!\n.\n(36)\nThis disjunction expresses the fact that P\nv∈V αv = 1 mod p (since n = 1 mod p). On the other\nhand, using Lemma 18, we can “sum together” all the equations (32) (for all {e, ̄e} ⊆E′ and all\ni ∈{0, . . . , p −1}), to obtain a disjunction expressing the statement that\nX\n{e, ̄e}⊆E′\ni∈{0,...,p−1}\n(i · xe,i + (p −i) · x ̄e,p−i) = 0\nmod p .\nBy Equation (35), we then obtain the desired contradiction. This idea is formalized in the proof of\nthe following theorem:\nTheorem 19. Let G = (V, E) be an r-regular graph with n vertices, where r is a constant. Fix\nsome modulus p. Then, there are polynomial-size (in n) R0(lin) refutations of ¬TseitinG,p.\nProof: First, use Lemma 17 to derive\n(r−1)·n\n_\nl=0\n X\nv∈V\nαv = n + l· p\n!\n.\n(37)\nSecond, use Lemma 18 to derive\n(i · xe,i + (p −i) · x ̄e,p−i = p) ∨(i · xe,i + (p −i) · x ̄e,p−i = 0) ,\n(38)\nfor every pair of opposite directed edges in G′ = (V, E′) (as in Definition 6.2) and every residue\ni ∈{0, . . . , p −1}.\n22"},{"paragraph_id":"p33","order":33,"text":"We now reason inside R0(lin).\nPick a pair of opposite directed edges e, ̄e and a residue i ∈\n{0, . . . , p −1}. If i · xe,i + (p −i) · x ̄e,p−i = 0, then subtract this equation successively from every\nequation in (37). We thus obtain a new disjunction, similar to that of (37), but which does not\ncontain the xe,i and x ̄e,p−i variables, and with the same free-terms.\nOtherwise, i·xe,i+(p−i)·x ̄e,p−i = p, then subtract this equation successively from every equation\nin (37). Again, we obtain a new disjunction, similar to that of (37), but which does not contain\nthe xe,i and x ̄e,p−i variables, and such that p is subtracted from every free-term in every equation.\nSince, by assumption, n ≡1 mod p, the free-terms in every equation are (still) equal 1 mod p.\nSo overall, in both cases (i · xe,i + (p −i) · x ̄e,p−i = 0 and i · xe,i + (p −i) · x ̄e,p−i = p) we obtained\na new disjunction with all the free-terms in equations equal 1 mod p.\nWe now continue the same process for every pair e, ̄e of opposite directed edges in G′ and\nevery residue i. Eventually, we discard all the variables xe,i in the equations, for every e ∈E′\nand i ∈{0, . . . , p −1}, while all the free-terms in every equation remain to be equal 1 mod p.\nTherefore, we arrive at a disjunction of equations of the form (0 = γ) for some γ = 1 mod p.\nBy using the Simplification rule we can cut-offall such equations, and arrive finally at the empty\ndisjunction.\n6.3. The Clique-Coloring Principle in R(lin). In this section we observe that there are\npolynomial-size R(lin) proofs of the clique-coloring principle (for certain, weak, parameters). This\nimplies, in particular, that R(lin) does not possess the feasible monotone interpolation property\n(see more details on the interpolation method in Section 7).\nAtserias, Bonet & Esteban [ABE02] demonstrated polynomial-size Res(2) refutations of the\nclique-coloring formulas (for certain weak parameters; Theorem 20). Thus, it is sufficient to show\nthat R(lin) polynomially-simulates Res(2) proofs (Proposition 2). This can be shown in a straight-\nforward manner. As noted in the first paragraph of Section 6, because the proofs of the clique-\ncoloring formula we discuss here only follow the proofs inside Res(2), then in fact these proofs do\nnot take any advantage of the capacity “to count” inside R(lin) (this capacity is exemplified, for\ninstance, in Section 4.2).\nWe start with the clique-coloring formulas (these formulas will also be used in Section 8). These\nformulas express the clique-coloring principle that has been widely used in the proof complexity\nliterature (cf., [BPR97], [Pud97], [Kra97], [Kra98], [ABE02], [Kra07]). This principle is based on\nthe following basic combinatorial idea. Let G = (V, E) be an undirected graph with n vertices and\nlet k′ < k be two integers. Then, one of the following must hold:\n(i): The graph G does not contain a clique with k vertices;\n(ii): The graph G is not a complete k′-partite graph.\nIn other words, there is no way to\npartition G into k′ subgraphs G1, . . . , Gk′, such that every Gi is an independent set, and\nfor all i ̸= j ∈[k′], all the vertices in Gi are connected by edges (in E) to all the vertices in\nGj.\nObviously, if Item (ii) above is false (that is, if G is a complete k′-partite graph), then there\nexists a k′-coloring of the vertices of G; hence the name clique-coloring for the principle.\nThe propositional formulation of the (negation of the) clique-coloring principle is as follows.\nEach variable pi,j, for all i ̸= j ∈[n], is an indicator variable for the fact that there is an edge in\nG between vertex i and vertex j. Each variable ql,i, for all l∈[k] and all i ∈[n], is an indicator\nvariable for the fact that the vertex i in G is the lth vertex in the k-clique. Each variable rl,i, for\nall l∈[k′] and all i ∈[n], is an indicator variable for the fact that the vertex i in G pertains to the\nindependent set Gl.\nDefinition 6.3. The negation of the clique-coloring principle consists of the following unsatisfiable\ncollection of clauses (as translated to disjunctions of linear equations), denoted ¬cliquen\nk,k′:\n23"},{"paragraph_id":"p34","order":34,"text":"(i) (ql,1 = 1) ∨· · · ∨(ql,n = 1), for all l∈[k]\n(expresses that there exists at least one vertex in G which constitutes the lth vertex of\nthe k-clique);\n(ii) (ql,i = 0) ∨(ql,j = 0), for all i ̸= j ∈[n], l∈[k]\n(expresses that there exists at most one vertex in G which constitutes the lth vertex of\nthe k-clique);\n(iii) (ql,i = 0) ∨(ql′,i = 0), for all i ∈[n], l̸= l′ ∈[k]\n(expresses that the ith vertex of G cannot be both the lth and the l′th vertex of the\nk-clique);\n(iv) (ql,i = 0) ∨(ql′,j = 0) ∨(pi,j = 1), for all l̸= l′ ∈[k], i ̸= j ∈[n]\n(expresses that if both the vertices i and j in G are in the k-clique, then there is an edge\nin G between i and j);\n(v) (r1,i = 1) ∨· · · ∨(rk′,i = 1), for all i ∈[n]\n(expresses that every vertex of G pertains to at least one independent set);\n(vi) (rl,i = 0) ∨(rl′,i = 0), for all l̸= l∈[k′], i ∈[n]\n(expresses that every vertex of G pertains to at most one independent set);\n(vii) (pi,j = 0) ∨(rt,i = 0) ∨(rt,j = 0), for all i ̸= j ∈[n], t ∈[k′]\n(expresses that if there is an edge between vertex i and j in G, then i and j cannot be\nin the same independent set);\nRemark 2. Our formulation of the clique-coloring formulas above is similar to the one used by\n[BPR97], except that we consider also the pi,j variables (we added the (iv) clauses and changed\naccordingly the (vii) clauses). This is done for the sake of clarity of the contradiction itself, and\nalso to make it clear that the formulas are in the appropriate form required by the interpolation\nmethod (see Section 7 for details on the interpolation method). By resolving over the pi,j variables\nin (iv) and (vii), one can obtain precisely the collection of clauses in [BPR97].\nAtserias, Bonet & Esteban [ABE02] demonstrated polynomial-size (in n) Res(2) refutations of\n¬cliquen\nk,k′, when k = √n and k′ = (log n)2/8 log log n. These are rather weak parameters, but\nthey suffice to establish the fact that Res(2) does not possess the feasible monotone interpolation\nproperty.\nThe Res(2) proof system (also called 2-DNF resolution), first considered in [Kra01], is resolution\nextended to operate with 2-DNF formulas, defined as follows.\nA 2-term is a conjunction of up to two literals. A 2-DNF is a disjunction of 2-terms. The size\nof a 2-term is the number of literals in it (that is, either 1 or 2). The size of a 2-DNF is the total\nsize of all the 2-terms in it.\nDefinition 6.4 (Res(2)). A Res(2) proof of a 2-DNF D from a collection K of 2-DNFs is a\nsequence of 2-DNFs D1, D2, . . . , Ds , such that Ds = D, and every Dj is either from K or was\nderived from previous line(s) in the sequence by the following inference rules:\nCut: Let A, B be two 2-DNFs.\nFrom A∨V2\ni=1 li and B∨W2\ni=1 ¬li derive A∨B, where the li’s are (not necessarily distinct)\nliterals (and ¬li is the negation of the literal li).\nAND-introduction: Let A, B be two 2-DNFs and l1, l2 two literals.\nFrom A ∨l1 and B ∨l2 derive A ∨B ∨V2\ni=1 li.\nWeakening: From a 2-DNF A derive A ∨V2\ni=1 li , where the li’s are (not necessarily dis-\ntinct) literals.\nA Res(2) refutation of a collection of 2-DNFs K is a Res(2) proof of the empty disjunction ✷from\nK (the empty disjunction stands for false). The size of a Res(2) proof is the total size of all the\n2-DNFs in it.\n24"},{"paragraph_id":"p35","order":35,"text":"Given a collection K of 2-DNFs we translate it into a collection of disjunctions of linear equations\nvia the following translation scheme. For a literal l, denote by bl the translation that maps a variable\nxi into xi, and ¬xi into 1−xi. A 2-term l1 ∧l2 is first transformed into the equation bl1 +bl2 = 2, and\nthen moving the free-terms in the left hand side of bl1 + bl2 = 2 (in case there are such free-terms)\nto the right hand side; So that the final translation of l1 ∧l2 has only a single free-term in the\nright hand side. A disjunction of 2-terms (that is, a 2-DNF) D = W\ni∈I(li,1 ∧li,2) is translated into\nthe disjunction of the translations of the 2-terms, denoted by bD. It is clear that every assignment\nsatisfies a 2-DNF D if and only if it satisfies bD.\nProposition 2. R(lin) polynomially simulates Res(2). In other words, if π is a Res(2) proof of D\nfrom a collection of 2-DNFs K1, . . . , Kt, then there is an R(lin) proof of bD from bK1, . . . , bKt whose\nsize is polynomial in the size of π.\nThe proof of Proposition 2 proceeds by induction on the length (that is, the number of proof-\nlines) in the Res(2) proof. This is pretty straightforward and similar to the simulation of resolution\nby R(lin), as illustrated in the proof of Proposition 1. We omit the details.\nTheorem 20 ([ABE02]). Let k = √n and k′ = (log n)2/8 log log n. Then ¬cliquen\nk,k′ has Res(2)\nrefutations of size polynomial in n.\nThus, Proposition 2 yields the following:\nCorollary 21. Let k, k′ be as in Theorem 20. Then ¬cliquen\nk,k′ has R(lin) refutations of size\npolynomial in n.\nThe following corollary is important (we refer the reader to Section A in the Appendix for the\nnecessary relevant definitions concerning the feasible monotone interpolation property and to Section\n7 for explanation and definitions concerning the general [non-monotone] interpolation method).\nCorollary 22. R(lin) does not possess the feasible monotone interpolation property.\nRemark 3. The proof of ¬cliquen\nk,k′ inside Res(2) demonstrated in [ABE02] (and hence, also\nthe corresponding proof inside R(lin)) proceeds along the following lines. First reduce ¬cliquen\nk,k′\nto the k to k′ pigeonhole principle.\nFor the appropriate values of the parameters k and k′ —\nand specifically, for the values in Theorem 20 — there is a short resolution proof of the k to k′\npigeonhole principle (this was shown by Buss & Pitassi [BP97]); (this resolution proof is polynomial\nin the number of pigeons k, but not in the number of holes k′, which is exponentially smaller than\nk).9 Therefore, in order to conclude the refutation of ¬cliquen\nk,k′ inside Res(2) (or inside R(lin)), it\nsuffices to simulate the short resolution refutation of the k to k′ pigeonhole principle. It is important\nto emphasize this point: After reducing, inside R(lin), ¬cliquen\nk,k′ to the pigeonhole principle, one\nsimulates the resolution refutation of the pigeonhole principle, and this has nothing to do with\nthe small-size R0(lin) refutations of the pigeonhole principle demonstrated in Section 6.1. This is\nbecause, the reduction (inside R(lin)) of ¬cliquen\nk,k′ to the k to k′ pigeonhole principle, results in\na substitution instance of the pigeonhole principle formulas; in other words, the reduction results\nin a collection of disjunctions that are similar to the pigeonhole principle disjunctions where each\noriginal pigeonhole principle variable is substituted by some big formula (and, in particular, these\ndisjunctions are not R0(lin)-lines at all). (Note that R0(lin) does not admit short proofs of the\nclique-coloring formulas as we show in Section 8.)\n9Whenever k ≥2k′ the k to k′ pigeonhole principle is referred to as the weak pigeonhole principle.\n25"},{"paragraph_id":"p36","order":36,"text":"7. Interpolation Results for R0(lin)\nIn this section we study the applicability of the feasible (non-monotone) interpolation technique\nto R0(lin) refutations. In particular, we show that R0(lin) admits a polynomial (in terms of the\nR0(lin)-proofs) upper bound on the (non-monotone) circuit-size of interpolants. In the next section\nwe shall give a polynomial upper bound on the monotone circuit-size of interpolants, but only in the\ncase that the interpolant corresponds to the clique-coloring formulas (whereas, in this section we are\ninterested in the general case; that is, upper bounding circuit-size of interpolants corresponding to\nany formula [of the prescribed type; see below]). First, we shortly describe the feasible interpolation\nmethod and explain how this method can be applied to obtain (sometime, conditional) lower bounds\non proof size. Explicit usage of the interpolation method in proof complexity goes back to [Kra94].\nLet Ai(⃗p, ⃗q), i ∈I, and Bj(⃗p,⃗r), j ∈J, (I and J are sets of indices) be a collection of formulas (for\ninstance, a collection of disjunctions of linear equations) in the displayed variables only. Denote by\nA(⃗p, ⃗q) the conjunction of all Ai(⃗p, ⃗q), i ∈I, and by B(⃗p,⃗r), the conjunction of all Bj(⃗p,⃗r), j ∈J.\nAssume that ⃗p, ⃗q,⃗r are pairwise disjoint sets of distinct variables, and that there is no assignment\nthat satisfies both A(⃗p, ⃗q) and B(⃗p,⃗r). Fix an assignment ⃗α to the variables in ⃗p. The ⃗p variables\nare the only common variables of the Ai’s and the Bj’s. Therefore, either A(⃗α, ⃗q) is unsatisfiable\nor B(⃗α,⃗r) is unsatisfiable.\nThe interpolation technique transforms a refutation of A(⃗p, ⃗q) ∧B(⃗p,⃗r), in some proof system,\ninto a circuit (usually a Boolean circuit) separating those assignments ⃗α (for ⃗p) for which A(⃗α, ⃗q)\nis unsatisfiable, from those assignments ⃗α for which B(⃗α,⃗r) is unsatisfiable (the two cases are not\nnecessarily exclusive, so if both cases hold for an assignment, the circuit can output either that the\nfirst case holds or that the second case holds). In other words, given a refutation of A(⃗p, ⃗q)∧B(⃗p,⃗r),\nwe construct a circuit C(⃗p), called the interpolant, such that\nC(⃗α) = 1\n=⇒\nA(⃗α, ⃗q) is unsatisfiable, and\nC(⃗α) = 0\n=⇒\nB(⃗α,⃗r) is unsatisfiable.\n(39)\n(Note that if U denotes the set of those assignments ⃗α for which A(⃗α, ⃗q) is satisfiable, and V\ndenotes the set of those assignments ⃗α for which B(⃗α,⃗r) is satisfiable, then U and V are disjoint\n[since A(⃗p, ⃗q) ∧B(⃗p,⃗r) is unsatisfiable], and C(⃗p) separates U from V ; see Definition 7.2 below.)\nAssume that for a proof system P the transformation from refutations of A(⃗p, ⃗q), B(⃗p,⃗r) into\nthe corresponding interpolant circuit C(⃗p) results in a circuit whose size is polynomial in the size\nof the refutation. Then, an exponential lower bound on circuits for which (39) holds, implies an\nexponential lower bound on P-refutations of A(⃗p, ⃗q), B(⃗p,⃗r).\n7.1. Interpolation for Semantic Refutations. We now lay out the basic concepts needed to\nformally describe the feasible interpolation technique.\nWe use the general notion of semantic\nrefutations (which generalizes any standard propositional refutation system). We shall use a close\nterminology to that in [Kra97].\nDefinition 7.1 (Semantic refutation). Let N be a fixed natural number and let E1, . . . , Ek ⊆\n{0, 1}N, where Tk\ni=1 Ei = ∅. A semantic refutation from E1, . . . , Ek is a sequence D1, . . . , Dm ⊆\n{0, 1}N with Dm = ∅and such that for every i ∈[m], Di is either one of the Ej’s or is deduced\nfrom two previous Dj, Dl, 1 ≤j, l< i, by the following semantic inference rule:\n• From A, B ⊆{0, 1}N deduce any C, such that C ⊇(A ∩B).\nObserve that any standard propositional refutation (with inference rules that derive from at\nmost two proof-lines, a third line) can be regarded as a semantic refutation: just substitute each\nrefutation-line by the set of its satisfying assignments; and by the soundness of the inference rules\napplied in the refutation, it is clear that each refutation-line (considered as the set of assignments\nthat satisfy it) is deduced by the semantic inference rule from previous refutation-lines.\n26"},{"paragraph_id":"p37","order":37,"text":"Definition 7.2 (Separating circuit). Let U, V ⊆{0, 1}n, where U ∩V = ∅, be two disjoint sets. A\nBoolean circuit C with n input variables is said to separate U from V if C(⃗x) = 1 for every ⃗x ∈U,\nand C(⃗x) = 0 for every ⃗x ∈V. In this case we also say that U and V are separated by C.\nConvention: In what follows we sometime identify a Boolean formula with the set of its satisfying\nassignments.\nNotation: For two (or more) binary strings u, v ∈{0, 1}∗, we write (u, v) to denote the concate-\nnation of the u with v (where v comes to the right of u, obviously).\nLet N = n+s+t be fixed from now on. Let A1, . . . , Ak ⊆{0, 1}n+s and let B1, . . . , Bl⊆{0, 1}n+t.\nDefine the following two sets of assignments of length n (formally, 0, 1 strings of length n) that\ncan be extended to satisfying assignments of A1, . . . , Ak and B1, . . . , Bl, respectively (formally,\nthose 0, 1 string of length n + s and n + t, that are contained in all A1, . . . , Ak and B1, . . . , Bl,\nrespectively):\nUA :=\n(\nu ∈{0, 1}n\n ∃q ∈{0, 1}s , (u, q) ∈\nk\\\ni=1\nAi\n)\n,\nVB :=\n(\nv ∈{0, 1}n\n ∃r ∈{0, 1}t , (v, r) ∈\nl\\\ni=1\nBi\n)\n.\nDefinition 7.3 (polynomial upper bounds on interpolants). Let P be a propositional\nrefutation system.\nAssume that ⃗p, ⃗q,⃗r are pairwise disjoint sets of distinct variables, where\n⃗p has n variables, ⃗q has s variables and ⃗r has t variables.\nLet A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and\nB1(⃗p,⃗r), . . . , Bl(⃗p,⃗r) be two collections of formulas with the displayed variables only.\nAssume\nthat for any such A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r), if there exists a P-refutation of\nsize S for A1(⃗p, ⃗q) ∧· · · ∧Ak(⃗p, ⃗q) ∧B1(⃗p,⃗r) ∧. . . ∧Bl(⃗p,⃗r) then there exists a Boolean circuit\nseparating UA from VB of size polynomial in S.10 In this case we say that P has a polynomial upper\nbound on interpolant circuits.\n7.1.1. The Communication Game Technique. The feasible interpolation via communication game\ntechnique is based on transforming proofs into Boolean circuits, where the size of the resulting\ncircuit depends on the communication complexity of each proof-line. This technique goes back to\n[IPU94] and [Razb95] and was subsequently applied and extended in [BPR97] and [Kra97] ([IPU94]\nand [BPR97] did not use explicitly the notion of interpolation of tautologies or contradictions). We\nshall employ the interpolation theorem of Kraj ́ıˇcek in [Kra97], that demonstrates how to transform\na small semantic refutation with each proof-line having low communication complexity into a small\nBoolean circuit separating the corresponding sets.\nThe underlying idea of the interpolation via communication game technique is that a (semantic)\nrefutation, where each proof-line is of small (that is, logarithmic) communication complexity, can be\ntransformed into an efficient communication protocol for the Karchmer-Wigderson game (following\n[KW88]) for two players. In the Karchmer-Wigderson game the first player knows some binary\nstring u ∈U and the second player knows some different binary string v ∈V , where U and V are\ndisjoint sets of strings. The two players communicate by sending information bits to one another\n(following a protocol previously agreed on). The goal of the game is for the two players to decide\non an index i such that the ith bit of u is different from the ith bit of v. An efficient Karchmer-\nWigderson protocol (by which we mean a protocol that requires the players to exchange at most a\nlogarithmic number of bits in the worst-case) can then be transformed into a small circuit separating\n10Here UA and VB are defined as above, by identifying the Ai(⃗p, ⃗q)’s and the Bi(⃗p,⃗r)’s with the sets of assignments\nthat satisfy them.\n27"},{"paragraph_id":"p38","order":38,"text":"U from V (see Definition 7.2). This efficient transformation from protocols for Karchmer-Wigderson\ngames (described in a certain way) into circuits, was demonstrated by Razborov in [Razb95]. So\noverall, given a semantic refutation with proof-lines of low communication complexity, one can\nobtain a small circuit for separating the corresponding sets.\nFirst, we need to define the concept of communication complexity in a suitable way for the\ninterpolation theorem.\nDefinition 7.4 (Communication complexity). Let N = n + s + t and A ⊆{0, 1}N. Let u, v ∈\n{0, 1}n, qu ∈{0, 1}s, rv ∈{0, 1}t. Denote by ui, vi the ith bit of u, v, respectively, and let (u, qu, rv)\nand (v, qu, rv) denote the concatenation of strings u, qu, rv and v, qu, rv, respectively. Consider the\nfollowing three tasks:\n(1) Decide whether (u, qu, rv) ∈A;\n(2) Decide whether (v, qu, rv) ∈A;\n(3) If one of the following holds:\n(i) (u, qu, rv) ∈A and (v, qu, rv) ̸∈A; or\n(ii) (u, qu, rv) ̸∈A and (v, qu, rv) ∈A,\nthen find an i ∈[n], such that ui ̸= vi;\nConsider a game between two players, Player I and Player II, where Player I knows u ∈{0, 1}n , qu ∈\n{0, 1}s and Player II knows v ∈{0, 1}n , rv ∈{0, 1}t. The two players communicate by exchanging\nbits of information between them (following a protocol previously agreed on). The communication\ncomplexity of A, denoted CC(A), is the minimal (over all protocols) number of bits that players I\nand II need to exchange in the worst-case in solving each of Tasks 1, 2 and 3 above.11\nFor A ⊆{0, 1}n+s define\n ̇A :="},{"paragraph_id":"p39","order":39,"text":"(a, b, c)\n (a, b) ∈A and c ∈{0, 1}t \n,\nwhere a and b range over {0, 1}n and {0, 1}s, respectively. Similarly, for B ⊆{0, 1}n+t define\n ̇B :="},{"paragraph_id":"p40","order":40,"text":"(a, b, c)\n (a, c) ∈B and b ∈{0, 1}t \n,\nwhere a and c range over {0, 1}n and {0, 1}t, respectively.\nTheorem 23 ([Kra97]). Let A1, . . . , Ak ⊆{0, 1}n+s and B1, . . . , Bl⊆{0, 1}n+t. Let D1, . . . , Dm\nbe a semantic refutation from ̇A1, . . . , ̇Ak and ̇B1, . . . , ̇Bl. Assume that CC(Di) ≤ζ, for all i ∈[m].\nThen, the sets UA and VB (as defined above) can be separated by a Boolean circuit of size (m +\nn)2O(ζ).\nIn light of Theorem 23, to demonstrate that a certain propositional refutation system P possesses\na polynomial upper bound on interpolant circuits (see Definition 7.3) it suffices to show that any\nproof-line of P induces a set of assignments with at most a logarithmic (in the number of variables)\ncommunication complexity (Definition 7.4).\n7.2. Polynomial Upper Bounds on Interpolants for R0(lin). Here we apply Theorem 23 to\nshow that R0(lin) has polynomial upper bounds on its interpolant circuits. Again, in what follows\nwe sometime identify a disjunction of linear equations with the set of its satisfying assignments.\nTheorem 24. R0(lin) has a polynomial upper bounds on interpolant circuits (Definition 7.3).\nAccording to the paragraph after Theorem 23, all we need in order to establish Theorem 24 is\nthe following lemma:\n11In other words, CC(A) is the minimal number ζ, for which there exists a protocol, such that for every input\n(u, qu to Player I and v, rv to Player II) and every task (from Tasks 1, 2 and 3), the players need to exchange at most\nζ bits in order to solve the task.\n28"},{"paragraph_id":"p41","order":41,"text":"Lemma 25. Let D be an R0(lin)-line with N variables and let eD be the set of assignments that\nsatisfy D.12Then, CC( eD) ≤O(log N).\nProof: Let N = n + s + t (and so eD ∈{0, 1}n+s+t). For the sake of convenience we shall assume\nthat the N variables in D are partitioned into (pairwise disjoint) three groups ⃗p := (p1 . . . , pn),\n⃗q := (q1, . . . , qs) and ⃗r := (r1, . . . , rt). Let u, v ∈{0, 1}n, qu ∈{0, 1}s, rv ∈{0, 1}t. Assume that\nPlayer I knows u, qu and Player II knows v, rv.\nBy the definition of an R0(lin)-line (see Definition 3.2) we can partition the disjunction D into\na constant number of disjuncts, where one disjunct is a (possibly empty, translation of a) clause in\nthe ⃗p, ⃗q,⃗r variables (see Section 3.1), and all other disjuncts have the following form:\n_\ni∈I"},{"paragraph_id":"p42","order":42,"text":"⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r = li"},{"paragraph_id":"p43","order":43,"text":",\n(40)\nwhere I is (an unbounded) set of indices, li are integer numbers, for all i ∈I, and ⃗a,⃗b,⃗c denote\nvectors of n, s and t constant coefficients, respectively.\nLet us denote the (translation of the) clause from D in the ⃗p, ⃗q,⃗r variables by\nP ∨Q ∨R ,\nwhere P, Q and R denote the (translated) sub-clauses consisting of the ⃗p, ⃗q and ⃗r variables,\nrespectively.\nWe need to show that by exchanging O(log N) bits, the players can solve each of Tasks 1, 2 and\n3 from Definition 7.4, correctly.\nTask 1: The players need to decide whether (u, qu, rv) ∈eD. Player II, who knows rv, computes\nthe numbers ⃗c · rv, for every ⃗c pertaining to every disjunct of the form shown in Equation (40)\nabove. Then, Player II sends the (binary representation of) these numbers to Player I. Since there\nare only a constantly many such numbers and the coefficients in every ⃗c are also constants, this\namounts to O(log t) ≤O(log N) bits that Player II sends to Player I. Player II also computes the\ntruth value of the sub-clause R, and sends this (single-bit) value to Player I.\nNow, it is easy to see that Player I has sufficient data to compute by herself/himself whether\n(u, qu, rv) ∈eD (Player I can then send a single bit informing Player II whether (u, qu, rv) ∈eD).\nTask 2: This is analogous to Task 1.\nTask 3: Assume that (u, qu, rv) ∈eD and (v, qu, rv) ̸∈eD (the case (u, qu, rv) ̸∈eD and (v, qu, rv) ∈eD\nis analogous).\nThe first rounds of the protocol are completely similar to that described in Task 1 above: Player\nII, who knows rv, computes the numbers ⃗c · rv, for every ⃗c pertaining to every disjunct of the form\nshown in Equation (40) above. Then, Player II sends the (binary representation of) these numbers\nto Player I. Player II also computes the truth value of the sub-clause R, and sends this (single-bit)\nvalue to Player I. Again, this amounts to O(log N) bits that Player II sends to Player I.\nBy assumption (that (u, qu, rv) ∈eD and (v, qu, rv) ̸∈eD) the players need to deal only with the\nfollowing two cases:\nCase 1: The assignment (u, qu, rv) satisfies the clause P ∨Q∨R while (v, qu, rv) falsifies P ∨Q∨R.\nThus, it must be that ⃗u satisfies the sub-clause P while ⃗v falsifies P. This means that for any i ∈[n]\nsuch that ui sets to 1 a literal in P (there ought to exist at least one such i), it must be that ui ̸= vi.\nTherefore, all that Player I needs to do is to send the (binary representation of) index i to Player\nII. (This amounts to O(log N) bits that Player I sends to Player II.)\n12The notation eD has nothing to do with the same notation used in Section 3.\n29"},{"paragraph_id":"p44","order":44,"text":"Case 2: There is some linear equation\n⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r = l\n(41)\nin D, such that ⃗a · u +⃗b · qu + ⃗c · rv = l. Note that (by assumption that (v, qu, rv) ̸∈eD) it must\nalso hold that: ⃗a · v +⃗b · qu + ⃗c · rv ̸= l(and so there is an i ∈[n], such that ui ̸= vi). Player I can\nfind linear equation (41), as he/she already received from Player II all the possible values of ⃗c · ⃗r\n(for all possible ⃗c ’s in D).\nRecall that the left hand side of a linear equation ⃗d · ⃗x = lis called the linear form of the\nequation. By the definition of an R0(lin)-line there are only constant many distinct linear forms\nin D. Since both players know these linear forms, we can assume that each linear form has some\nindex associated to it by both players. Player I sends to Player II the index of the linear form\n⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r from (41) in D. Since there are only constantly many such linear forms in D, it\ntakes only constant number of bits to send this index.\nNow both players need to apply a protocol for finding an i ∈[n] such that ui ̸= vi, where\n⃗a · ⃗u +⃗b · qu + ⃗c · rv = land ⃗a · ⃗v +⃗b · qu + ⃗c · rv ̸= l. Thus, it remains only to prove the following\nclaim:\nClaim 4. There is a communication protocol in which Player I and Player II need at most O(log N)\nbits of communication in order to find an i ∈[n] such that ui ̸= vi (under the above conditions).\nProof of claim: We invoke the well-known connection between Boolean circuit-depth and com-\nmunication complexity. Let f : {0, 1}N →{0, 1} be a Boolean function. Denote by dp(f) the\nminimal depth of a Boolean circuit computing f. Consider a game between two players: Player I\nknows some ⃗x ∈{0, 1}N and Player II knows some other ⃗y ∈{0, 1}N, such that f(⃗x) = 1 while\nf(⃗y) = 0. The goal of the game is to find an i ∈[N] such that xi ̸= yi. Denote by CC′(f) the\nminimal number of bits needed for the two players to communicate (in the worst case13) in order\nto solve this game.14 Then, for any function f it is known that dp(f) = CC′(f) (see [KW88]).\nTherefore, to conclude the proof of the claim it is enough to establish that the function f :\n{0, 1}N →{0, 1} that receives the input variables ⃗p, ⃗q,⃗r and computes the truth value of ⃗a · ⃗p +⃗b ·\n⃗q + ⃗c · ⃗r = lhas Boolean circuit of depth O(log N). In case all the coefficients in ⃗a,⃗b,⃗c are 1, it is\neasy to show15 that there is a Boolean circuit of depth O(log N) that computes the function f. In\nthe case that the coefficients in ⃗a,⃗b,⃗c are all constants, it is easy to show, by a reduction to the\ncase where all coefficients are 1,16 that there is a Boolean circuit of depth O(log N) that computes\nthe function f. We omit the details.\n8. Size Lower Bounds\nIn this section we establish an exponential-size lower bound on R0(lin) refutations of the clique-\ncoloring formulas. We shall employ the theorem of Bonet, Pitassi & Raz in [BPR97] that provides\nexponential-size lower bounds for any semantic refutation of the clique-coloring formulas, having\nlow communication complexity in each refutation-line.\n13Over all inputs ⃗x, ⃗y such that f(⃗x) = 1 and f(⃗y) = 0.\n14The measure CC′ is basically the same as CC defined earlier.\n15Using the known O(log N)-depth Boolean circuits for the threshold functions.\n16For instance, consider the simple case where we have only a single variable. That is, let c be a constant and\nassume that we wish to construct a circuit that computes c · x = l, for some integer l. Then, we take a circuit that\ncomputes the function f : {0, 1}c →{0, 1} that outputs the truth value of y1 + . . . + yc = l(thus, in f all coefficients\nare 1’s); and to compute c · x = lwe only have to substitute each yi in the circuit with the variable x.\n30"},{"paragraph_id":"p45","order":45,"text":"First we recall the strong lower bound obtained by Alon & Boppana [AB87] (improving over\n[Razb85]; see also [And85]) for the (monotone) clique separator functions, defined as follows (a\nfunction f : {0, 1}n →{0, 1} is called monotone if for all α ∈{0, 1}n, α′ ≥α implies f(α′) ≥f(α)):\nDefinition 8.1 (Clique separator). A monotone boolean function Qn\nk,k′ is called a clique separator\nif it interprets its inputs as the edges of a graph on n vertices, and outputs 1 on every input\nrepresenting a k-clique, and 0 on every input representing a complete k′-partite graph (see Section\n6.3).\nRecall that a monotone Boolean circuit is a circuit that uses only monotone Boolean gates (for\ninstance, only the fan-in two gates ∧, ∨).\nTheorem 26 ([AB87]). Let k, k′ be integers such that 3 ≤k′ < k and k\n√\nk′ ≤n/(8 log n), then\nevery monotone Boolean circuit that computes a clique separator function Qn\nk,k′ requires size at least\n1\n8"},{"paragraph_id":"p46","order":46,"text":"n\n4k\n√\nk′ log n\n (\n√\nk′+1)/2\n.\nFor the next theorem, we need a slightly different (and weaker) version of communication com-\nplexity, than that in Definition 7.4.\nDefinition 8.2 (Communication complexity (second definition)). Let X denote n Boolean variables\nx1, . . . , xn, and let S1, S2 be a partition of X into two disjoint sets of variables. The communication\ncomplexity of a Boolean function f : {0, 1}n →{0, 1} is the number of bits needed to be exchanged\nby two players, one knowing the values given to the S1 variables and the other knowing the values\ngiven to S2 variables, in the worst-case, over all possible partitions S1 and S2.\nTheorem 27 ([BPR97]). Every semantic refutation of ¬cliquen\nk,k′ (for k′ < k) with m refutation-\nlines and where each refutation-line (considered as a the characteristic function of the line) has\ncommunication complexity (as in Definition 8.2) ζ, can be transformed into a monotone circuit of\nsize m · 23ζ+1 that computes a separating function Qn\nk,k′.\nIn light of Theorem 26, in order to be able to apply Theorem 27 to R0(lin), and arrive at an\nexponential-size lower bound for R0(lin) refutations of the clique-coloring formulas, it suffices to\nshow that R0(lin) proof-lines have logarithmic communication complexity:\nLemma 28. Let D be an R0(lin)-line with N variables. Then, the communication complexity (as\nin Definition 8.2) of D is at most O(log N) (where D is identified here with the characteristic\nfunction of D).\nProof: The proof is similar to the proof of Lemma 25 for solving Task 1 (and the analogous Task\n2) in Definition 7.4.\nBy direct calculations we obtain the following lower bound from Theorems 26, 27 and Lemma\n28:\nCorollary 29. Let k be an integer such that 3 ≤k′ = k −1 and assume that 1\n2 · n/(8 log n) ≤\nk\n√\nk ≤n/(8 log n). Then, for all ε < 1/3, every R0(lin) refutation of ¬cliquen\nk,k′ is of size at least\n2Ω(nε).\nWhen considering the parameters of Theorem 20, we obtain a super-polynomial separation be-\ntween R0(lin) refutations and R(lin) refutations, as described below.\nFrom Theorems 26,27 and Lemma 28 we have (by direct calculations):\n31"},{"paragraph_id":"p47","order":47,"text":"Corollary 30. Let k = √n and k′ = (log n)2/8 log log n.\nThen, every R0(lin) refutation of\n¬cliquen\nk,k′ has size at least nΩ\n“\nlog n\n√log log n\n”\n.\nBy Corollary 21, R(lin) admits polynomial-size in n refutations of ¬cliquen\nk,k′ under the param-\neters in Corollary 30. Thus we obtain the following separation result:\nCorollary 31. R(lin) is super-polynomially stronger than R0(lin).\nComment 1. Note that we do not need to assume that the coefficients in R0(lin)-lines are constants\nfor the lower bound argument. If the coefficients in R0(lin)-lines are only polynomially bounded\n(in the number of variables) then the same lower bound as in Corollary 30 also applies. This is\nbecause R0(lin)-lines in which coefficients are polynomially bounded integers, still have low (that\nis, logarithmic) communication complexity (as in Definition 8.2).\n9. Applications to Multilinear Proofs\nIn this section we arrive at one of the main benefits of the work we have done so far; Namely,\napplying results on resolution over linear equations in order to obtain new results for multilinear\nproof systems. Subsection 9.1 that follows, contains definitions, sufficient for the current paper,\nconcerning the notion of multilinear proofs introduced in [RT06].\n9.1. Background on Algebraic and Multilinear Proofs.\n9.1.1. Arithmetic and Multilinear Formulas.\nDefinition 9.1 (Arithmetic formula). Fix a field F. An arithmetic formula is a tree, with edges\ndirected from the leaves to the root, and with unbounded (finite) fan-in. Every leaf of the tree\n(namely, a node of fan-in 0) is labeled with either an input variable or a field element. A field\nelement can also label an edge of the tree. Every other node of the tree is labeled with either + or\n× (in the first case the node is a plus gate and in the second case a product gate). We assume that\nthere is only one node of out-degree zero, called the root. The size of an arithmetic formula F is\nthe total number of nodes in its graph and is denoted by |F|. An arithmetic formula computes a\npolynomial in the ring of polynomials F[x1, . . . , xn] in the following way. A leaf just computes the\ninput variable or field element that labels it. A field element that labels an edge means that the\npolynomial computed at its tail (namely, the node where the edge is directed from) is multiplied\nby this field element. A plus gate computes the sum of polynomials computed by the tails of all\nincoming edges. A product gate computes the product of the polynomials computed by the tails of\nall incoming edges. (Subtraction is obtained using the constant −1.) The output of the formula is\nthe polynomial computed by the root. The depth of a formula F is the maximal number of edges\nin a path from a leaf to the root of F.\nWe say that an arithmetic formula has a plus (resp., product) gate at the root if the root of the\nformula is labeled with a plus (resp., product) gate.\nA polynomial is multilinear if in each of its monomials the power of every input variable is at\nmost one.\nDefinition 9.2 (Multilinear formula). An arithmetic formula is a multilinear formula (or equiva-\nlently, multilinear arithmetic formula) if the polynomial computed by each gate of the formula is\nmultilinear (as a formal polynomial, that is, as an element of F[x1, . . . , xn]).\nAn additional definition we shall need is the following linear operator, called the multilinearization\noperator:\n32"},{"paragraph_id":"p48","order":48,"text":"Definition 9.3 (Multilinearization operator). Given a field F and a polynomial q ∈F[x1, . . . , xn],\nwe denote by M[q] the unique multilinear polynomial equal to q modulo the ideal generated by all\nthe polynomials x2\ni −xi, for all variables xi.\nFor example, if q = x2\n1x2 + ax3\n4 (for some a ∈F) then M[q] = x1x2 + ax4 .\nThe simulation of R0(lin) by multilinear proofs will rely heavily on the fact that multilinear\nsymmetric polynomials have small depth-3 multilinear formulas over fields of characteristic 0 (see\n[SW01] for a proof of this fact). To this end we define precisely the concept of symmetric polyno-\nmials.\nA renaming of the variables x1, . . . , xn is a permutation σ ∈Sn (the symmetric group on [n])\nsuch that xi is mapped to xσ(i) for every 1 ≤i ≤n.\nDefinition 9.4 (Symmetric polynomial). Given a set of variables X = {x1, . . . , xn}, a symmetric\npolynomial f over X is a polynomial in (all the variables of) X such that renaming of variables\ndoes not change the polynomial (as a formal polynomial).\n9.1.2. Polynomial Calculus with Resolution. Here we define the PCR proof system, introduced by\nAlekhnovich et al. in [ABSRW02].\nDefinition 9.5 (Polynomial Calculus with Resolution (PCR)). Let F be some fixed field and\nlet Q := {Q1, . . . , Qm} be a collection of multivariate polynomials from the ring of polynomials\nF[x1, . . . , xn, ̄x1, . . . , ̄xn]. The variables ̄x1, . . . , ̄xn are treated as new formal variables. Call the\nset of polynomials x2 −x, for x ∈{x1, . . . , xn, ̄x1, . . . , ̄xn}, plus the polynomials xi + ̄xi −1, for\nall 1 ≤i ≤n, the set of Boolean axioms of PCR. A PCR proof from Q of a polynomial g is\na finite sequence π = (p1, ..., pl) of multivariate polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] (each\npolynomial pi is interpreted as the polynomial equation pi = 0), where pl= g and for each i ∈[l],\neither pi = Qj for some j ∈[m], or pi is a Boolean axiom, or pi was deduced from pj, pk , where\nj, k < i, by one of the following inference rules:\nProduct: From p deduce xi · p , for some variable xi ;\nFrom p deduce ̄xi · p , for some variable ̄xi ;\nAddition: From p and q deduce α · p + β · q, for some α, β ∈F.\nA PCR refutation of Q is a proof of 1 (which is interpreted as 1 = 0) from Q. The number of\nsteps in a PCR proof is the number of proof-lines in it (that is, lin the case of π above).\nNote that the Boolean axioms of PCR have only 0, 1 solutions, where ̄xi = 0 if xi = 1 and ̄xi = 1\nif xi = 0.\n9.1.3. Multilinear Proof Systems. In [RT06] the authors introduced a natural (semantic) algebraic\nproof system that operates with multilinear arithmetic formulas denoted fMC (which stands for\nformula multilinear calculus), defined as follows:\nDefinition 9.6 (Formula Multilinear Calculus (fMC)). Fix a field F and let Q := {Q1, . . . , Qm} be\na collection of multilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] (the variables ̄x1, . . . , ̄xn are\ntreated as formal variables). Call the set of polynomials consisting of xi + ̄xi −1 and xi · ̄xi for\n1 ≤i ≤n , the Boolean axioms of fMC. An fMC proof from Q of a polynomial g is a finite sequence\nπ = (p1, ..., pl) of multilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] , such that pl= g and for\neach i ∈[l], either pi = Qj for some j ∈[m], or pi is a Boolean axiom of fMC, or pi was deduced\nby one of the following inference rules using pj, pk for j, k < i:\nProduct: from p deduce q · p , for some polynomial q ∈F[x1, . . . , xn, ̄x1, . . . , ̄xn] such that\np · q is multilinear;\nAddition: from p, q deduce α · p + β · q, for some α, β ∈F.\n33"},{"paragraph_id":"p49","order":49,"text":"All the polynomials in an fMC proof are represented as multilinear formulas. (A polynomial pi in\nan fMC proof is interpreted as the polynomial equation pi = 0.) An fMC refutation of Q is a proof\nof 1 (which is interpreted as 1 = 0) from Q. The size of an fMC proof π is defined as the total\nsum of all the formula sizes in π and is denoted by |π|.\nNote that the Boolean axioms have only 0, 1 solutions, where ̄xi = 0 if xi = 1 and ̄xi = 1 if\nxi = 0, for each 1 ≤i ≤n .\nDefinition 9.7 (Depth-k Formula Multilinear Calculus (depth-k fMC)). For a natural number k,\ndepth-k fMC denotes a restriction of the fMC proof system, in which proofs consist of multilinear\npolynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] represented as multilinear formulas of depth at most k.\n9.2. From R(lin) Proofs to PCR Proofs. We now demonstrate a general and straightforward\ntranslation from R(lin) proofs into PCR proofs over fields of characteristic 0. We use the term\n“translation” in order to distinguish it from a simulation; since here we are not interested in the\nsize of PCR proofs. In fact we have not defined the size of PCR proofs at all. We shall be interested\nonly in the number of steps in PCR proofs.\nFrom now on, all polynomials and arithmetic formulas are considered over some fix field F of\ncharacteristic 0.\nRecall that any field of characteristic 0 contains (an isomorphic copy of) the\ninteger numbers, and so we can use integer coefficients in the field.\nDefinition 9.8 (Polynomial translation of R(lin) proof-lines). Let D be a disjunction of linear\nequations:"},{"paragraph_id":"p50","order":50,"text":"a(1)\n1 x1 + . . . + a(1)\nn xn = a(1)\n0"},{"paragraph_id":"p51","order":51,"text":"∨· · · ∨"},{"paragraph_id":"p52","order":52,"text":"a(t)\n1 x1 + . . . + a(t)\nn xn = a(t)\n0"},{"paragraph_id":"p53","order":53,"text":".\n(42)\nWe denote by bD its translation into the following polynomial:17"},{"paragraph_id":"p54","order":54,"text":"a(1)\n1 x1 + . . . + a(1)\nn xn −a(1)\n0"},{"paragraph_id":"p55","order":55,"text":"· · ·"},{"paragraph_id":"p56","order":56,"text":"a(t)\n1 x1 + . . . + a(t)\nn xn −a(t)\n0"},{"paragraph_id":"p57","order":57,"text":".\n(43)\nIf D is the empty disjunction, we define bD to be the polynomial 1.\nIt is clear that every 0, 1 assignment to the variables in D, satisfies D, if and only if bD evaluates\nto 0 under the assignment.\nProposition 3. Let π = (D1, . . . , Dl) be an R(lin) proof sequence of Dl, from some collection\nof initial disjunctions of linear equations Q1, . . . , Qm. Then, there exists a PCR proof of bDlfrom\nbQ1, . . . , bQm with at most a polynomial in |π| number of steps.\nProof: We proceed by induction on the number of lines in π.\nThe base case is the translation of the axioms of R(lin) via the translation scheme in Definition\n9.8. An R(lin) Boolean axiom (xi = 0) ∨(xi = 1) is translated into xi · (xi −1) which is already a\nBoolean axiom of PCR.\nFor the induction step, we translate every R(lin) inference rule application into a polynomial-size\nPCR proof sequence as follows. We use the following simple claim:\nClaim 5. Let p and q be two polynomials and let s be the minimal size of an arithmetic formula\ncomputing q. Then one can derive in PCR, with only a polynomial in s number of steps, from p\nthe product q · p.18\nProof of claim: By induction on s.\n17This notation should not be confused with the same notation in Section 6.3.\n18Again, note that we only require that the number of steps in the proof is polynomial. We do not consider here\nthe size of the PCR proof.\n34"},{"paragraph_id":"p58","order":58,"text":"Assume that Di = Dj ∨L was derived from Dj using the Weakening inference rule of R(lin),\nwhere j < i ≤land L is some linear equation. Then, by Claim 5, bDi = bDj · bL can be derived from\nbDj with a derivation of at most polynomial in |Dj ∨L| many steps.\nAssume that Di was derived from Dj where Dj is Di ∨(0 = k), using the Simplification inference\nrule of R(lin), where j < i ≤land k is a non-zero integer.\nThen, bDi can be derived from\nbDj = bDi · −k by multiplying with −k−1 (via the Addition rule of PCR).\nThus, it remains to simulate the resolution rule application of R(lin). Let A, B be two disjunc-\ntions of linear equations and assume that A ∨B ∨((⃗a + ⃗b) · ⃗x = a0 + b0) was derived in π from\nA ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0) (the case where A ∨B ∨((⃗a −⃗b) · ⃗x = a0 −b0) was derived from\nA ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0), is similar).\nWe need to derive bA · bB · ((⃗a +⃗b) · ⃗x −a0 −b0) from bA · (⃗a · ⃗x −a0) and bB · (⃗b · ⃗x −b0). This is\ndone by multiplying bA · (⃗a · ⃗x −a0) with bB and multiplying bB · (⃗b · ⃗x −b0) with bA (using Claim 5),\nand then adding the resulted polynomials together.\nRemark 4. When translating R(lin) proofs into PCR proofs we actually do not make any use of\nthe “negative” variables ̄x1, . . . , ̄xn. Nevertheless, the multilinear proof systems make use of these\nvariables in order to polynomially simulate PCR proofs (see Theorem 33 and its proof in [RT06]).\nWe shall need the following corollary in the sequel:\nCorollary 32. Let π = D1, . . . , Dlbe an R0(lin) proof of Dl, and let s be the maximal size of an\nR0(lin)-line in π. Then there is a PCR proof π′ of bDlwith polynomial-size in |π| number of steps\nand such that every line of π′ is a translation (via Definition 9.8) of an R0(lin)-line (Definition\n3.2), where the size of the R0(lin)-line is polynomial in s.\nProof: The simulation of R(lin) by PCR shown above, can be thought of as, first, considering\nbD1, . . . , bDlas the “skeleton” of a PCR proof of bDl. And second, for each Di that was deduced by\none of R(lin)’s inference rules from previous lines, one inserts the corresponding PCR proof sequence\nthat simulates the appropriate inference rule application (as described in the proof of Proposition\n3). By definition, those PCR proof-lines that correspond to lines in the skeleton bD1, . . . , bDlare\ntranslations of R0(lin)-lines (with size at most polynomial in s). Thus, to conclude the proof of\nthe corollary, one needs only to check that for any R0(lin)-line Di that was deduced by one of\nR(lin)’s inference rules from previous R0(lin)-lines (as demonstrated in the proof of Proposition 3),\nthe inserted corresponding PCR proof sequence uses only translations of R0(lin)-lines (with size\npolynomial in s). This can be verified by a straightforward inspection.\n9.3. From PCR Proofs to Multilinear Proofs. We now recall the general simulation result\nproved in [RT06] stating the following: Let π be a PCR refutation of some initial collection of\nmultilinear polynomials Q over some fixed field. Assume that π has polynomially many steps (that\nis, the number of proof lines in the PCR proof sequence is polynomial). If the ‘multilinearization’\n(namely, the result of applying the M[·] operator – see Definition 9.3) of each of the polynomials\nin π has a polynomial-size depth d multilinear formula (with a plus gate at the root), then there is\na polynomial-size depth-d fMC refutation of Q. More formally, we have:\nTheorem 33 ([RT06]). Fix a field F (not necessarily of characteristic 0) and let Q be a set of\nmultilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn]. Let π = (p1, . . . , pm) be a PCR refutation\nof Q. For each pi ∈π, let Φi be a multilinear formula for the polynomial M[pi]. Let s be the total\nsize of all formulas Φi, that is, s = Σm\ni=1|Φi|, and let d ≥2 be the maximal depth of all formulas\nΦi. Assume that the depth of all the formulas Φi that have a product gate at the root is at most\nd −1. Then there is a depth-d fMC refutation of Q of size polynomial in s.\n35"},{"paragraph_id":"p59","order":59,"text":"9.3.1. Depth-3 Multilinear Proofs. Here we show that multilinear proofs operating with depth-3\nmultilinear formulas (that is, depth-3 fMC) over fields of characteristic 0 polynomially simulate\nR0(lin) proofs. In light of Proposition 32 and Theorem 33, to this end it suffices to show that any\nR0(lin)-line D translates into a corresponding polynomial p (via the translation in Definition 9.8)\nsuch that M[p] has a multilinear formula of size polynomial (in the number of variables) and depth\nat most 3 (with a plus gate at the root) over fields of characteristic 0.\nWe need the following proposition from [RT06]:\nProposition 4 ([RT06]). Let F be a field of characteristic 0. For a constant c, let X1, . . . , Xc\nbe c finite sets of variables (not necessarily disjoint), where Σc\ni=1|Xi| = n . Let f1, . . . , fc be c\nsymmetric polynomials over X1, . . . , Xc (over the field F), respectively. Then, there is a depth-3\nmultilinear formula for M[f1 · · · fc] of size polynomial (in n), with a plus gate at the root.\nThe following is the key lemma of the simulation:\nLemma 34. Let D be an R0(lin)-line with n variables and let p = bD (see Definition 9.8). Then,\nM[p] has a depth-3 multilinear formula over fields of characteristic 0, with a plus gate at the root\nand size at most polynomial in the size of D.\nProof: Assume that the underlying variables of D are ⃗x = x1 . . . , xn. By the definition of an\nR0(lin)-line (see Definition 3.2) we can partition the disjunction D into a constant number of\ndisjuncts, where one disjunct is a (possibly empty, translation of a) clause C,19 and all other\ndisjuncts have the following form:\nm\n_\ni=1\n(⃗a · ⃗x = li) ,\n(44)\nwhere the li’s are integers, m is not necessarily bounded and ⃗a denotes a vector of n constant\ninteger coefficients.\nLet us denote by q the polynomial representing the clause C.20\nConsider a disjunct as shown in (44). Since the coefficients ⃗a are constants, ⃗a · ⃗x can be written\nas a sum of constant number of linear forms, each with the same constant coefficient. In other\nwords, ⃗a · ⃗x can be written as z1 + . . . + zd, for some constant d, where for all i ∈[d]:\nzi := b ·\nX\nj∈J\nxj ,\n(45)\nfor some J ⊆[n] and some constant integer b. We shall assume without loss of generality that d is\nthe same constant for every disjunct of the form (44) inside D (otherwise, take d to be the maximal\nsuch d).\nThus, (44) is translated (via the translation scheme in Definition 9.8) into:\nm\nY\ni=1\n(z1 + ... + zd −li) .\n(46)\nBy fully expanding the product in (46), we arrive at:\nX\nr1+...+rd+1=m"},{"paragraph_id":"p60","order":60,"text":"αrd+1 ·\nd\nY\nk=1\nzrk\nk\n!\n,\n(47)\n19If there is more than one clause in D, we simply combine all the clauses into a single clause.\n20C is a translation of a clause (that is, disjunction of literals) into a disjunction of linear equations, as defined\nin Section 3.1. The polynomial q is then the polynomial translation of this disjunction of linear equations, as in\nDefinition 9.8.\n36"},{"paragraph_id":"p61","order":61,"text":"where the ri’s are non-negative integers, and where the αr’s, for every 0 ≤r ≤m are just integer\ncoefficients, formally defined as follows (this definition is not essential; we present it only for the\nsake of concreteness):\nαr :=\nX\nU⊆[m]\n| U|=r\nY\nj∈U\n(−lj) .\n(48)\nClaim 6. The polynomial bD (the polynomial translation of D) is a linear combination (over F) of\npolynomially (in |D|) many terms, such that each term can be written as\nq ·\nY\nk∈K\nzrk\nk ,\nwhere K is a collection of a constant number of indices, rk’s are non-negative integers, and the zk’s\nand q are as above (that is, the zk’s are linear forms, where each zk has a single coefficient for all\nvariables in it, as in (45), and q is a polynomial translation of a clause).\nProof of claim: Denote the total number of disjuncts of the form (44) in D by h. By definition\n(of R0(lin)-line), h is a constant. Consider the polynomial (47) above. In bD, we actually need to\nmultiply h many polynomials of the form shown in (47) and the polynomial q.\nFor every j ∈[h] we write the (single) linear form in the jth disjunct as a sum of constantly\nmany linear forms zj,1 + . . . + zj,d, where each linear form zj,k has the same coefficient for every\nvariable in it. Thus, bD can be written as:\nq ·\nh\nY\nj=1"},{"paragraph_id":"p62","order":62,"text":"X\nr1+...+rd+1=mj"},{"paragraph_id":"p63","order":63,"text":"α(j)\nrd+1 ·\nd\nY\nk=1\nzrk\nj,k\n!\n|\n{z\n}\n(⋆)"},{"paragraph_id":"p64","order":64,"text":",\n(49)\n(where the mj’s are not bounded, and the coefficients α(j)\nrd+1 are as defined in (48) except that here\nwe add the index (j) to denote that they depend on the jth disjunct in D). Denote the maximal\nmj, for all j ∈[h], by m0. The size of D, denoted |D|, is at least m0. Note that since d is a\nconstant, the number of summands in each (middle) sum in (49) is polynomial in m0, which is at\nmost polynomial in |D|. Thus, by expanding the outermost product in (49), we arrive at a sum of\npolynomially in |D| many summands. Each summand in this sum is a product of h terms of the\nform (⋆) multiplied by q.\nIt remains to apply the multilinearization operator (Definition 9.3) on bD, and verify that the re-\nsulting polynomial has a depth-3 multilinear formula with a plus gate at the root and of polynomial-\nsize (in |D|). Since M[·] is a linear operator, it suffices to show that when applying M[·] on each\nsummand in bD, as described in Claim 6, one obtains a (multilinear) polynomial that has a depth-3\nmultilinear formula with a plus gate at the root, and of polynomial-size in the number of variables\nn (note that clearly n ≤|D|). This is established in the following claim:\nClaim 7. The polynomial M"},{"paragraph_id":"p65","order":65,"text":"q · Q\nk∈K zrk\nk"},{"paragraph_id":"p66","order":66,"text":"has a depth-3 multilinear formula of polynomial-size\nin n (the overall number of variables) and with a plus gate at the root (over fields of characteristic\n0), under the same notation as in Claim 6.\nProof of claim: Recall that a power of a symmetric polynomial is a symmetric polynomial in itself.\nSince each zk (for all k ∈K) is a symmetric polynomial, then its power zrk\nk is also symmetric. The\npolynomial q is a translation of a clause, hence it is a product of two symmetric polynomials: the\n37"},{"paragraph_id":"p67","order":67,"text":"symmetric polynomial that is the translation of the disjunction of literals with positive signs, and\nthe symmetric polynomial that is the translation of the disjunction of literals with negative signs.\nTherefore, q ·Q\nk∈K zrk\nk is a product of constant number of symmetric polynomials. By Proposition\n4, M"},{"paragraph_id":"p68","order":68,"text":"q · Q\nk∈K zrk\nk"},{"paragraph_id":"p69","order":69,"text":"(where here the M[·] operator operates on the ⃗x variables in the zk’s and q) is\na polynomial for which there is a polynomial-size (in n) depth-3 multilinear formula with a plus\ngate at the root (over fields of characteristic 0).\nWe now come to the main corollary of this section.\nCorollary 35. Multilinear proofs operating with depth-3 multilinear formulas (that is, depth-3 fMC\nproofs) polynomially-simulate R0(lin) proofs.\nProof: Immediate from Corollary 32, Theorem 33 and Proposition 34.\nFor the sake of clarity we repeat the chain of transformations needed to prove the simulation.\nGiven an R0(lin) proof π, we first use Corollary 32 to transform π into a PCR proof π′, with number\nof steps that is at most polynomial in |π|, and where each line in π′ is a polynomial translation of\nsome R0(lin)-line with size at most polynomial in the maximal line in π (which is clearly at most\npolynomial in |π|). Thus, by Proposition 34 each polynomial in π′ has a corresponding multilinear\npolynomial with a polynomial-size in |π| depth-3 multilinear formula (and a plus gate at the root).\nTherefore, by Theorem 33, we can transform π′ into a depth-3 fMC proof with only a polynomial\n(in |π|) increase in size.\n9.4. Small Depth-3 Multilinear Proofs. Since R0(lin) admits polynomial-size (in n) refutations\nof the m to n pigeonhole principle (for any m > n) (as defined in 6.1), Corollary 35 and Theorem\n15 yield:\nTheorem 36. For any m > n there are polynomial-size (in n) depth-3 fMC refutations of the m\nto n pigeonhole principle PHPm\nn (over fields of characteristic 0).\nThis improves over the result in [RT06] that demonstrated a polynomial-size (in n) depth-3 fMC\nrefutations of a weaker principle, namely the m to n functional pigeonhole principle.\nFurthermore, corollary 35 and Theorem 19 yield:\nTheorem 37. Let G be an r-regular graph with n vertices, where r is a constant, and fix some\nmodulus p. Then there are polynomial-size (in n) depth-3 fMC refutations of Tseitin mod p formulas\n¬TseitinG,p (over fields of characteristic 0).\nThe polynomial-size refutations of Tseitin graph tautologies here are different than those demon-\nstrated in [RT06]. Theorem 37 establishes polynomial-size refutations over any field of characteristic\n0 of Tseitin mod p formulas, whereas [RT06] required the field to contain a primitive pth root of\nunity. On the other hand, the refutations in [RT06] of Tseitin mod p formulas do not make any use\nof the semantic nature of the fMC proof system, in the sense that they do not utilize the fact that\nthe base field is of characteristic 0 (which in turn enables one to efficiently represent any symmetric\n[multilinear] polynomial by a depth-3 multilinear formula).\n10. Relations with Extensions of Cutting Planes\nIn this section we tie some loose ends by showing that, in full generality, R(lin) polynomially\nsimulates R(CP) with polynomially bounded coefficients, denoted R(CP*). First we define the\nR(CP*) proof system – introduced in [Kra98] – which is a common extension of resolution and\nCP* (the latter is cutting planes with polynomially bounded coefficients). The system R(CP*),\nthus, is essentially resolution operating with disjunctions of linear inequalities (with polynomially\nbounded integral coefficients) augmented with the cutting planes inference rules.\n38"},{"paragraph_id":"p70","order":70,"text":"A linear inequality is written as\n⃗a · ⃗x ≥a0 ,\n(50)\nwhere ⃗a is a vector of integral coefficients a1, . . . , an, ⃗x is a vector of variables x1, . . . , xn, and a0\nis an integer. The size of the linear inequality (50) is the sum of all a0, . . . , an written in unary\nnotation (this is similar to the size of linear equations in R(lin)). A disjunction of linear inequalities\nis just a disjunction of inequalities of the form in (50). The semantics of a disjunction of inequalities\nis the natural one, that is, a disjunction is true under an assignment of integral values to ⃗x if and\nonly if at least one of the inequalities is true under the assignment. The size of a disjunction of\nlinear inequalities is the total size of all linear inequalities in it. We can also add in the obvious\nway linear inequalities, that is, if L1 is the linear inequality ⃗a·⃗x ≥a0 and L2 is the linear inequality\n⃗b · ⃗x ≥b0, then L1 + L2 is the linear inequality (⃗a +⃗b) · ⃗x ≥a0 + b0.\nThe proof system R(CP*) operates with disjunctions of linear inequalities with integral coeffi-\ncients (written in unary representation), and is defined as follows (our formulation is similar to\nthat in [Koj07]):21\nDefinition 10.1 (R(CP*)). Let K := {K1, . . . , Km} be a collection of disjunctions of linear in-\nequalities (whose coefficients are written in unary representation). An R(CP*)-proof from K of a\ndisjunction of linear inequalities D is a finite sequence π = (D1, ..., Dl) of disjunctions of linear\ninequalities, such that Dl= D and for each i ∈[l]: either Di = Kj for some j ∈[m]; or Di is one\nof the following R(CP*)-axioms:\n(1) xi ≥0, for any variable xi;\n(2) −xi ≥−1, for any variable xi;\n(3) (⃗a · ⃗x ≥a0) ∨(−⃗a · ⃗x ≥1 −a0), where all coefficients (including a0) are integers;\nor Di was deduced from previous lines by one of the following R(CP*)-inference rules:\n(1) Let A, B be two disjunctions of linear inequalities and let L1, L2 be two linear inequalities.22\nFrom A ∨L1 and B ∨L2 derive A ∨B ∨(L1 + L2).\n(2) Let L be some linear equation.\nFrom a disjunction of linear equations A derive A ∨L.\n(3) Let A be a disjunction of linear equations\nFrom A ∨(0 ≥1) derive A.\n(4) Let c be a non-negative integer.\nFrom (⃗a · ⃗x ≥a0) ∨A derive (c⃗a · ⃗x ≥ca0) ∨A.\n(5) Let A be a disjunction of linear inequalities, and let c ≥1 be an integer.\nFrom (c⃗a · ⃗x ≥a0) ∨A derive (a · ⃗x ≥⌈a0/c⌉) ∨A.\nAn R(CP*) refutation of a collection of disjunctions of linear inequalities K is a proof of the empty\ndisjunction from K. The size of a proof π in R(CP*) is the total size of all the disjunctions of\nlinear inequalities in π, denoted |π|.\nIn order for R(lin) to simulate R(CP*) proofs, we need to fix the following translation scheme.\nEvery linear inequality L of the form ⃗a·⃗x ≥a0 is translated into the following disjunction, denoted\nbL:\n(⃗a · ⃗x = a0) ∨(⃗a · ⃗x = a0 + 1) ∨· · · ∨(⃗a · ⃗x = a0 + k) ,\n(51)\nwhere k is such that a0 + k equals the sum of all positive coefficients in ⃗a, that is, a0 + k =\nmax\n⃗x∈{0,1}n (⃗a · ⃗x) (in case the sum of all positive coefficients in ⃗a is less than a0, then we put k = 0).\nAn inequality with no variables of the form 0 ≥a0 is translated into 0 = a0 in case it is false (that\n21When we allow coefficients to be written in binary representation, instead of unary representation, the resulting\nproof system is denoted R(CP).\n22In all R(CP*)-inference rules, A, B are possibly the empty disjunctions.\n39"},{"paragraph_id":"p71","order":71,"text":"is, in case 0 < a0), and into 0 = 0 in case it is true (that is, in case 0 ≥a0). Note that since the\ncoefficients of linear inequalities (and linear equations) are written in unary representation, any\nlinear inequality of size s translates into a disjunction of linear equations of size O(s2). Clearly,\nevery 0, 1 assignment to the variables ⃗x satisfies L if and only if it satisfies its translation bL. A\ndisjunction of linear inequalities D is translated into the disjunction of the translations of all the\nlinear inequalities in it, denoted bD.\nA collection K := {K1, . . . , Km} of disjunctions of linear\ninequalities, is translated into the collection\nn\nbK1, . . . , bKm\no\n.\nTheorem 38. R(lin) polynomially-simulates R(CP*). In other words, if π is an R(CP*) proof of\na linear inequality D from a collection of disjunctions of linear inequalities K1, . . . , Kt, then there\nis an R(lin) proof of bD from bK1, . . . , bKt whose size is polynomial in |π|.\nProof: By induction on the number of proof-lines in π.\nBase case: Here we only need to show that the axioms of R(CP*) translates into axioms of\nR(lin), or can be derived with polynomial-size (in the size of the original R(CP*) axiom) R(lin)\nderivations (from R(lin)’s axioms).\nR(CP*) axiom number (1): xi ≥0 translates into the R(lin) axiom (xi = 0) ∨(xi = 1).\nR(CP*) axiom number (2): −xi ≥−1, translates into (−xi = −1) ∨(−xi = 0).\nFrom the\nBoolean axiom (xi = 1) ∨(xi = 0) of R(lin), one can derive with a constant-size R(lin) proof the\nline (−xi = −1)∨(−xi = 0) (for instance, by subtracting twice each equation in (xi = 1)∨(xi = 0)\nfrom itself).\nR(CP*) axiom number (3): (⃗a ·⃗x ≥a0) ∨(−⃗a ·⃗x ≥1−a0). The inequality (⃗a ·⃗x ≥a0) translates\ninto\nh_\nb=a0\n(⃗a · ⃗x = b) ,\nwhere h is the maximal value of ⃗a · ⃗x over 0, 1 assignments to ⃗x (that is, h is just the sum of all\npositive coefficients in ⃗a). The inequality (−⃗a · ⃗x ≥1 −a0) translates into\nf_\nb=1−a0\n(−⃗a · ⃗x = b) ,\nwhere f is the maximal value of −⃗a · ⃗x over 0, 1 assignments to ⃗x (that is, f is just the sum of\nall negative coefficients in ⃗a). Note that one can always flip the sign of any equation ⃗a · ⃗x = b in\nR(lin). This is done, for instance, by subtracting twice ⃗a · ⃗x = b from itself. So overall R(CP*)\naxiom number (3) translates into\nh_\nb=a0\n(⃗a · ⃗x = b) ∨\nf_\nb=1−a0\n(−⃗a · ⃗x = b) ,\nthat can be converted inside R(lin) into\na0−1\n_\nb=−f\n(⃗a · ⃗x = b) ∨\nh_\nb=a0\n(⃗a · ⃗x = b) .\n(52)\nLet A′ := {−f, −f + 1, . . . , a0 −1, a0, a0 + 1, . . . , h} and let A be the set of all possible values that\n⃗a · ⃗x can get over all possible Boolean assignments to ⃗x. Notice that A ⊆A′. By Lemma 8, for any\n⃗a · ⃗x, there is a polynomial-size (in the size of the linear form ⃗a · ⃗x) derivation of W\nα∈A(⃗a · ⃗x = α).\nBy using the R(lin) Weakening rule we can then derive W\nα∈A′(⃗a · ⃗x = α) which is equal to (52).\nInduction step: Here we simply need to show how to polynomially simulate inside R(lin) every\ninference rule application of R(CP*).\n40"},{"paragraph_id":"p72","order":72,"text":"Rule (1): Let A, B be two disjunctions of linear inequalities and let L1, L2 be two linear inequalities.\nAssume we already have a R(lin) proofs of bA ∨bL1 and bB ∨bL2. We need to derive bA ∨bB ∨\\\nL1 + L2.\nCorollary 7 shows that there is a polynomial-size (in the size of bL1 and bL2; which is polynomial\nin the size of L1 and L2) derivation of\n\\\nL1 + L2 from bL1 and bL2, from which the desired derivation\nimmediately follows.\nRule (2): The simulation of this rule in R(lin) is done using the R(lin) Weakening rule.\nRule (3): The simulation of this rule in R(lin) is done using the R(lin) Simplification rule (remem-\nber that 0 ≥1 translates into 0 = 1 under our translation scheme).\nRule (4): Let c be a non-negative integer. We need to derive\n\\\n(c⃗a · ⃗x ≥ca0)∨bA from\n\\\n(⃗a · ⃗x ≥a0)∨bA\nin R(lin).\nThis amounts only to “adding together” c times the disjunction\n\\\n(⃗a · ⃗x ≥a0) in\n\\\n(⃗a · ⃗x ≥a0) ∨bA. This can be achieved by c many applications of Corollary 7. We omit the details.\nRule (5): We need to derive\n\\\n(⃗a · ⃗x ≥⌈a0/c⌉) ∨bA, from\n\\\n(c⃗a · ⃗x ≥a0) ∨bA. Consider the disjunction\nof linear equations\n\\\n(c⃗a · ⃗x ≥a0), which can be written as:\n(c⃗a · ⃗x = a0) ∨(c⃗a · ⃗x = a0 + 1) ∨. . . ∨(c⃗a · ⃗x = a0 + r) ,\n(53)\nwhere a0 + r is the maximal value c⃗a · ⃗x can get over 0, 1 assignments to ⃗x. By Lemma 8 there is a\npolynomial-size (in the size of ⃗a · ⃗x) R(lin) proof of\n_\nα∈A\n(⃗a · ⃗x = α) ,\n(54)\nwhere A is the set of all possible values of ⃗a · ⃗x over 0, 1 assignments to ⃗x.\nWe now use (53) to cut-offfrom (54) all equations (⃗a · ⃗x = β) for all β < ⌈a0/c⌉(this will give\nus the desired disjunction of linear equations). Consider the equation (⃗a · ⃗x = β) in (54) for some\nfixed β < ⌈a0/c⌉. Use the resolution rule of R(lin) to add this equation to itself c times inside (54).\nWe thus obtain\n(c⃗a · ⃗x = cβ) ∨\n_\nα∈A\\{β}\n(⃗a · ⃗x = α) .\n(55)\nSince β is an integer and β < ⌈a0/c⌉, we have cβ < a0. Thus, the equation (c⃗a · ⃗x = cβ) does\nnot appear in (53). We can then successively resolve (c⃗a · ⃗x = cβ) in (55) with each equation\n(c⃗a · ⃗x = a0), . . . , (c⃗a · ⃗x = a0 + r) in (53). Hence, we arrive at W\nα∈A\\{β} (⃗a · ⃗x = α). Overall, we\ncan cut-offall equations (⃗a · ⃗x = β), for β < ⌈a0/c⌉, from (54). We then get the disjunction\n_\nα∈A′\n(⃗a · ⃗x = α) ,\nwhere A′ is the set of all elements of A greater or equal to ⌈a0/c⌉(in other words, all values greater\nor equal to ⌈a0/c⌉that ⃗a·⃗x can get over 0, 1 assignments to ⃗x). Using the Weakening rule of R(lin)\n(if necessary) we can arrive finally at the desired disjunction\n\\\n(⃗a · ⃗x ≥⌈a0/c⌉), which concludes the\nR(lin) simulation of R(CP*)’s inference Rule (5).\nAppendix A. Feasible Monotone Interpolation\nHere we formally define the feasible monotone interpolation property. The definition is taken\nmainly from [Kra97].\nRecall that for two binary strings of length n (or equivalently, Boolean assignments for n propo-\nsitional variables) α, α′, we denote by α′ ≥α that α′ is bitwise greater than α, that is, that for all\ni ∈[n], α′\ni ≥αi (where α′\ni and αi are the ith bits of α′ and α, respectively). Let A(⃗p, ⃗q), B(⃗p,⃗r)\nbe two collections of formulas in the displayed variables only, where ⃗p, ⃗q,⃗r are pairwise disjoint\n41"},{"paragraph_id":"p73","order":73,"text":"sequences of distinct variables (similar to the notation at the beginning of Section 7). Assume\nthat there is no assignment that satisfies both A(⃗p, ⃗q) and B(⃗p,⃗r). We say that A(⃗p, ⃗q), B(⃗p,⃗r) are\nmonotone if one of the following conditions hold:\n(1) If ⃗α is an assignment to ⃗p and ⃗β is an assignment to ⃗q such that A(⃗α, ⃗β) = 1, then for any\nassignment ⃗α′ ≥⃗α it holds that A(⃗α′, ⃗β) = 1.\n(2) If ⃗α is an assignment to ⃗p and ⃗β is an assignment to ⃗r such that B(⃗α, ⃗β) = 1, then for any\nassignment ⃗α′ ≤⃗α it holds that B(⃗α′, ⃗β) = 1.\nFix a certain proof system P. Recall the definition of the interpolant function (corresponding\nto a given unsatisfiable A(⃗p, ⃗q) ∧B(⃗p,⃗r); that is, functions for which (39) in Section 7 hold).\nAssume that for every monotone A(⃗p, ⃗q), B(⃗p,⃗r) there is a transformation from every P-refutation\nof A(⃗p, ⃗q)∧B(⃗p,⃗r) into the corresponding interpolant monotone Boolean circuit C(⃗p) (that is, C(⃗p)\nuses only monotone gates23) and whose size is polynomial in the size of the refutation (note that\nfor every monotone A(⃗p, ⃗q), B(⃗p,⃗r) the corresponding interpolant circuit must compute a monotone\nfunction;24 the interpolant circuit itself, however, might not be monotone, namely, it may use non-\nmonotone gates). In such a case, we say that P has the feasible monotone interpolation property.\nThis means that, if a proof system P has the feasible monotone interpolation property, then an\nexponential lower bound on monotone circuits that compute the interpolant function corresponding\nto A(⃗p, ⃗q) ∧B(⃗p,⃗r) implies an exponential-size lower bound on P-refutations of A(⃗p, ⃗q) ∧B(⃗p,⃗r).\nDefinition A.1 (Feasible monotone interpolation property). Let P be a propositional refu-\ntation system. Let A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r) be two collections of formulas\nwith the displayed variables only (where ⃗p has n variables, ⃗q has s variables and ⃗r has t variables),\nsuch that either (the set of satisfying assignments of) A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) meet condition 1 above\nor (the set of satisfying assignments of) B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r) meet condition 2 above. Assume\nthat for any such A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r), if there exists a P-refutation for\nA1(⃗p, ⃗q) ∧· · · ∧Ak(⃗p, ⃗q) ∧B1(⃗p,⃗r) ∧. . . ∧Bl(⃗p,⃗r) of size S then there exists a monotone Boolean\ncircuit separating UA from VB (as defined in Section 7.1) of size polynomial in S. In this case we\nsay that P possesses the feasible monotone interpolation property.\nAcknowledgments\nWe wish to thank Arist Kojevnikov for useful correspondence on his paper.\nThis work was\ncarried out in partial fulfillment of the requirements for the Ph.D. degree of the second author.\nReferences\n[AB87]\nNoga Alon and Ravi B. Boppana. The monotone circuit complexity of boolean functions. Combinatorica,\n7(1):1–22, 1987. 8, 26\n[ABE02]\nAlbert Atserias, Maria L. Bonet, and Juan L. Esteban. Lower bounds for the weak pigeonhole principle\nand random formulas beyond resolution. 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To appear in The Journal of Symbolic Logic. Preliminary version available in\nElectronic Colloquium on Computational Complexity, ECCC, January 2007. Report No. TR07-007. 6.3\n[KW88]\nMauricio Karchmer and Avi Wigderson. Monotone circuits for connectivity require super-logarithmic\ndepth. In Proceedings of the 20th Annual ACM Symposium on Theory of Computing, pages 539–550.\nACM, 1988. 7.1.1, 7.2\n[Pud97]\nPavel Pudl ́ak. Lower bounds for resolution and cutting plane proofs and monotone computations. The\nJournal of Symbolic Logic, 62(3):981–998, Sept. 1997. 6.3, 23\n[Razb85]\nAlexander A. Razborov. Lower bounds on the monotone complexity of some Boolean functions. Dokl.\nAkad. Nauk SSSR (in Russian), 281(4):798–801, 1985. [English translation in Sov. Math. Dokl., vol . 31\n(1985), pp. 354-357.]. 8\n[Razb95]\nAlexander A. Razborov. Unprovability of lower bounds on circuit size in certain fragments of bounded\narithmetic. Izv. Ross. Akad. Nauk Ser. Mat., 59(1):201–224, 1995. 7.1.1\n[Razb02]\nAlexander A. Razborov. Proof complexity of pigeonhole principles. In Developments in language theory\n(Vienna, 2001), volume 2295 of Lecture Notes in Comput. Sci., pages 110–116. Springer, Berlin, 2002. 1\n[Raz04]\nRan Raz. Multi-linear formulas for permanent and determinant are of super-polynomial size. In Pro-\nceedings of the 36th Annual ACM Symposium on the Theory of Computing, pages 633–641, Chicago, IL,\n2004. ACM. 1\n[Raz06]\nRan Raz. Separation of multilinear circuit and formula size. Theory of Computing, Vol. 2, article 6,\n2006. 1\n[RT06]\nRan Raz and Iddo Tzameret. The strength of multilinear proofs. Comput. Complexity (to appear). Pre-\nliminary version in Electronic Colloquium on Computational Complexity, ECCC, January 2006. Report\nNo. TR06-001. (document), 1, 1.1, 1.2, 1.2, 1, 2, 6.2, 9, 9.1.3, 4, 9.3, 33, 9.3.1, 4, 9.4, 9.4\n[SW01]\nAmir Shpilka and Avi Wigderson. Depth-3 arithmetic circuits over fields of characteristic zero. Comput.\nComplexity, 10:1–27, 2001. 9.1.1\n[Tse68]\nG. C. Tseitin. On the complexity of derivations in propositional calculus. Studies in constructive math-\nematics and mathematical logic Part II. Consultants Bureau, New-York-London, 1968. 6.2\n43"},{"paragraph_id":"p75","order":75,"text":"Department of Applied Mathematics and Computer Science, Weizmann Institute, Rehovot 76100,\nIsrael\nE-mail address: ranraz@wisdom.weizmann.ac.il\nSchool of Computer Science, Tel Aviv University, Tel Aviv 69978, Israel\nE-mail address: tzameret@tau.ac.il\n44"}],"pages":[{"page":1,"text":"arXiv:0708.1529v1 [cs.CC] 10 Aug 2007\nRESOLUTION OVER LINEAR EQUATIONS\nAND MULTILINEAR PROOFS\nRAN RAZ AND IDDO TZAMERET\nAbstract. We develop and study the complexity of propositional proof systems of varying strength\nextending resolution by allowing it to operate with disjunctions of linear equations instead of\nclauses. We demonstrate polynomial-size refutations for hard tautologies like the pigeonhole prin-\nciple, Tseitin graph tautologies and the clique-coloring tautologies in these proof systems. Using\nthe (monotone) interpolation by a communication game technique we establish an exponential-size\nlower bound on refutations in a certain, considerably strong, fragment of resolution over linear\nequations, as well as a general polynomial upper bound on (non-monotone) interpolants in this\nfragment.\nWe then apply these results to extend and improve previous results on multilinear proofs (over\nfields of characteristic 0), as studied in [RT06]. Specifically, we show the following:\n• Proofs operating with depth-3 multilinear formulas polynomially simulate a certain, consid-\nerably strong, fragment of resolution over linear equations.\n• Proofs operating with depth-3 multilinear formulas admit polynomial-size refutations of the\npigeonhole principle and Tseitin graph tautologies. The former improve over a previous result\nthat established small multilinear proofs only for the functional pigeonhole principle. The\nlatter are different than previous proofs, and apply to multilinear proofs of Tseitin mod p\ngraph tautologies over any field of characteristic 0.\nWe conclude by connecting resolution over linear equations with extensions of the cutting planes\nproof system.\nContents\n1.\nIntroduction\n2\n1.1.\nComparison to Earlier Work\n4\n1.2.\nSummary of Results\n5\n2.\nNotation and Background on Propositional Proof Systems\n7\n3.\nResolution over Linear Equations and its Subsystems\n8\n3.1.\nDisjunctions of Linear Equations\n8\n3.2.\nResolution over Linear Equations – R(lin)\n9\n3.3.\nFragment of Resolution over Linear Equations – R0(lin)\n10\n4.\nReasoning and Counting inside R(lin) and its Subsystems\n11\n4.1.\nBasic Reasoning inside R(lin) and its Subsystems\n12\n4.2.\nBasic Counting inside R(lin) and R0(lin)\n13\n5.\nImplicational Completeness of R(lin) and its Subsystems\n16\n6.\nShort Proofs for Hard Tautologies\n17\n6.1.\nThe Pigeonhole Principle Tautologies in R0(lin)\n17\n6.2.\nTseitin mod p Tautologies in R0(lin)\n19\n2000 Mathematics Subject Classification.\n03F20, 68Q17, 68Q15.\nKey words and phrases. proof complexity, resolution, algebraic proof systems, multilinear proofs, cutting planes,\nfeasible monotone interpolation.\nThe first author was supported by The Israel Science Foundation and The Minerva Foundation.\nThe second author was supported by The Israel Science Foundation (grant no. 250/05).\n1"},{"page":2,"text":"6.3.\nThe Clique-Coloring Principle in R(lin)\n23\n7.\nInterpolation Results for R0(lin)\n26\n7.1.\nInterpolation for Semantic Refutations\n26\n7.2.\nPolynomial Upper Bounds on Interpolants for R0(lin)\n28\n8.\nSize Lower Bounds\n30\n9.\nApplications to Multilinear Proofs\n32\n9.1.\nBackground on Algebraic and Multilinear Proofs\n32\n9.2.\nFrom R(lin) Proofs to PCR Proofs\n34\n9.3.\nFrom PCR Proofs to Multilinear Proofs\n35\n9.4.\nSmall Depth-3 Multilinear Proofs\n38\n10.\nRelations with Extensions of Cutting Planes\n38\nAppendix A.\nFeasible Monotone Interpolation\n41\nAcknowledgments\n42\nReferences\n42\n1. Introduction\nThis paper considers two kinds of proof systems. The first kind are extensions of resolution\nthat operate with disjunctions of linear equations with integral coefficients instead of clauses. The\nsecond kind are algebraic proof systems operating with multilinear arithmetic formulas. Proofs in\nboth kinds of systems establish the unsatisfiability of formulas in conjunctive normal form (CNF).\nWe are primarily concerned with connections between these two families of proof systems and with\nextending and improving previous results on multilinear proofs.\nThe resolution system is a popular propositional proof system that establishes the unsatisfiability\nof CNF formulas (or equivalently, the truth of tautologies in disjunctive normal form) by operating\nwith clauses (a clause is a disjunction of propositional variables and their negations). It is well\nknown that resolution cannot provide small (that is, polynomial-size) proofs for many basic count-\ning arguments. The most notable example of this are the strong exponential lower bounds on the\nresolution refutation size of the pigeonhole principle and its different variants (Haken [Hak85] was\nthe first to establish such a lower bound; see also [Razb02] for a survey on the proof complexity of\nthe pigeonhole principle). Due to the popularity of resolution both in practice, as the core of many\nautomated theorem provers, and as a theoretical case-study in propositional proof complexity, it\nis natural to consider weak extensions of resolution that can overcome its inefficiency in provid-\ning proofs of counting arguments. The proof systems we present in this paper are extensions of\nresolution, of various strength, that are suited for this purpose.\nPropositional proof systems of a different nature that also attracted much attention in proof\ncomplexity theory are algebraic proof systems, which are proof systems operating with (multivariate)\npolynomials over a field. In this paper, we are particularly interested in algebraic proof systems\nthat operate with multilinear polynomials represented as multilinear arithmetic formulas, called by\nthe generic name multilinear proofs (a polynomial is multilinear if the power of each variable in its\nmonomials is at most one). The investigation into such proof systems was initiated in [RT06], and\nhere we continue this line of research. This research is motivated on the one hand by the apparent\nconsiderable strength of such systems; and on the other hand, by the known super-polynomial\nsize lower bounds on multilinear formulas computing certain important functions [Raz04, Raz06],\ncombined with the general working assumption that establishing lower bounds on the size of objects\na proof system manipulates (in this case, multilinear formulas) is close to establishing lower bounds\non the size of the proofs themselves.\n2"},{"page":3,"text":"The basic proof system we shall study is denoted R(lin). The proof-lines1 in R(lin) proofs are\ndisjunctions of linear equations with integral coefficients over the variables ⃗x = x1, . . . , xn.\nIt\nturns out that (already proper subsystems of) R(lin) can handle very elegantly basic counting\narguments.\nThe following defines the R(lin) proof system. Given an initial CNF, we translate\nevery clause W\ni∈I xi ∨W\nj∈J ¬xj (where I are the indices of variables with positive polarities and\nJ are the indices of variables with negative polarities) pertaining to the CNF, into the disjunction\nW\ni∈I(xi = 1)∨W\nj∈J(xj = 0). Let A and B be two disjunctions of linear equations, and let ⃗a·⃗x = a0\nand ⃗b · ⃗x = b0 be two linear equations (where ⃗a,⃗b are two vectors of n integral coefficients, and\n⃗a · ⃗x is the scalar product Pn\ni=1 aixi; and similarly for ⃗b · ⃗x). The rules of inference belonging to\nR(lin) allow to derive A ∨B ∨((⃗a +⃗b) · ⃗x = a0 + b0) from A ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0) (or\nsimilarly, to derive A ∨B ∨((⃗a −⃗b) ·⃗x = a0 −b0) from A ∨(⃗a ·⃗x = a0) and B ∨(⃗b ·⃗x = b0)). We can\nalso simplify disjunctions by discarding (unsatisfiable) equations of the form (0 = k), for k ̸= 0.\nIn addition, for every variable xi, we shall add an axiom (xi = 0) ∨(xi = 1), which forces xi to\ntake on only Boolean values. A derivation of the empty disjunction (which stands for false) from\nthe (translated) clauses of a CNF is called an R(lin) refutation of the given CNF. This way, every\nunsatisfiable CNF has an R(lin) refutation (this can be proved by a straightforward simulation of\nresolution by R(lin)).\nThe basic idea connecting resolution operating with disjunctions of linear equations and multilin-\near proofs is this: Whenever a disjunction of linear equations is simple enough — and specifically,\nwhen it is close to a symmetric function, in a manner made precise — then it can be represented\nby a small size and small depth multilinear arithmetic formula over fields of characteristic 0. This\nidea was already used (somewhat implicitly) in [RT06] to obtain polynomial-size multilinear proofs\noperating with depth-3 multilinear formulas of the functional pigeonhole principle (this principle\nis weaker than the pigeonhole principle). In the current paper we generalize previous results on\nmultilinear proofs by fully using this idea: We show how to polynomially simulate with multilinear\nproofs, operating with small depth multilinear formulas, certain short proofs carried inside resolu-\ntion over linear equations. This enables us to provide new polynomial-size multilinear proofs for\ncertain hard tautologies, improving results from [RT06].\nMore specifically, we introduce a certain fragment of R(lin), which can be polynomially simu-\nlated by depth-3 multilinear proofs (that is, multilinear proofs operating with depth-3 multilinear\nformulas). On the one hand this fragment of resolution over linear equations already is sufficient\nto formalize in a transparent way basic counting arguments, and so it admits small proofs of the\npigeonhole principle and the Tseitin mod p formulas (which yields some new upper bounds on\nmultilinear proofs); and on the other hand we can use the (monotone) interpolation technique to\nestablish an exponential-size lower bound on refutations in this fragment as well as demonstrating a\ngeneral (non-monotone) polynomial upper bound on interpolants for this fragment. The possibility\nthat multilinear proofs (possibly, operating with depth-3 multilinear formulas) possess the feasible\nmonotone interpolation property (and hence, admit exponential-size lower bounds) remains open.\nAnother family of propositional proof systems we discuss in relation to the systems mentioned\nabove are the cutting planes system and its extensions. The cutting planes proof system operates\nwith linear inequalities with integral coefficients, and this system is very close to the extensions\nof resolution we present in this paper. In particular, the following simple observation can be used\nto polynomially simulate cutting planes proofs with polynomially bounded coefficients (and some\nof its extensions) inside resolution over linear equations: The truth value of a linear inequality\n⃗a · ⃗x ≥a0 (where ⃗a is a vector of n integral coefficients and ⃗x is a vector of n Boolean variables) is\n1Each element (usually a formula) of a proof-sequence is referred to as a proof-line.\n3"},{"page":4,"text":"equivalent to the truth value of the following disjunction of linear equalities:\n(⃗a · ⃗x = a0) ∨(⃗a · ⃗x = a0 + 1) ∨· · · ∨(⃗a · ⃗x = a0 + k) ,\nwhere a0 + k equals the sum of all positive coefficients in ⃗a (that is, a0 + k =\nmax\n⃗x∈{0,1}n (⃗a · ⃗x)).\nNote on terminology. All the proof systems considered in this paper intend to prove the unsatis-\nfiability over 0, 1 values of collections of clauses (possibly, of translation of the clauses to disjunctions\nof linear equations). In other words, proofs in such proof systems intend to refute the collections\nof clauses, which is to validate their negation. Therefore, throughout this paper we shall sometime\nspeak about refutations and proofs interchangeably, always intending refutations, unless otherwise\nstated.\n1.1. Comparison to Earlier Work. To the best of our knowledge this paper is the first that\nconsiders resolution proofs operating with disjunctions of linear equations. Previous works consid-\nered extensions of resolution over linear inequalities augmented with the cutting planes inference\nrules (the resulting proof system denoted R(CP)). In full generality, we show that resolution over\nlinear equations can polynomially simulate R(CP) when the coefficients in all the inequalities are\npolynomially bounded (however, the converse is not known to hold). On the other hand, we shall\nconsider a certain fragment of resolution over linear equations, in which we do not even know how to\npolynomially simulate cutting planes proofs with polynomially bounded coefficients in inequalities\n(let alone R(CP) with polynomially bounded coefficients in inequalities). We now shortly discuss\nthe previous work on R(CP) and related proof systems.\nExtensions of resolution to disjunctions of linear inequalities were first considered by Kraj ́ıˇcek\n[Kra98] who developed the proof systems LK(CP) and R(CP). The LK(CP) system is a first-order\n(Gentzen-style) sequent calculus that operates with linear inequalities instead of atomic formulas\nand augments the standard first-order sequent calculus inference rules with the cutting planes\ninference rules. The R(CP) proof system is essentially resolution over linear inequalities, that is,\nresolution that operates with disjunctions of linear inequalities instead of clauses.\nThe main motivation of [Kra98] is to extend the feasible interpolation technique and consequently\nthe lower bounds results, from cutting planes and resolution to stronger proof systems. That paper\nestablishes an exponential-size lower bound on a restricted version of R(CP) proofs, namely, when\nthe number of inequalities in each proof-line is O(nε), where n is the number of variables of the\ninitial formulas, ε is a small enough constant and the coefficients in the cutting planes inequalities\nare polynomially bounded.\nOther papers considering extensions of resolution over linear inequalities are the more recent\npapers by Hirsch & Kojevnikov [HK06] and Kojevnikov [Koj07]. The first paper [HK06] considers\na combination of resolution with LP (an incomplete subsystem of cutting planes based on simple\nlinear programming reasoning), with the ‘lift and project’ proof system (L&P), and with the cutting\nplanes proof system. The second paper [Koj07] deals with improving the parameters of the tree-like\nR(CP) lower-bounds obtained in [Kra98].\nWhereas previous results concerned primarily with extending the cutting planes proof system,\nour foremost motivation is to extend and improve previous results on algebraic proof systems\noperating with multilinear formulas obtained in [RT06]. In that paper the concept of multilinear\nproofs was introduced and several basic results concerning multilinear proofs were proved.\nIn\nparticular, polynomial-size proofs of two important combinatorial principles were demonstrated:\nthe functional pigeonhole principle and the Tseitin (mod p) graph tautologies. In the current paper\nwe improve both these results.\nAs mentioned above, motivated by relations with multilinear proofs operating with depth-3 mul-\ntilinear formulas, we shall consider a certain subsystem of resolution over linear equations. For\nthis subsystem we apply twice the interpolation by a communication game technique. The first\n4"},{"page":5,"text":"application is of the non-monotone version of the technique, and the second application is of the\nmonotone version. Namely, the first application provides a general (non-monotone) interpolation\ntheorem that demonstrates a polynomial (in the size of refutations) upper bound on interpolants;\nThe proof uses the general method of transforming a refutation into a Karchmer-Wigderson com-\nmunication game for two players, from which a Boolean circuit is then attainable. In particular,\nwe shall apply the interpolation theorem of Kraj ́ıˇcek from [Kra97]. The second application of the\n(monotone) interpolation by a communication game technique is implicit and proceeds by using\nthe lower bound criterion of Bonet, Pitassi & Raz in [BPR97]. This criterion states that (semantic)\nproof systems (of a certain natural and standard kind) whose proof-lines (considered as Boolean\nfunctions) have low communication complexity cannot prove efficiently a certain tautology (namely,\nthe clique-coloring tautologies).\n1.2. Summary of Results. This paper introduces and connects several new concepts and ideas\nwith some known ones. It identifies new extensions of resolution operating with linear equations,\nand relates (a certain) such extension to multilinear proofs. The upper bounds for the pigeonhole\nprinciple and Tseitin mod p formulas in fragments of resolution over linear equations are new. By\ngeneralizing the machinery developed in [RT06], these upper bounds yield new and improved re-\nsults concerning multilinear proofs. The lower bound for the clique-coloring formulas in a fragment\nof resolution over linear equations employs the standard monotone interpolation by a communica-\ntion game technique, and specifically utilizes the theorem of Bonet, Pitassi & Raz from [BPR97].\nThe general (non-monotone) interpolation result for a fragment of resolution over linear equations\nemploys the theorem of Kraj ́ıˇcek from [Kra97]. The upper bound in (the stronger variant of –\nas described in the introduction) resolution over linear equations of the clique-coloring formulas\nfollows that of Atserias, Bonet & Esteban [ABE02]. We now give a detailed outline of the results\nin this paper.\nThe proof systems. In Section 3 we formally define two extensions of resolution of decreasing\nstrength allowing resolution to operate with disjunctions of linear equations. The size of a linear\nequation a1x1 + . . . + anxn = a0 is the sum of all a0, . . . , an written in unary notation. The size of\na disjunction of linear equations is the total size of all linear equations in the disjunction. The size\nof a proof operating with disjunctions of linear equations is the total size of all the disjunctions in\nit.\nR(lin): This is the stronger proof system (described in the introduction) that operates with\ndisjunctions of linear equations with integer coefficients.\nR0(lin): This is a (provably proper) fragment of R(lin). It operates with disjunctions of (arbi-\ntrarily many) linear equations whose variables have constant coefficients, under the restriction that\nevery disjunction can be partitioned into a constant number of sub-disjunctions, where each sub-\ndisjunction either consists of linear equations that differ only in their free-terms or is a (translation\nof a) clause.\nNote that any single linear inequality with Boolean variables can be represented by a disjunction\nof linear equations that differ only in their free-terms (see the example in the introduction section).\nSo the R0(lin) proof system is close to a proof system operating with disjunctions of constant\nnumber of linear inequalities (with constant integral coefficients). In fact, disjunctions of linear\nequations varying only in their free-terms, have more (expressive) strength than a single inequality.\nFor instance, the parity function can be easily represented by a disjunction of linear equations,\nwhile it cannot be represented by a single linear inequality (or even by a disjunction of linear\ninequalities).\nAs already mentioned, the motivation to consider the restricted proof system R0(lin) comes from\nits relation to multilinear proofs operating with depth-3 multilinear formulas (in short, depth-3\n5"},{"page":6,"text":"multilinear proofs): R0(lin) corresponds roughly to the subsystem of R(lin) that we know how\nto simulate by depth-3 multilinear proofs via the technique in [RT06] (the technique is based on\nconverting disjunctions of linear forms into symmetric polynomials, which are known to have small\ndepth-3 multilinear formulas). This simulation is then applied in order to improve over known\nupper bounds for depth-3 multilinear proofs, as R0(lin) is already sufficient to efficiently prove\ncertain “hard tautologies”.\nMoreover, we are able to establish an exponential lower bound on\nR0(lin) refutations size (see below for both upper and lower bounds on R0(lin) proofs). We also\nestablish a super-polynomial separation of R(lin) from R0(lin) (via the clique-coloring principle, for\na certain choice of parameters; see below).\nShort refutations. We demonstrate the following short refutations in R0(lin) and R(lin):\n(1) Polynomial-size refutations of the pigeonhole principle in R0(lin);\n(2) Polynomial-size refutations of Tseitin mod p graph formulas in R0(lin);\n(3) Polynomial-size refutations of the clique-coloring formulas in R(lin) (for certain parameters).\nThe refutations here follow by direct simulation of the Res(2) refutations of clique-coloring\nformulas from [ABE02].\nAll the three families of formulas above are prominent “hard tautologies” in proof complexity\nliterature, which means that strong size lower bounds on proofs in various proof systems are known\nfor them (for the exact formulation of these families of formulas see Section 6).\nInterpolation results. We provide a polynomial upper-bound on (non-monotone) interpolants\ncorresponding to R0(lin) refutations; Namely, we show that any R0(lin)-refutation of a given formula\ncan be transformed into a (non-monotone) Boolean circuit computing the corresponding interpolant\nfunction of the formula (if there exists such a function), with at most a polynomial increase in size.\nWe employ the general interpolation theorem of Kraj ́ıˇcek [Kra97] for semantic proof systems.\nLower bounds. We provide the following exponential lower bound:\nTheorem 1. R0(lin) does not have sub-exponential refutations for the clique-coloring formulas.\nThis result is proved by applying a result of Bonet, Pitassi & Raz [BPR97], that (implicitly) use\nthe monotone interpolation by a communication game technique for establishing an exponential-\nsize lower bound on refutations of general semantic proof systems operating with proof-lines of low\ncommunication complexity.\nApplications to multilinear proofs. Multilinear proof systems are (semantic) refutation sys-\ntems operating with multilinear polynomials over a fixed field, where every multilinear polynomial\nis represented by a multilinear arithmetic formula.\nIn this paper we shall consider multilinear\nformulas over fields of characteristic 0 only. The size of a multilinear proof (that is, a proof in\na multilinear proof system) is the total size of all multilinear formulas in the proof (for formal\ndefinitions concerning multilinear proofs see Section 9).\nWe shall first connect multilinear proofs with resolution over linear equations by the following\nresult:\nTheorem 2. Multilinear proofs operating with depth-3 multilinear formulas over characteristic 0\npolynomially-simulate R0(lin).\nAn immediate corollary of this theorem and the upper bounds in R0(lin) described above are\npolynomial-size multilinear proofs for the pigeonhole principle and the Tseitin mod p formulas.\n(1) Polynomial-size depth-3 multilinear refutations for the pigeonhole principle over fields of\ncharacteristic 0. This improves over [RT06] that shows a similar upper bound for a weaker\nprinciple, namely, the functional pigeonhole principle.\n(2) Polynomial-size depth-3 multilinear refutations for the Tseitin mod p graph formulas over\nfields of characteristic 0. These refutations are different than those demonstrated in [RT06],\n6"},{"page":7,"text":"and further they establish short multilinear refutations of the Tseitin mod p graph formulas\nover any field of characteristic 0 (the proof in [RT06] showed how to refute the Tseitin mod\np formulas by multilinear refutations only over fields that contain a primitive pth root of\nunity).\nRelations with cutting planes proofs. As mentioned in the introduction, a proof system com-\nbining resolution with cutting planes was presented by Kraj ́ıˇcek in [Kra98]. The resulting system\nis denoted R(CP) (see Section 10 for a definition). When the coefficients in the linear inequalities\ninside R(CP) proofs are polynomially bounded, the resulting proof system is denoted R(CP*). We\nestablish the following simulation result:\nTheorem 3. R(lin) polynomially simulates resolution over cutting planes inequalities with polyno-\nmially bounded coefficients R(CP*).\nWe do not know if the converse also holds.\n2. Notation and Background on Propositional Proof Systems\nFor a natural number n, we use [n] to denote {1, . . . , n}. For a vector of n (integral) coefficients\n⃗a and a vector of n variables ⃗x, we denote by ⃗a · ⃗x the scalar product Pn\ni=1 aixi. If ⃗b is another\nvector (of length n), then ⃗a +⃗b denotes the addition of ⃗a and ⃗b as vectors, and c⃗a (for an integer\nc) denotes the product of the scalar c with ⃗a (where, −⃗a denotes −1⃗a). For two linear equations\nL1 : ⃗a · ⃗x = a0 and L2 : ⃗b · ⃗x = b0, their addition (⃗a +⃗b) · ⃗x = a0 + b0 is denoted L1 + L2 (and their\nsubtraction (⃗a −⃗b) · ⃗x = a0 −b0 is denoted L1 −L2). For two Boolean assignments (identified as\n0, 1 strings) α, α′ ∈{0, 1}n we write α′ ≥α if α′\ni ≥αi, for all i ∈[n] (where αi, α′\ni are the ith bits\nof α and α′, respectively).\nWe now recall some basic concepts on propositional proof systems. For background on algebraic\nproof systems (and specifically multilinear proofs) see Section 9.\nResolution. In order to put our work in context, we need to define the resolution refutation system.\nA CNF formula over the variables x1, . . . , xn is defined as follows. A literal is a variable xi or\nits negation ¬xi. A clause is a disjunction of literals. A CNF formula is a conjunction of clauses.\nThe size of a clause is the number of literals in it.\nResolution is a complete and sound proof system for unsatisfiable CNF formulas. Let C and D\nbe two clauses containing neither xi nor ¬xi, the resolution rule allows one to derive C ∨D from\nC ∨xi and D ∨¬xi. The clause C ∨D is called the resolvent of the clauses C ∨xi and D ∨¬xi on\nthe variable xi, and we also say that C ∨xi and D ∨¬xi were resolved over xi. The weakening rule\nallows to derive the clause C ∨D from the clause C, for any two clauses C, D.\nDefinition 2.1 (Resolution). A resolution proof of the clause D from a CNF formula K is a\nsequence of clauses D1, D2, . . . , Dl, such that: (1) each clause Dj is either a clause of K or a\nresolvent of two previous clauses in the sequence or derived by the weakening rule from a previous\nclause in the sequence; (2) the last clause Dl= D. The size of a resolution proof is the sum of all\nthe sizes of the clauses in it. A resolution refutation of a CNF formula K is a resolution proof of\nthe empty clause ✷from K (the empty clause stands for false; that is, the empty clause has no\nsatisfying assignments).\nA proof in resolution (or any of its extensions) is called also a derivation or a proof-sequence.\nEach sequence-element in a proof-sequence is called also a proof-line. A proof-sequence containing\nthe proof-lines D1, . . . , Dlis also said to be a derivation of D1, . . . , Dl.\n7"},{"page":8,"text":"Cook-Reckhow proof systems. Following [CR79], a Cook-Reckhow proof system is a polynomial-\ntime algorithm A that receives a Boolean formula F (for instance, a CNF) and a string π over some\nfinite alphabet (“the (proposed) refutation” of F), such that there exists a π with A(F, π) = 1 if\nand only if F is unsatisfiable. The completeness of a (Cook-Reckhow) proof system (with respect\nto the set of all unsatisfiable Boolean formulas; or for a subset of it, e.g. the set of unsatisfiable\nCNF formulas) stands for the fact that every unsatisfiable formula F has a string π (“the refutation\nof F”) so that A(F, π) = 1. The soundness of a (Cook-Reckhow) proof system stands for the fact\nthat every formula F so that A(F, π) = 1 for some string π is unsatisfiable (in other words, no\nsatisfiable formula has a refutation).\nFor instance, resolution is a Cook-Reckhow proof system, since it is complete and sound for the\nset of unsatisfiable CNF formulas, and given a CNF formula F and a string π it is easy to check in\npolynomial-time (in both F and π) whether π constitutes a resolution refutation of F.\nWe shall also consider proof systems that are not necessarily (that is, not known to be) Cook-\nReckhow proof systems. Specifically, multilinear proof systems (over large enough fields) meet the\nrequirements in the definition of Cook-Reckhow proof systems, except that the condition on A\nabove is relaxed: we allow A to be in probabilistic polynomial-time BPP (which is not known to\nbe equal to deterministic polynomial-time).\nPolynomial simulations of proof systems. When comparing the strength of different proof\nsystems we shall confine ourselves to CNF formulas only. That is, we consider propositional proof\nsystems as proof systems for the set of unsatisfiable CNF formulas. For that purpose, if a proof\nsystem does not operate with clauses directly, then we fix a (direct) translation from clauses to\nthe objects operated by the proof system. This is done for both resolution over linear equations\n(which operate with disjunctions of linear equations) and its fragments, and also for multilinear\nproofs (which operate with multilinear polynomials, represented as multilinear formulas); see for\nexample Subsection 3.1 for such a direct translation.\nDefinition 2.2. Let P1, P2 be two proof systems for the set of unsatisfiable CNF formulas (we\nidentify a CNF formula with its corresponding translation, as discussed above). We say that P2\npolynomially simulates P1 if given a P1 refutation π of a CNF formula F, then there exists a\nrefutation of F in P2 of size polynomial in the size of π. In case P2 polynomially simulates P1 while\nP1 does not polynomially simulates P2 we say that P2 is strictly stronger than P1.\n3. Resolution over Linear Equations and its Subsystems\nThe proof systems we consider in this section are extensions of resolution. Proof-lines in res-\nolution are clauses.\nInstead of this, the extensions of resolution we consider here operate with\ndisjunctions of linear equations with integral coefficients. For this section we use the convention\nthat all the formal variables in the propositional proof systems considered are taken from the set\nX := {x1, . . . , xn}.\n3.1. Disjunctions of Linear Equations. For L a linear equation a1x1 + . . . + anxn = a0, the\nright hand side a0 is called the free-term of L and the left hand side a1x1 + . . . + anxn is called the\nlinear form of L (the linear form can be 0). A disjunction of linear equations is of the following\ngeneral form:\n \na(1)\n1 x1 + . . . + a(1)\nn xn = a(1)\n0\n \n∨· · · ∨\n \na(t)\n1 x1 + . . . + a(t)\nn xn = a(t)\n0\n \n,\n(1)\nwhere t ≥0 and the coefficients a(j)\ni\nare integers (for all 0 ≤i ≤n, 1 ≤j ≤t). We discard duplicate\nlinear equations from a disjunction of linear equations. The semantics of such a disjunction is the\nnatural one: We say that an assignment of integral values to the variables x1, ..., xn satisfies (1)\n8"},{"page":9,"text":"if and only if there exists j ∈[t] so that the equation a(j)\n1 x1 + . . . + a(j)\nn xn = a(j)\n0\nholds under the\ngiven assignment.\nThe symbol |= denotes the semantic implication relation, that is, for every collection D1, . . . , Dm\nof disjunctions of linear equations,\nD1, . . . , Dm |= D0\nmeans that every assignment of 0, 1 values that satisfies all D1, . . . , Dm also satisfies D0.2 In this\ncase we also say that D1, . . . , Dm semantically imply D0.\nThe size of a linear equation a1x1 + . . . + anxn = a0 is Pn\ni=0 |ai|, i.e., the sum of the bit sizes\nof all ai written in unary notation. Accordingly, the size of the linear form a1x1 + . . . + anxn is\nPn\ni=1 |ai|. The size of a disjunction of linear equations is the total size of all linear equations in it.\nSince all linear equations considered in this paper are of integral coefficients, we shall speak\nof linear equations when we actually mean linear equations with integral coefficients. Similar to\nresolution, the empty disjunction is unsatisfiable and stands for the truth value false.\nTranslation of clauses. As described in the introduction, we can translate any CNF formula to\na collection of disjunctions of linear equations in a direct manner: Every clause W\ni∈I xi ∨W\nj∈J ¬xj\n(where I and J are sets of indices of variables) pertaining to the CNF is translated into the\ndisjunction W\ni∈I(xi = 1) ∨W\nj∈J(xj = 0). For a clause D we denote by eD its translation into a\ndisjunction of linear equations. It is easy to verify that any Boolean assignment to the variables\nx1, . . . , xn satisfies a clause D if and only if it satisfies eD (where true is treated as 1 and false as\n0).\n3.2. Resolution over Linear Equations – R(lin). Defined below is our basic proof system\nR(lin) that enables resolution to reason with disjunctions of linear equations. As we wish to reason\nabout Boolean variables we augment the system with the axioms (xi = 0)∨(xi = 1), for all i ∈[n],\ncalled the Boolean axioms.\nDefinition 3.1 (R(lin)). Let K := {K1, . . . , Km} be a collection of disjunctions of linear equations.\nAn R(lin)-proof from K of a disjunction of linear equations D is a finite sequence π = (D1, ..., Dl)\nof disjunctions of linear equations, such that Dl= D and for every i ∈[l], either Di = Kj for\nsome j ∈[m], or Di is a Boolean axiom (xh = 0) ∨(xh = 1) for some h ∈[n], or Di was deduced\nby one of the following R(lin)-inference rules, using Dj, Dk for some j, k < i:\nResolution: Let A, B be two disjunctions3of linear equations and let L1, L2 be two linear\nequations.\nFrom A ∨L1 and B ∨L2 derive A ∨B ∨(L1 + L2).\nSimilarly, from A ∨L1 and B ∨L2 derive A ∨B ∨(L1 −L2).\nWeakening: From a disjunction of linear equations A derive A∨L , where L is an arbitrary\nlinear equation over X.\nSimplification: From A ∨(0 = k) derive A, where A is a disjunction of linear equations\nand k ̸= 0.\nAn R(lin) refutation of a collection of disjunctions of linear equations K is a proof of the empty\ndisjunction from K. The size of an R(lin)-proof π is the total size of all the disjunctions of linear\nequations in π, denoted |π|.\nSimilar to resolution, in case A ∨B ∨(L1 + L2) is derived from A ∨L1 and B ∨L2 by the\nresolution rule, we say that A ∨L1 and B ∨L2 were resolved over L1 and L2, respectively, and we\n2Alternatively, we can consider assignments of any integral values (instead of only Boolean values) to the variables\nin D1, . . . , Dm, stipulating that the collection D1, . . . , Dm contains all disjunctions of the form (xj = 0) ∨(xj = 1)\nfor all the variables xj ∈X (these formulas force any satisfying assignment to give only 0, 1 values to the variables).\n3Possibly the empty disjunction. This remark also applies to the inference rules below.\n9"},{"page":10,"text":"call A ∨B ∨(L1 + L2) the resolvent of A ∨L1 and B ∨L2 (and similarly, when A ∨B ∨(L1 −L2)\nis derived from A ∨L1 and B ∨L2 by the resolution rule; we use the same terminology for both\naddition and subtraction, and it should be clear from the context which operation is actually\napplied). We also describe such an application of the resolution rule by saying that L1 was added\n(resp., subtracted) to (resp. from) L2 in A ∨L1 and B ∨L2.\nIn light of the direct translation between CNF formulas and collections of disjunctions of linear\nequations (described in the previous subsection), we can consider R(lin) to be a proof system for\nthe set of unsatisfiable CNF formulas:\nProposition 1. The R(lin) refutation system is a sound and complete Cook-Reckhow (see Sec-\ntion 2) refutation system for unsatisfiable CNF formulas (translated into unsatisfiable collection of\ndisjunctions of linear equations).\nProof: Completeness of R(lin) (for the set of unsatisfiable CNF formulas) stems from a straight-\nforward simulation of resolution, as we now show.\nClaim 1. R(lin) polynomially simulates resolution.\nProof of claim: Proceed by induction on the length of the resolution refutation to show that any\nresolution derivation of a clause A can be translated with only a linear increase in size into an R(lin)\nderivation of the corresponding disjunction of linear equations eA (see the previous subsection for\nthe definition of eA).\nThe base case: An initial clause A is translated into its corresponding disjunction of linear\nequations eA.\nThe induction step: If a resolution clause A ∨B was derived by the resolution rule from A ∨xi\nand B ∨¬xi, then in R(lin) we subtract (xi = 0) from (xi = 1) in eB ∨(xi = 0) and eA ∨(xi = 1),\nrespectively, to obtain eA ∨eB ∨(0 = 1). Then, using the Simplification rule, we can cut-off(0 = 1)\nfrom eA ∨eB ∨(0 = 1), and arrive at eA ∨eB.\nIf a clause A ∨B was derived in resolution from A by the Weakening rule, then we derive eA ∨eB\nfrom eA by the Weakening rule in R(lin).\nSoundness of R(lin) stems from the soundness of the inference rules (which means that: If D\nwas derived from C, B by the R(lin) resolution rule then any assignment that satisfies both C and\nB also satisfies D; and if D was derived from C by either the Weakening rule or the Simplification\nrule, then any assignment that satisfies C also satisfies D).\nThe R(lin) proof system is a Cook-Reckhow proof system, as it is easy to verify in polynomial-\ntime whether an R(lin) proof-line is inferred, by an application of one of R(lin)’s inference rules,\nfrom a previous proof-line (or proof-lines). Thus, any sequence of disjunctions of linear equations,\ncan be checked in polynomial-time (in the size of the sequence) to decide whether or not it is a\nlegitimate R(lin) proof-sequence.\nIn Section 5 we shall see that a stronger notion of completeness (that is, implicational complete-\nness) holds for R(lin) and its subsystems.\n3.3. Fragment of Resolution over Linear Equations – R0(lin). Here we consider a restriction\nof R(lin), denoted R0(lin). As discussed in the introduction section, R0(lin) is roughly the fragment\nof R(lin) we know how to polynomially simulate with depth-3 multilinear proofs.\nBy results established in the sequel (Sections 6.3 and 8) R(lin) is strictly stronger than R0(lin),\nwhich means that R(lin) polynomially simulates R0(lin), while the converse does not hold.\nR0(lin) operates with disjunctions of (arbitrarily many) linear equations with constant coefficients\n(excluding the free terms), under the following restriction: Every disjunction can be partitioned\n10"},{"page":11,"text":"into a constant number of sub-disjunctions, where each sub-disjunction either consists of linear\nequations that differ only in their free-terms or is a (translation of a) clause.\nAs mentioned in the introduction, every linear inequality with Boolean variables can be rep-\nresented by a disjunction of linear equations that differ only in their free-terms. So the R0(lin)\nproof system resembles, to some extent, a proof system operating with disjunctions of constant\nnumber of linear inequalities with constant integral coefficients (on the other hand, it is probable\nthat R0(lin) is stronger than such a proof system, as a disjunction of linear equations that differ\nonly in their free terms is [expressively] stronger than a linear inequality [or even a disjunction of\nlinear inequalities]: the former can define the parity function while the latter cannot).\nExample of an R0(lin)-line:\n(x1 + . . . + xl= 1) ∨· · · ∨(x1 + . . . + xl= l) ∨(xl+1 = 1) ∨· · · ∨(xn = 1),\nfor some 1 ≤l≤n. The next section contains other concrete (and natural) examples of R0(lin)-\nlines.\nLet us define formally what it means to be an R0(lin) proof-line, that is, a proof-line inside an\nR0(lin) proof, called R0(lin)-line:\nDefinition 3.2 (R0(lin)-line). Let D be a disjunction of linear equations whose variables have\nconstant integer coefficients (the free-terms are unbounded). Assume D can be partitioned into a\nconstant number k of sub-disjunctions D1, . . . , Dk, where each Di either consists of (an unbounded)\ndisjunction of linear equations that differ only in their free-terms, or is a translation of a clause (as\ndefined in Subsection 3.1). Then the disjunction D is called an R0(lin)-line.\nThus, any R0(lin)-line is of the following general form:\n_\ni∈I1\n \n⃗a(1) · ⃗x = l(1)\ni\n \n∨· · · ∨\n_\ni∈Ik\n \n⃗a(k) · ⃗x = l(k)\ni\n \n∨\n_\nj∈J\n(xj = bj) ,\n(2)\nwhere k and all at\nr (for r ∈[n] and t ∈[k]) are integer constants and bj ∈{0, 1} (for all j ∈J) (and\nI1, . . . , Ik, J are unbounded sets of indices). Note that a disjunction of clauses can be combined\ninto a single clause. Hence, without loss of generality we can assume that in any R0(lin)-line only\na single (translation of a) clause occurs. This is depicted in (2) (where in addition we have ignored\nin (2) the possibility that the single clause obtained by combining several clauses contains xj ∨¬xj,\nfor some j ∈[n]).\nDefinition 3.3 (R0(lin)). The R0(lin) proof system is a restriction of the R(lin) proof system in\nwhich each proof-line is an R0(lin)-line (as in Definition 3.2).\nFor a completeness proof of R0(lin) see Section 5.4\n4. Reasoning and Counting inside R(lin) and its Subsystems\nIn this section we illustrate a simple way to reason by case-analysis inside R(lin) and its subsys-\ntems. This kind of reasoning will simplify the presentation of proofs inside R(lin) (and R0(lin)) in\nthe sequel (essentially, a similar – though weaker – kind of reasoning is applicable already in reso-\nlution). We will then demonstrate efficient and transparent proofs for simple counting arguments\nthat will also facilitate us in the sequel.\n4The simulation of resolution inside R(lin) (in the proof of Proposition 1) is carried on with each R(lin) proof-line\nbeing in fact a translation of a clause, and hence, an R0(lin)-line (notice that the Boolean axioms of R(lin) are\nR0(lin)-lines). This already implies that R0(lin) is a complete refutation system for the set of unsatisfiable CNF\nformulas. In section 5 we give a proof of a stronger notion of completeness for R0(lin).\n11"},{"page":12,"text":"4.1. Basic Reasoning inside R(lin) and its Subsystems. Given K a collection of disjunc-\ntions of linear equations {K1, . . . , Km} and C a disjunction of linear equations, denote by K ∨C\nthe collection {K1 ∨C, . . . , Km ∨C}.\nRecall that the formal variables in our proof system are\nx1, . . . , xn.\nLemma 4. Let K be a collection of disjunctions of linear equations, and let z abbreviate some linear\nform with integer coefficients. Let E1, . . . , Elbe ldisjunctions of linear equations. Assume that for\nall i ∈[l] there is an R(lin) derivation of Ei from z = ai and K with size at most s where a1, . . . , al\nare distinct integers. Then, there is an R(lin) proof of Wl\ni=1 Ei from K and (z = a1)∨· · ·∨(z = al),\nwith size polynomial in s and l.\nProof: Denote by D the disjunction (z = a1) ∨· · · ∨(z = al) and by πi the R(lin) proof of Ei from\nK and z = ai (with size at most s), for all i ∈[l]. It is easy to verify that for all i ∈[l] the sequence\nπi ∨W\nj∈[l]\\{i}(z = aj) is an R(lin) proof of Ei ∨W\nj∈[l]\\{i}(z = aj) from K and D. So overall, given\nD and K as premises, there is an R(lin) derivation of size polynomial in s and lof the following\ncollection of disjunctions of linear equations:\nE1 ∨\n_\nj∈[l]\\{1}\n(z = aj), . . . , El∨\n_\nj∈[l]\\{l}\n(z = aj) .\n(3)\nWe now use the Resolution rule to cut-offall the equations (z = ai) inside all the disjunctions\nin (3). Formally, we prove that for every 1 ≤k ≤lthere is a polynomial-size (in s and l) R(lin)\nderivation from (3) of\nE1 ∨· · · ∨Ek ∨\n_\nj∈[l]\\[k]\n(z = aj) ,\n(4)\nand so putting k = l, will conclude the proof of the lemma.\nWe proceed by induction on k.\nThe base case for k = 1 is immediate (from (3)).\nFor the\ninduction case, assume that for some 1 ≤k < lwe already have an R(lin) proof of (4), with size\npolynomial in s and l.\nConsider the line\nEk+1 ∨\n_\nj∈[l]\\{k+1}\n(z = aj) .\n(5)\nWe can now cut-offthe disjunctions W\nj∈[l]\\[k](z = aj) and W\nj∈[l]\\{k+1}(z = aj) from (4) and (5),\nrespectively, using the Resolution rule (since the aj’s in (4) and in (5) are disjoint).\nWe will\ndemonstrate this derivation in some detail now, in order to exemplify a proof carried inside R(lin).\nWe shall be less formal sometime in the sequel.\nResolve (4) with (5) over (z = ak+1) and (z = a1), respectively, to obtain\n(0 = a1 −ak+1) ∨E1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,k+1}\n(z = aj) .\n(6)\nSince a1 ̸= ak+1, we can use the Simplification rule to cut-off(0 = a1 −ak+1) from (6), and we\narrive at\nE1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,k+1}\n(z = aj) .\n(7)\nNow, similarly, resolve (4) with (7) over (z = ak+1) and (z = a2), respectively, and use Simplification\nto obtain\nE1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,2,k+1}\n(z = aj) .\n12"},{"page":13,"text":"Continue in a similar manner until you arrive at\nE1 ∨· · · ∨Ek ∨Ek+1 ∨\n_\nj∈[l]\\{1,2,...,k,k+1}\n(z = aj) ,\nwhich is precisely what we need.\nUnder the appropriate conditions, Lemma 4 also holds for R0(lin) proofs. This is stated in the\nfollowing lemma.\nLemma 5. Let K be a collection of disjunctions of linear equations, and let z abbreviate a linear\nform with integer coefficients. Let E1, . . . , Elbe ldisjunctions of linear equations. Assume that for\nall i ∈[l] there is an R0(lin) derivation of Ei from z = ai and K with size at most s, where the\nai’s are distinct integers. Then, assuming Wl\ni=1 Ei is an R0(lin)-line, there is an R0(lin) proof of\nWl\ni=1 Ei from K and (z = a1) ∨· · · ∨(z = al), with size polynomial in s and l.\nProof: It can be verified by simple inspection that, under the conditions spelled out in the state-\nment of the lemma, each proof-line in the R(lin) derivations in the proof of Lemma 4 is actually\nan R0(lin)-line.5\nAbbreviations. Lemmas 4 and 5 will sometime facilitate us to proceed inside R(lin) and R0(lin)\nwith a slightly less formal manner. For example, the situation in Lemma 4 above can be depicted\nby saying that “if z = ai implies Ei (with a polynomial-size proof) for all i ∈[l], then Wl\ni=1(z = ai)\nimplies Wl\ni=1 Ei (with a polynomial-size proof)”.\nIn case Wl\ni=1(z = ai) above is just the Boolean axiom (xi = 0) ∨(xi = 1), for some i ∈[n], and\nxi = 0 implies E0 and xi = 1 implies E1 (both with polynomial-size proofs), then to simplify the\nwriting we shall sometime not mention the Boolean axiom at all. For example, the latter situation\ncan be depicted by saying that “if xi = 0 implies E0 with a polynomial-size proof and xi = 1 implies\nE1 with a polynomial-size proof, then we can derive E0 ∨E1 with a polynomial-size proof”.\n4.2. Basic Counting inside R(lin) and R0(lin). In this subsection we illustrate how to effi-\nciently prove several basic counting arguments inside R(lin) and R0(lin). This will facilitate us in\nshowing short proofs for hard tautologies in the sequel. In accordance with the last paragraph in\nthe previous subsection, we shall carry the proofs inside R(lin) and R0(lin) with a slightly less rigor.\nLemma 6. Let z1 abbreviate ⃗a · ⃗x and z2 abbreviate ⃗b · ⃗x. Let D1 be W\nα∈A(z1 = α) and let D2 be\nW\nβ∈B (z2 = β), where A, B are two (finite) sets of integers. Then there is a polynomial-size (in the\nsize of D1, D2) R(lin) proof from D1, D2 of:\n_\nα∈A,β∈B\n(z1 + z2 = α + β) .\n(8)\nMoreover, if ⃗a and ⃗b consist of constant integers (which means that D1, D2 are R0(lin)-lines), then\nthere is a polynomial-size (in the size of D1, D2) R0(lin) proof of (8) from D1, D2.\nProof: Denote the elements of A by α1, . . . , αk. In case z1 = αi, for some i ∈[k] then we can add\nz1 = αi to every equation in W\nβ∈B (z2 = β) to get W\nβ∈B(z1 + z2 = αi + β). Therefore, there exist\nk R(lin) proofs, each with polynomial-size (in |D1| and |D2|), of\n_\nβ∈B\n(z1 + z2 = α1 + β) ,\n_\nβ∈B\n(z1 + z2 = α2 + β) ,\n. . .\n,\n_\nβ∈B\n(z1 + z2 = αk + β)\n5Note that when the proofs of Ei from z = ai, for all i ∈[l], are all done inside R0(lin), then the linear form z\nought to have constant coefficients.\n13"},{"page":14,"text":"from z1 = α1, z1 = α2 ,. . . ,z1 = αk, respectively.\nThus, by Lemma 4, we can derive\n_\nα∈A,β∈B\n(z1 + z2 = α + β)\n(9)\nfrom D1 and D2 in a polynomial-size (in |D1| and |D2|) R(lin)-proof. This concludes the first part\nof the lemma.\nAssume that ⃗a and ⃗b consist of constant coefficients only. Then by inspecting the R(lin)-proof\nof (9) from D1 and D2 demonstrated above (and by using Lemma 5 instead of Lemma 4), one can\nverify that this proof is in fact carried inside R0(lin).\nAn immediate corollary of Lemma 6 is the efficient formalization in R(lin) of the following obvious\ncounting argument: If a linear form equals some value in the interval (of integer numbers) [a0, a1]\nand another linear form equals some value in [b0, b1] (for some a0 ≤a1 and b0 ≤b1), then their\naddition equals some value in [a0 + b0, a1 + b1]. More formally:\nCorollary 7. Let z1 abbreviate ⃗a·⃗x and z2 abbreviate ⃗b·⃗x. Let D1 be (z1 = a0)∨(z1 = a0 +1) . . .∨\n(z1 = a1), and let D2 be (z2 = b0) ∨(z2 = b0 + 1) . . . ∨(z2 = b1). Then there is a polynomial-size\n(in the size of D1, D2) R(lin) proof from D1, D2 of\n(z1 + z2 = a0 + b0) ∨(z1 + z2 = a0 + b0 + 1) ∨. . . ∨(z1 + z2 = a1 + b1) .\n(10)\nMoreover, if ⃗a and ⃗b consist of constant integers (which means that D1, D2 are R0(lin)-lines), then\nthere is a polynomial-size (in the size of D1, D2) R0(lin) proofs of (10) from D1, D2.\nLemma 8. Let ⃗a · ⃗x be a linear form with n variables, and let A := {⃗a · ⃗x | ⃗x ∈{0, 1}n} be the set\nof all possible values of ⃗a ·⃗x over Boolean assignments to ⃗x. Then there is a polynomial-size, in the\nsize of the linear form ⃗a · ⃗x,6 R(lin) proof of\n_\nα∈A\n(⃗a · ⃗x = α) .\n(11)\nMoreover, if the coefficients in ⃗a are constants, then there is a polynomial-size (in the size of ⃗a · ⃗x)\nR0(lin) proof of (11).\nProof: Without loss of generality, assume that all the coefficients in ⃗a are nonzero. Consider the\nBoolean axiom (x1 = 0) ∨(x1 = 1) and the (first) coefficient a1 from ⃗a. Assume that a1 ≥1. Add\n(x1 = 0) to itself a1 times, and arrive at (a1x1 = 0) ∨(x1 = 1). Then, in the resulted line, add\n(x1 = 1) to itself a1 times, until the following is reached:\n(a1x1 = 0) ∨(a1x1 = a1) .\nSimilarly, in case a1 ≤−1 we can subtract (|a1| + 1 many times) (x1 = 0) from itself in (x1 =\n0) ∨(x1 = 1), and then subtract (|a1| + 1 many times) (x1 = 1) from itself in the resulted line.\nIn the same manner, we can derive the disjunctions: (a2x2 = 0) ∨(a2x2 = a2), . . . , (anxn =\n0) ∨(anxn = an).\nConsider (a1x1 = 0) ∨(a1x1 = a1) and (a2x2 = 0) ∨(a2x2 = a2). From these two lines, by\nLemma 6, there is a polynomial-size in |a1| + |a2| derivation of:\n(a1x1 + a2x2 = 0) ∨(a1x1 + a2x2 = a1) ∨(a1x1 + a2x2 = a2) ∨(a1x1 + a2x2 = a1 + a2) .\n(12)\nIn a similar fashion, now consider (a3x3 = 0) ∨(a3x3 = a3) and apply again Lemma 6, to obtain\n_\nα∈A′\n(a1x1 + a2x2 + a3x3 = α) ,\n(13)\n6Recall that the size of ⃗a · ⃗x is Pn\ni=1 |ai|, that is, the size of the unary representation of ⃗a.\n14"},{"page":15,"text":"where A′ are all possible values to a1x1 + a2x2 + a3x3 over Boolean assignments to x1, x2, x3. The\nderivation of (13) is of size polynomial in |a1| + |a2| + |a3|.\nContinue to consider, successively, all other lines (a4x4 = 0) ∨(a4x4 = a4), . . . , (anxn = 0) ∨\n(anxn = an), and apply the same reasoning. Each step uses a derivation of size at most polynomial\nin Pn\ni=1 |ai|. And so overall we reach the desired line (11), with a derivation of size polynomial in\nthe size of ⃗a · ⃗x. This concludes the first part of the lemma.\nAssume that ⃗a consists of constant coefficients only. Then by inspecting the R(lin)-proof demon-\nstrated above (and by using the second part of Lemma 6), one can see that this proof is in fact\ncarried inside R0(lin).\nLemma 9. There is a polynomial-size (in n) R0(lin) proof from\n(x1 = 1) ∨· · · ∨(xn = 1)\n(14)\nof\n(x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n) .\n(15)\nProof: We show that for every i ∈[n], there is a polynomial-size (in n) R0(lin) proof from (xi = 1)\nof (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n). This concludes the proof since, by Lemma 5,\nwe then can derive from (14) (with a polynomial-size (in n) R0(lin) proof) the disjunction (14) in\nwhich each (xi = 1) (for all i ∈[n]) is replace by (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n),\nwhich is precisely the disjunction (15) (note that (15) is an R0(lin)-line).\nClaim 2. For every i ∈[n], there is a a polynomial-size (in n) R0(lin) proof from (xi = 1) of\n(x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n).\nProof of claim: By Lemma 8, for every i ∈[n] there is a polynomial-size (in n) R0(lin) proof\n(using only the Boolean axioms) of\n(x1 + . . . + xi−1 + xi+1 + . . . + xn = 0) ∨· · · ∨(x1 + . . . + xi−1 + xi+1 + . . . + xn = n −1) .\n(16)\nNow add successively (xi = 1) to every equation in (16) (note that this can be done in R0(lin)).\nWe obtain precisely (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n).\nLemma 10. There is a polynomial-size (in n) R0(lin) proof of (x1+. . .+xn = 0)∨(x1+. . .+xn = 1)\nfrom the collection of disjunctions consisting of (xi = 0) ∨(xj = 0), for all 1 ≤i < j ≤n.\nProof: We proceed by induction on n. The base case for n = 1 is immediate from the Boolean\naxiom (x1 = 0) ∨(x1 = 1). Assume we already have a polynomial-size proof of\n(x1 + . . . + xn = 0) ∨(x1 + . . . + xn = 1).\n(17)\nIf xn+1 = 0 we add xn+1 = 0 to both of the equations in (17), and reach:\n(x1 + . . . + xn+1 = 0) ∨(x1 + . . . + xn+1 = 1).\n(18)\nOtherwise, xn+1 = 1, and so we can cut-off(xn+1 = 0) in all the initial disjunctions (xi = 0) ∨\n(xn+1 = 0), for all 1 ≤i ≤n. We thus obtain (x1 = 0), . . . , (xn = 0). Adding together (x1 =\n0), . . . , (xn = 0) and (xn+1 = 1) we arrive at\n(x1 + . . . + xn+1 = 1) .\n(19)\nSo overall, either (18) holds or (19) holds; and so (using Lemma 5) we arrive at the disjunction of\n(19) and (18), which is precisely (18).\n15"},{"page":16,"text":"5. Implicational Completeness of R(lin) and its Subsystems\nIn this section we provide a proof of the implicational completeness of R(lin) and its subsystems.\nWe shall need this property in the sequel (see Section 6.2). The implicational completeness of a\nproof system is a stronger property than mere completeness. Essentially, a system is implicationally\ncomplete if whenever something is semantically implied by a set of initial premises, then it is\nalso derivable from the initial premises. In contrast to this, mere completeness means that any\ntautology (or in case of a refutation system, any unsatisfiable set of initial premises) has a proof in\nthe system (respectively, a refutation in the system). As a consequence, the proof of implicational\ncompleteness in this section establishes an alternative completeness proof to that obtained via\nsimulating resolution (see Proposition 1). Note that we are not concerned in this section with the\nsize of the proofs, but only with their existence.\nRecall the definition of the semantic implication relation |= from Section 3.1. Formally, we say\nthat R(lin) is implicationally complete if for every collection of disjunctions of linear equations\nD0, D1, . . . , Dm, it holds that D1, . . . , Dm |= D0 implies that there is an R(lin) proof of D0 from\nD1, . . . , Dm.\nTheorem 11. R(lin) is implicationally complete.\nProof: We proceed by induction on n, the number of variables x1, . . . , xn in D0, D1, . . . , Dm.\nThe base case n = 0. We need to show that D1, . . . , Dm |= D0 implies that there is an R(lin)\nproof of D0 from D1, . . . , Dm, where all Di’s (for 0 ≤i ≤m) have no variables but only constants.\nThis means that each Di is a disjunction of equations of the form (0 = a0) for some integer a0 (if\na linear equation have no variables, then the left hand side of this equation must be 0; see Section\n3.1).\nThere are two cases to consider. In the first case D0 is satisfiable. Since D0 has no variables,\nthis means precisely that D0 is the equation (0 = 0). Thus, D0 can be derived easily from any\naxiom in R(lin) (for instance, by subtracting each equation in (x1 = 0) ∨(x1 = 1) from itself, to\nreach (0 = 0) ∨(0 = 0), which is equal to (0 = 0), since we discard duplicate equations inside\ndisjunctions).\nIn the second case D0 is unsatisfiable. Thus, since D1, . . . , Dm |= D0, there is no assignment sat-\nisfying all D1, . . . , Dm. Hence, there must be at least one unsatisfiable disjunction Di in D1, . . . , Dm\n(as a disjunction with no variables is either tautological or unsatisfiable). Such an unsatisfiable Di\nis a disjunction of zero or more unsatisfiable equations of the form (0 = a0), for some integer a0 ̸= 0.\nWe can then use Simplification to cut-offall the unsatisfiable equations in Di to reach the empty\ndisjunction. By the Weakening rule, we can now derive D0 from the empty disjunction.\nThe induction step. Assume that the theorem holds for disjunctions with n variables. Let the\nunderlying variables of D0, D1, . . . , Dm be x1, . . . , xn+1, and assume that\nD1, . . . , Dm |= D0 .\n(20)\nWe write the disjunction D0 as:\nt_\nj=1\n n\nX\ni=1\na(j)\ni xi + a(j)\nn+1xn+1 = a(j)\n0\n!\n,\n(21)\nwhere the a(j)\ni ’s are integer coefficients. We need to show that there is an R(lin) proof of D0 from\nD1, . . . , Dm.\nLet D be a disjunction of linear equations, let xi be a variable and let b ∈{0, 1}. We shall\ndenote by D↾xi=b the disjunction D, where in every equation in D the variable xi is substituted by\nb, and the constant terms in the left hand sides of all resulting equations (after substituting b for\n16"},{"page":17,"text":"xi) switch sides (and change signs, obviously) to the right hand sides of the equations (we have to\nswitch sides of constant terms, as by definition linear equations in R(lin) proofs have all constant\nterms appearing only on the right hand sides of equations).\nWe now reason (slightly) informally inside R(lin) (as illustrated in Section 4.1). Fix some b ∈\n{0, 1}, and assume that xn+1 = b. Then, from D1, . . . , Dm we can derive (inside R(lin)):\nD1↾xn+1=b, . . . , Dm↾xn+1=b .\n(22)\nThe only variables occurring in (22) are x1, . . . , xn. From assumption (20) we clearly have D1↾xn+1=b\n, . . . , Dm↾xn+1=b |= D0↾xn+1=b. And so by the induction hypothesis there is an R(lin) derivation of\nD0↾xn+1=b from D1↾xn+1=b, . . . , Dm↾xn+1=b. So overall, assuming that xn+1 = b, there is an R(lin)\nderivation of D0↾xn+1=b from D1, . . . , Dm.\nWe now consider the two possible cases: xn+1 = 0 and xn+1 = 1.\nIn case xn+1 = 0, by the above discussion, we can derive D0↾xn+1=0 from D1, . . . , Dm. For every\nj ∈[t], add successively (a(j)\nn+1 times) the equation xn+1 = 0 to the jth equation in D0↾xn+1=0 (see\n(21)). We thus obtain precisely D0.\nIn case xn+1 = 1, again, by the above discussion, we can derive D0↾xn+1=1 from D1, . . . , Dm. For\nevery j ∈[t], add successively (a(j)\nn+1 times) the equation xn+1 = 1 to the jth equation in D0↾xn+1=1\n(recall that we switch sides of constant terms in every linear equation after the substitution of xn+1\nby 1 is performed in D0↾xn+1=1). Again, we obtain precisely D0.\nBy inspecting the proof of Theorem 11, it is possible to verify that if all the disjunctions\nD0, , . . . , Dm are R0(lin)-lines (see Definition 3.2), then the proof of D0 in R(lin) uses only R0(lin)-\nlines as well. Therefore, we have:\nCorollary 12. R0(lin) is implicationally complete.\nRemark 1. Corollary 12 states that any R0(lin)-line that is semantically implied by a set of initial\nR0(lin)-lines, is in fact derivable in R0(lin) from the initial R0(lin)-lines. On the other hand, it is\npossible that a certain proof of the same R0(lin)-line inside R(lin) will be significantly shorter than\nthe proof inside R0(lin). Indeed, we shall see in Section 8 that for certain CNF formulas R(lin) has\na super-polynomial speed-up over R0(lin).\n6. Short Proofs for Hard Tautologies\nIn this section we show that R0(lin) is already enough to admit small proofs for “hard” counting\nprinciples like the pigeonhole principle and the Tseitin graph formulas for constant degree graphs.\nOn the other hand, as we shall see in Section 8, R0(lin) inherits the same weakness that cutting\nplanes proofs have with respect to the clique-coloring tautologies. Nevertheless, we can efficiently\nprove the clique-coloring principle in (the stronger system) R(lin), but not by using R(lin) “ability\nto count”, rather by using its (straightforward) ability to simulate Res(2) proofs (that is, resolution\nproofs extended to operate with 2-DNF formulas, instead of clauses).\n6.1. The\nPigeonhole\nPrinciple\nTautologies\nin\nR0(lin). This\nsubsection\nillustrates\npolynomial-size R0(lin) proofs of the pigeonhole principle.\nThis will allow us to establish\npolynomial-size multilinear proofs operating with depth-3 multilinear formulas of the pigeonhole\nprinciple (in Section 9).\nThe m to n pigeonhole principle states that m pigeons cannot be mapped one-to-one into n < m\nholes. The negation of the pigeonhole principle, denoted ¬PHPm\nn , is formulated as an unsatisfiable\nCNF formula as follows (where clauses are translated to disjunctions of linear equations):\nDefinition 6.1. The ¬PHPm\nn is the following set of clauses:\n(1) Pigeons axioms:\n(xi,1 = 1) ∨· · · ∨(xi,n = 1), for all 1 ≤i ≤m;\n17"},{"page":18,"text":"(2) Holes axioms:\n(xi,k = 0) ∨(xj,k = 0),\nfor all 1 ≤i < j ≤m and for all 1 ≤k ≤n.\nThe intended meaning of each propositional variable xi,j is that the ith pigeon is mapped to the\njth hole.\nWe now describe a polynomial-size in n refutation of ¬PHPm\nn inside R0(lin). For this purpose it\nis sufficient to prove a polynomial-size refutation of the pigeonhole principle when the number of\npigeons m equals n + 1 (because the set of clauses pertaining to ¬PHPn+1\nn\nis already contained in\nthe set of clauses pertaining to ¬PHPm\nn , for any m > n). Thus, we fix m = n+1. In this subsection\nwe shall say a proof in R0(lin) is of polynomial-size, always intending polynomial-size in n (unless\notherwise stated).\nBy Lemma 9, for all i ∈[m] we can derive from the Pigeon axiom (for the ith pigeon):\n(xi,1 + . . . + xi,n = 1) ∨· · · ∨(xi,1 + . . . + xi,n = n)\n(23)\nwith a polynomial-size R0(lin) proof.\nBy Lemma 10, from the Hole axioms we can derive, with a polynomial-size R0(lin) proof\n(x1,j + . . . + xm,j = 0) ∨(x1,j + . . . + xm,j = 1),\n(24)\nfor all j ∈[n].\nLet S abbreviate the sum of all formal variables xi,j. In other words,\nS :=\nX\ni∈[m],j∈[n]\nxi,j .\nLemma 13. There is a polynomial-size R0(lin) proof from (23) (for all i ∈[m]) of\n(S = m) ∨(S = m + 1) · · · ∨(S = m · n).\nProof: For every i ∈[m] fix the abbreviation zi := xi,1 + . . . + xi,n.\nThus, by (23) we have\n(zi = 1) ∨· · · ∨(zi = n).\nConsider (z1 = 1) ∨· · · ∨(z1 = n) and (z2 = 1) ∨· · · ∨(z2 = n). By Corollary 7, we can derive\nfrom these two lines\n(z1 + z2 = 2) ∨(z1 + z2 = 3) ∨· · · ∨(z1 + z2 = 2n)\n(25)\nwith a polynomial-size R0(lin) proof.\nNow, consider (z3 = 1) ∨· · · ∨(z3 = n) and (25). By Corollary 7 again, from these two lines we\ncan derive with a polynomial-size R0(lin) proof:\n(z1 + z2 + z3 = 3) ∨(z1 + z2 + z3 = 4) ∨· · · ∨(z1 + z2 + z3 = 3n) .\n(26)\nContinuing in the same way, we eventually arrive at\n(z1 + . . . + zm = m) ∨(z1 + . . . + zm = m + 1) ∨· · · ∨(z1 + . . . + zm = m · n) ,\nwhich concludes the proof, since S equals z1 + . . . + zm.\nLemma 14. There is a polynomial-size R0(lin) proof from (24) of\n(S = 0) ∨· · · ∨(S = n).\nProof: For all j ∈[n], fix the abbreviation yj := x1,j + . . . + xm,j.\nThus, by (24) we have\n(yj = 0) ∨(yj = 1), for all j ∈[n]. Now the proof is similar to the proof of Lemma 8, except that\nhere single variables are abbreviations of linear forms.\nIf y1 = 0 then we can add y1 to the two sums in (y2 = 0) ∨(y2 = 1), and reach (y1 + y2 =\n0) ∨(y1 + y2 = 1) and if y1 = 1 we can do the same and reach (y1 + y2 = 1) ∨(y1 + y2 = 2). So, by\nLemma 5, we can derive with a polynomial-size R0(lin) proof\n(y1 + y2 = 0) ∨(y1 + y2 = 1) ∨(y1 + y2 = 2) .\n(27)\n18"},{"page":19,"text":"Now, we consider the three cases in (27): y1 +y2 = 0 or y1 +y2 = 1 or y1 +y2 = 2, and the clause\n(y3 = 0) ∨(y3 = 1). We arrive in a similar manner at (y1 + y2 + y3 = 0) ∨· · · ∨(y1 + y2 + y3 = 3).\nWe continue in the same way until we arrive at (S = 0) ∨· · · ∨(S = n).\nTheorem 15. There is a polynomial-size R0(lin) refutation of the m to n pigeonhole principle\n¬PHPm\nn .\nProof: By Lemmas 13 and 14 above, all we need is to show a polynomial-size refutation of (S =\nm) ∨· · · ∨(S = m · n) and (S = 0) ∨· · · ∨(S = n).\nSince n < m, for all 0 ≤k ≤n, if S = k then using the Resolution and Simplification rules we\ncan cut-offall the sums in (S = m) ∨· · · ∨(S = m · n) and arrive at the empty clause. Thus, by\nLemma 5, there is a polynomial-size R0(lin) proof of the empty clause from (S = 0) ∨· · · ∨(S = n)\nand (S = m) ∨· · · ∨(S = m · n).\n6.2. Tseitin mod p Tautologies in R0(lin). This subsection establishes polynomial-size R0(lin)\nproofs of Tseitin graph tautologies (for constant degree graphs). This will allow us (in Section 9)\nto extend the multilinear proofs of the Tseitin mod p tautologies to any field of characteristic 0\n(the proofs in [RT06] required working over a field containing a primitive pth root of unity when\nproving the Tseitin mod p tautologies; for more details see Section 9).\nTseitin mod p tautologies (introduced in [BGIP01]) are generalizations of the (original, mod 2)\nTseitin graph tautologies (introduced in [Tse68]). To build the intuition for the generalized version,\nwe start by describing the (original) Tseitin mod 2 principle.\nLet G = (V, E) be a connected\nundirected graph with an odd number of vertices n. The Tseitin mod 2 tautology states that there\nis no sub-graph G′ = (V, E′), where E′ ⊆E, so that for every vertex v ∈V , the number of edges\nfrom E′ incident to v is odd. This statement is valid, since otherwise, summing the degrees of all\nthe vertices in G′ would amount to an odd number (since n is odd), whereas this sum also counts\nevery edge in E′ twice, and so is even.\nAs mentioned above, the Tseitin mod 2 principle was generalized by Buss et al. [BGIP01] to\nobtain the Tseitin mod p principle. Let p ≥2 be some fixed integer and let G = (V, E) be a\nconnected undirected r-regular graph with n vertices and no double edges. Let G′ = (V, E′) be the\ncorresponding directed graph that results from G by replacing every (undirected) edge in G with\ntwo opposite directed edges. Assume that n ≡1 (mod p). Then, the Tseitin mod p principle states\nthat there is no way to assign to every edge in E′ a value from {0, . . . , p −1}, so that:\n(i): For every pair of opposite directed edges e, ̄e in E′, with assigned values a, b, respectively,\na + b ≡0 (mod p); and\n(ii): For every vertex v in V , the sum of the values assigned to the edges in E′ coming out of\nv is congruent to 1 (mod p).\nThe Tseitin mod p principle is valid, since if we sum the values assigned to all edges of E′ in\npairs we obtain 0 (mod p) (by (i)), where summing them by vertices we arrive at a total value of 1\n(mod p) (by (ii) and since n ≡1 (mod p)). We shall see in what follows, that this simple counting\nargument can be carried on in a natural (and efficient) way already inside R0(lin).\nAs an unsatisfiable propositional formula (in CNF form) the negation of the Tseitin mod p\nprinciple is formulated by assigning a variable xe,i for every edge e ∈E′ and every residue i modulo\np. The variable xe,i is an indicator variable for the fact that the edge e has an associated value i.\nThe following are the clauses of the Tseitin mod p CNF formula (as translated to disjunctions of\nlinear equations).\nDefinition 6.2 (Tseitin mod p formulas (¬TseitinG,p)). Let p ≥2 be some fixed integer and\nlet G = (V, E) be a connected undirected r-regular graph with n vertices and no double edges, and\n19"},{"page":20,"text":"assume that n ≡1 (mod p). Let G′ = (V, E′) be the corresponding directed graph that results\nfrom G by replacing every (undirected) edge in G with two opposite directed edges.\nGiven a vertex v ∈V , denote the edges in E′ coming out of v by e[v, 1], . . . , e[v, r] and define the\nfollowing set of (translation of) clauses:\nMODp,1(v):=\n( r_\nk=1\n(xe[v,k],ik = 0)\n i1, . . . , ir ∈{0, . . . , p −1} and\nr\nX\nk=1\nik ̸≡1 mod p\n)\n.\nThe Tseitin mod p formula, denoted ¬TseitinG,p, consists of the following (translation) of clauses:\n1.\np−1\nW\ni=0\n(xe,i = 1) , for all e ∈E′\n(expresses that every edge is assigned at least one value from 0, . . . , p −1);\n2. (xe,i = 0) ∨(xe,j = 0) , for all i ̸= j ∈{0, . . . , p −1} and all e ∈E′\n(expresses that every edge is assigned at most one value from 0, . . . , p −1);\n3. (xe,i = 1) ∨(x ̄e,p−i = 0) and (xe,i = 0) ∨(x ̄e,p−i = 1), 7\nfor all two opposite directed edges e, ̄e ∈E′ and all i ∈{0, . . . , p −1}\n(expresses condition (i) of the Tseitin mod p principle above);\n4. MODp,1(v) , for all v ∈V\n(expresses condition (ii) of the Tseitin mod p principle above).\nNote that for every edge e ∈E′, the polynomials of (1,2) in Definition 6.2, combined with the\nBoolean axioms of R0(lin), force any collection of edge-variables xe,0, . . . , xe,p−1 to contain exactly\none i ∈{0, . . . , p −1} so that xe,i = 1.\nAlso, it is easy to verify that, given a vertex v ∈V ,\nany assignment σ of 0, 1 values (to the relevant variables) satisfies both the disjunctions of (1,2)\nand the disjunctions of MODp,1(v) if and only if σ corresponds to an assignment of values from\n{0, . . . , p −1} to the edges coming out of v that sums up to 1 (mod p).\nUntil the rest of this subsection we fix an integer p ≥2 and a connected undirected r-regular\ngraph G = (V, E) with n vertices and no double edges, such that n ≡1 mod p and r is a constant.\nAs in Definition 6.2, we let G′ = (V, E′) be the corresponding directed graph that results from G\nby replacing every (undirected) edge in G with two opposite directed edges. We now proceed to\nrefute ¬TseitinG,p inside R0(lin) with a polynomial-size (in n) refutation.\nGiven a vertex v ∈V , and the edges in E′ coming out of v, denoted e[v, 1], . . . , e[v, r], define the\nfollowing abbreviation:\nαv :=\nr\nX\nj=1\np−1\nX\ni=0\ni · xe[v,j],i .\n(28)\nLemma 16. Let v ∈V be any vertex in G′. Then there is a constant-size R0(lin) proof from\n¬TseitinG,p of the following disjunction:\nr−1\n_\nl=0\n(αv = 1 + l· p) .\n(29)\nProof: Let Tv ⊆¬TseitinG,p be the set of all disjunctions of the form (1,2,4) from Definition 6.2\nthat contain only variables pertaining to vertex v (that is, all the variables xe,i, where e ∈E′ is an\nedge coming out of v, and i ∈{0, . . . , p −1}).\n7If i = 0 then x ̄e,p−i denotes x ̄e,0.\n20"},{"page":21,"text":"Claim 3. Tv semantically implies (29), that is:8\nTv |=\nr−1\n_\nl=0\n(αv = 1 + l· p) .\nProof of claim: Let σ be an assignment of 0, 1 values to the variables in Tv that satisfies both the\ndisjunctions of (1,2) and the disjunctions of MODp,1(v) in Definition 6.2. As mentioned above (the\ncomment after Definition 6.2), such a σ corresponds to an assignment of values from {0, . . . , p −1}\nto the edges coming out of v, that sums up to 1 mod p. This means precisely that αv = 1 mod p\nunder the assignment σ. Thus, there exists a nonnegative integer k, such that αv = 1 + kp under\nσ.\nIt remains to show that k ≤r −1 (and so the only possible values that αv can get under σ\nare 1, 1 + p, 1 + 2p, . . . , 1 + (r −1)p). Note that because σ gives the value 1 to only one variable\nfrom xe[v,j],0, . . . , xe[v,j],p−1 (for every j ∈[r]), then the maximal value that αv can have under σ is\nr(p −1). Thus, 1 + kp ≤rp −r and so k ≤r −1.\nFrom Claim 3 and from the implicational completeness of R0(lin) (Corollary 12), there exists an\nR0(lin) derivation of (29) from Tv. It remains to show that this derivation is of constant-size.\nSince the degree r of G′ and the modulus p are both constants, both Tv and (29) have constant\nnumber of variables and constant coefficients (including the free-terms). Thus, there is a constant-\nsize R0(lin) derivation of (29) from Tv.\nLemma 17. There is a polynomial-size (in n) R0(lin) derivation from ¬TseitinG,p of the following\ndisjunction:\n(r−1)·n\n_\nl=0\n X\nv∈V\nαv = n + l· p\n!\n.\nProof: Simply add successively all the equations pertaining to disjunctions (29), for all vertices\nv ∈V .\nFormally, we show that for every subset of vertices V ⊆V , with |V| = k, there is a\npolynomial-size (in n) R0(lin) derivation from ¬TseitinG,p of\n(r−1)·k\n_\nl=0\n X\nv∈V\nαv = k + l· p\n!\n,\n(30)\nand so putting V = V , will conclude the proof.\nWe proceed by induction on the size of V. The base case, |V| = 1, is immediate from Lemma 16.\nAssume that we already derived (30) with a polynomial-size (in n) R0(lin) proof, for some V ⊂V ,\nsuch that |V| = k < n. Let u ∈V \\ V. By Lemma 16, we can derive\nr−1\n_\nl=0\n(αu = 1 + l· p)\n(31)\nfrom ¬TseitinG,p with a constant-size proof. Now, by Lemma 6, each linear equation in (31) can\nbe added to each linear equation in (30), with a polynomial-size (in n) R0(lin) proof. This results\nin the following disjunction:\n(r−1)·(k+1)\n_\nl=0\n \n \nX\nv∈V∪{u}\nαv = k + 1 + l· p\n \n ,\n8Recall that we only consider assignments of 0, 1 values to variables when considering the semantic implication\nrelation |=.\n21"},{"page":22,"text":"which is precisely what we need to conclude the induction step.\nLemma 18. Let e, ̄e be any pair of opposite directed edges in G′ and let i ∈{0, . . . , p −1}. Let\nTe ⊆¬TseitinG,p be the set of all disjunctions of the form (1,2,3) from Definition 6.2 that contain\nonly variables pertaining to edges e, ̄e (that is, all the variables xe,j, x ̄e,j, for all j ∈{0, . . . , p −1}).\nThen, there is a constant-size R0(lin) proof from Te of the following disjunction:\n(i · xe,i + (p −i) · x ̄e,p−i = 0) ∨(i · xe,i + (p −i) · x ̄e,p−i = p) .\n(32)\nProof: First note that Te semantically implies\n(xe,i + x ̄e,p−i = 0) ∨(xe,i + x ̄e,p−i = 2) .\n(33)\nThe number of variables in Te and (33) is constant. Hence, there is a constant-size R0(lin)-proof\nof (32) from Te. Also note that\n(xe,i + x ̄e,p−i = 0) ∨(xe,i + x ̄e,p−i = 2) |=\n(i · xe,i + (p −i) · x ̄e,p−i = 0) ∨(i · xe,i + (p −i) · x ̄e,p−i = p) .\n(34)\nTherefore, there is also an R0(lin)-proof of constant-size from Te of the lower line in (34).\nWe are now ready to complete the polynomial-size R0(lin) refutation of ¬TseitinG,p. Using the\ntwo prior lemmas, the refutation idea is simple, as we now explain. Observe that\nX\nv∈V\nαv =\nX\n{e, ̄e}⊆E′\ni∈{0,...,p−1}\n(i · xe,i + (p −i) · x ̄e,p−i) ,\n(35)\nwhere by {e, ̄e} ⊆E′ we mean that e, ̄e is pair of opposite directed edges in G′.\nDerive by Lemma 17 the disjunction\n(r−1)·n\n_\nl=0\n X\nv∈V\nαv = n + l· p\n!\n.\n(36)\nThis disjunction expresses the fact that P\nv∈V αv = 1 mod p (since n = 1 mod p). On the other\nhand, using Lemma 18, we can “sum together” all the equations (32) (for all {e, ̄e} ⊆E′ and all\ni ∈{0, . . . , p −1}), to obtain a disjunction expressing the statement that\nX\n{e, ̄e}⊆E′\ni∈{0,...,p−1}\n(i · xe,i + (p −i) · x ̄e,p−i) = 0\nmod p .\nBy Equation (35), we then obtain the desired contradiction. This idea is formalized in the proof of\nthe following theorem:\nTheorem 19. Let G = (V, E) be an r-regular graph with n vertices, where r is a constant. Fix\nsome modulus p. Then, there are polynomial-size (in n) R0(lin) refutations of ¬TseitinG,p.\nProof: First, use Lemma 17 to derive\n(r−1)·n\n_\nl=0\n X\nv∈V\nαv = n + l· p\n!\n.\n(37)\nSecond, use Lemma 18 to derive\n(i · xe,i + (p −i) · x ̄e,p−i = p) ∨(i · xe,i + (p −i) · x ̄e,p−i = 0) ,\n(38)\nfor every pair of opposite directed edges in G′ = (V, E′) (as in Definition 6.2) and every residue\ni ∈{0, . . . , p −1}.\n22"},{"page":23,"text":"We now reason inside R0(lin).\nPick a pair of opposite directed edges e, ̄e and a residue i ∈\n{0, . . . , p −1}. If i · xe,i + (p −i) · x ̄e,p−i = 0, then subtract this equation successively from every\nequation in (37). We thus obtain a new disjunction, similar to that of (37), but which does not\ncontain the xe,i and x ̄e,p−i variables, and with the same free-terms.\nOtherwise, i·xe,i+(p−i)·x ̄e,p−i = p, then subtract this equation successively from every equation\nin (37). Again, we obtain a new disjunction, similar to that of (37), but which does not contain\nthe xe,i and x ̄e,p−i variables, and such that p is subtracted from every free-term in every equation.\nSince, by assumption, n ≡1 mod p, the free-terms in every equation are (still) equal 1 mod p.\nSo overall, in both cases (i · xe,i + (p −i) · x ̄e,p−i = 0 and i · xe,i + (p −i) · x ̄e,p−i = p) we obtained\na new disjunction with all the free-terms in equations equal 1 mod p.\nWe now continue the same process for every pair e, ̄e of opposite directed edges in G′ and\nevery residue i. Eventually, we discard all the variables xe,i in the equations, for every e ∈E′\nand i ∈{0, . . . , p −1}, while all the free-terms in every equation remain to be equal 1 mod p.\nTherefore, we arrive at a disjunction of equations of the form (0 = γ) for some γ = 1 mod p.\nBy using the Simplification rule we can cut-offall such equations, and arrive finally at the empty\ndisjunction.\n6.3. The Clique-Coloring Principle in R(lin). In this section we observe that there are\npolynomial-size R(lin) proofs of the clique-coloring principle (for certain, weak, parameters). This\nimplies, in particular, that R(lin) does not possess the feasible monotone interpolation property\n(see more details on the interpolation method in Section 7).\nAtserias, Bonet & Esteban [ABE02] demonstrated polynomial-size Res(2) refutations of the\nclique-coloring formulas (for certain weak parameters; Theorem 20). Thus, it is sufficient to show\nthat R(lin) polynomially-simulates Res(2) proofs (Proposition 2). This can be shown in a straight-\nforward manner. As noted in the first paragraph of Section 6, because the proofs of the clique-\ncoloring formula we discuss here only follow the proofs inside Res(2), then in fact these proofs do\nnot take any advantage of the capacity “to count” inside R(lin) (this capacity is exemplified, for\ninstance, in Section 4.2).\nWe start with the clique-coloring formulas (these formulas will also be used in Section 8). These\nformulas express the clique-coloring principle that has been widely used in the proof complexity\nliterature (cf., [BPR97], [Pud97], [Kra97], [Kra98], [ABE02], [Kra07]). This principle is based on\nthe following basic combinatorial idea. Let G = (V, E) be an undirected graph with n vertices and\nlet k′ < k be two integers. Then, one of the following must hold:\n(i): The graph G does not contain a clique with k vertices;\n(ii): The graph G is not a complete k′-partite graph.\nIn other words, there is no way to\npartition G into k′ subgraphs G1, . . . , Gk′, such that every Gi is an independent set, and\nfor all i ̸= j ∈[k′], all the vertices in Gi are connected by edges (in E) to all the vertices in\nGj.\nObviously, if Item (ii) above is false (that is, if G is a complete k′-partite graph), then there\nexists a k′-coloring of the vertices of G; hence the name clique-coloring for the principle.\nThe propositional formulation of the (negation of the) clique-coloring principle is as follows.\nEach variable pi,j, for all i ̸= j ∈[n], is an indicator variable for the fact that there is an edge in\nG between vertex i and vertex j. Each variable ql,i, for all l∈[k] and all i ∈[n], is an indicator\nvariable for the fact that the vertex i in G is the lth vertex in the k-clique. Each variable rl,i, for\nall l∈[k′] and all i ∈[n], is an indicator variable for the fact that the vertex i in G pertains to the\nindependent set Gl.\nDefinition 6.3. The negation of the clique-coloring principle consists of the following unsatisfiable\ncollection of clauses (as translated to disjunctions of linear equations), denoted ¬cliquen\nk,k′:\n23"},{"page":24,"text":"(i) (ql,1 = 1) ∨· · · ∨(ql,n = 1), for all l∈[k]\n(expresses that there exists at least one vertex in G which constitutes the lth vertex of\nthe k-clique);\n(ii) (ql,i = 0) ∨(ql,j = 0), for all i ̸= j ∈[n], l∈[k]\n(expresses that there exists at most one vertex in G which constitutes the lth vertex of\nthe k-clique);\n(iii) (ql,i = 0) ∨(ql′,i = 0), for all i ∈[n], l̸= l′ ∈[k]\n(expresses that the ith vertex of G cannot be both the lth and the l′th vertex of the\nk-clique);\n(iv) (ql,i = 0) ∨(ql′,j = 0) ∨(pi,j = 1), for all l̸= l′ ∈[k], i ̸= j ∈[n]\n(expresses that if both the vertices i and j in G are in the k-clique, then there is an edge\nin G between i and j);\n(v) (r1,i = 1) ∨· · · ∨(rk′,i = 1), for all i ∈[n]\n(expresses that every vertex of G pertains to at least one independent set);\n(vi) (rl,i = 0) ∨(rl′,i = 0), for all l̸= l∈[k′], i ∈[n]\n(expresses that every vertex of G pertains to at most one independent set);\n(vii) (pi,j = 0) ∨(rt,i = 0) ∨(rt,j = 0), for all i ̸= j ∈[n], t ∈[k′]\n(expresses that if there is an edge between vertex i and j in G, then i and j cannot be\nin the same independent set);\nRemark 2. Our formulation of the clique-coloring formulas above is similar to the one used by\n[BPR97], except that we consider also the pi,j variables (we added the (iv) clauses and changed\naccordingly the (vii) clauses). This is done for the sake of clarity of the contradiction itself, and\nalso to make it clear that the formulas are in the appropriate form required by the interpolation\nmethod (see Section 7 for details on the interpolation method). By resolving over the pi,j variables\nin (iv) and (vii), one can obtain precisely the collection of clauses in [BPR97].\nAtserias, Bonet & Esteban [ABE02] demonstrated polynomial-size (in n) Res(2) refutations of\n¬cliquen\nk,k′, when k = √n and k′ = (log n)2/8 log log n. These are rather weak parameters, but\nthey suffice to establish the fact that Res(2) does not possess the feasible monotone interpolation\nproperty.\nThe Res(2) proof system (also called 2-DNF resolution), first considered in [Kra01], is resolution\nextended to operate with 2-DNF formulas, defined as follows.\nA 2-term is a conjunction of up to two literals. A 2-DNF is a disjunction of 2-terms. The size\nof a 2-term is the number of literals in it (that is, either 1 or 2). The size of a 2-DNF is the total\nsize of all the 2-terms in it.\nDefinition 6.4 (Res(2)). A Res(2) proof of a 2-DNF D from a collection K of 2-DNFs is a\nsequence of 2-DNFs D1, D2, . . . , Ds , such that Ds = D, and every Dj is either from K or was\nderived from previous line(s) in the sequence by the following inference rules:\nCut: Let A, B be two 2-DNFs.\nFrom A∨V2\ni=1 li and B∨W2\ni=1 ¬li derive A∨B, where the li’s are (not necessarily distinct)\nliterals (and ¬li is the negation of the literal li).\nAND-introduction: Let A, B be two 2-DNFs and l1, l2 two literals.\nFrom A ∨l1 and B ∨l2 derive A ∨B ∨V2\ni=1 li.\nWeakening: From a 2-DNF A derive A ∨V2\ni=1 li , where the li’s are (not necessarily dis-\ntinct) literals.\nA Res(2) refutation of a collection of 2-DNFs K is a Res(2) proof of the empty disjunction ✷from\nK (the empty disjunction stands for false). The size of a Res(2) proof is the total size of all the\n2-DNFs in it.\n24"},{"page":25,"text":"Given a collection K of 2-DNFs we translate it into a collection of disjunctions of linear equations\nvia the following translation scheme. For a literal l, denote by bl the translation that maps a variable\nxi into xi, and ¬xi into 1−xi. A 2-term l1 ∧l2 is first transformed into the equation bl1 +bl2 = 2, and\nthen moving the free-terms in the left hand side of bl1 + bl2 = 2 (in case there are such free-terms)\nto the right hand side; So that the final translation of l1 ∧l2 has only a single free-term in the\nright hand side. A disjunction of 2-terms (that is, a 2-DNF) D = W\ni∈I(li,1 ∧li,2) is translated into\nthe disjunction of the translations of the 2-terms, denoted by bD. It is clear that every assignment\nsatisfies a 2-DNF D if and only if it satisfies bD.\nProposition 2. R(lin) polynomially simulates Res(2). In other words, if π is a Res(2) proof of D\nfrom a collection of 2-DNFs K1, . . . , Kt, then there is an R(lin) proof of bD from bK1, . . . , bKt whose\nsize is polynomial in the size of π.\nThe proof of Proposition 2 proceeds by induction on the length (that is, the number of proof-\nlines) in the Res(2) proof. This is pretty straightforward and similar to the simulation of resolution\nby R(lin), as illustrated in the proof of Proposition 1. We omit the details.\nTheorem 20 ([ABE02]). Let k = √n and k′ = (log n)2/8 log log n. Then ¬cliquen\nk,k′ has Res(2)\nrefutations of size polynomial in n.\nThus, Proposition 2 yields the following:\nCorollary 21. Let k, k′ be as in Theorem 20. Then ¬cliquen\nk,k′ has R(lin) refutations of size\npolynomial in n.\nThe following corollary is important (we refer the reader to Section A in the Appendix for the\nnecessary relevant definitions concerning the feasible monotone interpolation property and to Section\n7 for explanation and definitions concerning the general [non-monotone] interpolation method).\nCorollary 22. R(lin) does not possess the feasible monotone interpolation property.\nRemark 3. The proof of ¬cliquen\nk,k′ inside Res(2) demonstrated in [ABE02] (and hence, also\nthe corresponding proof inside R(lin)) proceeds along the following lines. First reduce ¬cliquen\nk,k′\nto the k to k′ pigeonhole principle.\nFor the appropriate values of the parameters k and k′ —\nand specifically, for the values in Theorem 20 — there is a short resolution proof of the k to k′\npigeonhole principle (this was shown by Buss & Pitassi [BP97]); (this resolution proof is polynomial\nin the number of pigeons k, but not in the number of holes k′, which is exponentially smaller than\nk).9 Therefore, in order to conclude the refutation of ¬cliquen\nk,k′ inside Res(2) (or inside R(lin)), it\nsuffices to simulate the short resolution refutation of the k to k′ pigeonhole principle. It is important\nto emphasize this point: After reducing, inside R(lin), ¬cliquen\nk,k′ to the pigeonhole principle, one\nsimulates the resolution refutation of the pigeonhole principle, and this has nothing to do with\nthe small-size R0(lin) refutations of the pigeonhole principle demonstrated in Section 6.1. This is\nbecause, the reduction (inside R(lin)) of ¬cliquen\nk,k′ to the k to k′ pigeonhole principle, results in\na substitution instance of the pigeonhole principle formulas; in other words, the reduction results\nin a collection of disjunctions that are similar to the pigeonhole principle disjunctions where each\noriginal pigeonhole principle variable is substituted by some big formula (and, in particular, these\ndisjunctions are not R0(lin)-lines at all). (Note that R0(lin) does not admit short proofs of the\nclique-coloring formulas as we show in Section 8.)\n9Whenever k ≥2k′ the k to k′ pigeonhole principle is referred to as the weak pigeonhole principle.\n25"},{"page":26,"text":"7. Interpolation Results for R0(lin)\nIn this section we study the applicability of the feasible (non-monotone) interpolation technique\nto R0(lin) refutations. In particular, we show that R0(lin) admits a polynomial (in terms of the\nR0(lin)-proofs) upper bound on the (non-monotone) circuit-size of interpolants. In the next section\nwe shall give a polynomial upper bound on the monotone circuit-size of interpolants, but only in the\ncase that the interpolant corresponds to the clique-coloring formulas (whereas, in this section we are\ninterested in the general case; that is, upper bounding circuit-size of interpolants corresponding to\nany formula [of the prescribed type; see below]). First, we shortly describe the feasible interpolation\nmethod and explain how this method can be applied to obtain (sometime, conditional) lower bounds\non proof size. Explicit usage of the interpolation method in proof complexity goes back to [Kra94].\nLet Ai(⃗p, ⃗q), i ∈I, and Bj(⃗p,⃗r), j ∈J, (I and J are sets of indices) be a collection of formulas (for\ninstance, a collection of disjunctions of linear equations) in the displayed variables only. Denote by\nA(⃗p, ⃗q) the conjunction of all Ai(⃗p, ⃗q), i ∈I, and by B(⃗p,⃗r), the conjunction of all Bj(⃗p,⃗r), j ∈J.\nAssume that ⃗p, ⃗q,⃗r are pairwise disjoint sets of distinct variables, and that there is no assignment\nthat satisfies both A(⃗p, ⃗q) and B(⃗p,⃗r). Fix an assignment ⃗α to the variables in ⃗p. The ⃗p variables\nare the only common variables of the Ai’s and the Bj’s. Therefore, either A(⃗α, ⃗q) is unsatisfiable\nor B(⃗α,⃗r) is unsatisfiable.\nThe interpolation technique transforms a refutation of A(⃗p, ⃗q) ∧B(⃗p,⃗r), in some proof system,\ninto a circuit (usually a Boolean circuit) separating those assignments ⃗α (for ⃗p) for which A(⃗α, ⃗q)\nis unsatisfiable, from those assignments ⃗α for which B(⃗α,⃗r) is unsatisfiable (the two cases are not\nnecessarily exclusive, so if both cases hold for an assignment, the circuit can output either that the\nfirst case holds or that the second case holds). In other words, given a refutation of A(⃗p, ⃗q)∧B(⃗p,⃗r),\nwe construct a circuit C(⃗p), called the interpolant, such that\nC(⃗α) = 1\n=⇒\nA(⃗α, ⃗q) is unsatisfiable, and\nC(⃗α) = 0\n=⇒\nB(⃗α,⃗r) is unsatisfiable.\n(39)\n(Note that if U denotes the set of those assignments ⃗α for which A(⃗α, ⃗q) is satisfiable, and V\ndenotes the set of those assignments ⃗α for which B(⃗α,⃗r) is satisfiable, then U and V are disjoint\n[since A(⃗p, ⃗q) ∧B(⃗p,⃗r) is unsatisfiable], and C(⃗p) separates U from V ; see Definition 7.2 below.)\nAssume that for a proof system P the transformation from refutations of A(⃗p, ⃗q), B(⃗p,⃗r) into\nthe corresponding interpolant circuit C(⃗p) results in a circuit whose size is polynomial in the size\nof the refutation. Then, an exponential lower bound on circuits for which (39) holds, implies an\nexponential lower bound on P-refutations of A(⃗p, ⃗q), B(⃗p,⃗r).\n7.1. Interpolation for Semantic Refutations. We now lay out the basic concepts needed to\nformally describe the feasible interpolation technique.\nWe use the general notion of semantic\nrefutations (which generalizes any standard propositional refutation system). We shall use a close\nterminology to that in [Kra97].\nDefinition 7.1 (Semantic refutation). Let N be a fixed natural number and let E1, . . . , Ek ⊆\n{0, 1}N, where Tk\ni=1 Ei = ∅. A semantic refutation from E1, . . . , Ek is a sequence D1, . . . , Dm ⊆\n{0, 1}N with Dm = ∅and such that for every i ∈[m], Di is either one of the Ej’s or is deduced\nfrom two previous Dj, Dl, 1 ≤j, l< i, by the following semantic inference rule:\n• From A, B ⊆{0, 1}N deduce any C, such that C ⊇(A ∩B).\nObserve that any standard propositional refutation (with inference rules that derive from at\nmost two proof-lines, a third line) can be regarded as a semantic refutation: just substitute each\nrefutation-line by the set of its satisfying assignments; and by the soundness of the inference rules\napplied in the refutation, it is clear that each refutation-line (considered as the set of assignments\nthat satisfy it) is deduced by the semantic inference rule from previous refutation-lines.\n26"},{"page":27,"text":"Definition 7.2 (Separating circuit). Let U, V ⊆{0, 1}n, where U ∩V = ∅, be two disjoint sets. A\nBoolean circuit C with n input variables is said to separate U from V if C(⃗x) = 1 for every ⃗x ∈U,\nand C(⃗x) = 0 for every ⃗x ∈V. In this case we also say that U and V are separated by C.\nConvention: In what follows we sometime identify a Boolean formula with the set of its satisfying\nassignments.\nNotation: For two (or more) binary strings u, v ∈{0, 1}∗, we write (u, v) to denote the concate-\nnation of the u with v (where v comes to the right of u, obviously).\nLet N = n+s+t be fixed from now on. Let A1, . . . , Ak ⊆{0, 1}n+s and let B1, . . . , Bl⊆{0, 1}n+t.\nDefine the following two sets of assignments of length n (formally, 0, 1 strings of length n) that\ncan be extended to satisfying assignments of A1, . . . , Ak and B1, . . . , Bl, respectively (formally,\nthose 0, 1 string of length n + s and n + t, that are contained in all A1, . . . , Ak and B1, . . . , Bl,\nrespectively):\nUA :=\n(\nu ∈{0, 1}n\n ∃q ∈{0, 1}s , (u, q) ∈\nk\\\ni=1\nAi\n)\n,\nVB :=\n(\nv ∈{0, 1}n\n ∃r ∈{0, 1}t , (v, r) ∈\nl\\\ni=1\nBi\n)\n.\nDefinition 7.3 (polynomial upper bounds on interpolants). Let P be a propositional\nrefutation system.\nAssume that ⃗p, ⃗q,⃗r are pairwise disjoint sets of distinct variables, where\n⃗p has n variables, ⃗q has s variables and ⃗r has t variables.\nLet A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and\nB1(⃗p,⃗r), . . . , Bl(⃗p,⃗r) be two collections of formulas with the displayed variables only.\nAssume\nthat for any such A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r), if there exists a P-refutation of\nsize S for A1(⃗p, ⃗q) ∧· · · ∧Ak(⃗p, ⃗q) ∧B1(⃗p,⃗r) ∧. . . ∧Bl(⃗p,⃗r) then there exists a Boolean circuit\nseparating UA from VB of size polynomial in S.10 In this case we say that P has a polynomial upper\nbound on interpolant circuits.\n7.1.1. The Communication Game Technique. The feasible interpolation via communication game\ntechnique is based on transforming proofs into Boolean circuits, where the size of the resulting\ncircuit depends on the communication complexity of each proof-line. This technique goes back to\n[IPU94] and [Razb95] and was subsequently applied and extended in [BPR97] and [Kra97] ([IPU94]\nand [BPR97] did not use explicitly the notion of interpolation of tautologies or contradictions). We\nshall employ the interpolation theorem of Kraj ́ıˇcek in [Kra97], that demonstrates how to transform\na small semantic refutation with each proof-line having low communication complexity into a small\nBoolean circuit separating the corresponding sets.\nThe underlying idea of the interpolation via communication game technique is that a (semantic)\nrefutation, where each proof-line is of small (that is, logarithmic) communication complexity, can be\ntransformed into an efficient communication protocol for the Karchmer-Wigderson game (following\n[KW88]) for two players. In the Karchmer-Wigderson game the first player knows some binary\nstring u ∈U and the second player knows some different binary string v ∈V , where U and V are\ndisjoint sets of strings. The two players communicate by sending information bits to one another\n(following a protocol previously agreed on). The goal of the game is for the two players to decide\non an index i such that the ith bit of u is different from the ith bit of v. An efficient Karchmer-\nWigderson protocol (by which we mean a protocol that requires the players to exchange at most a\nlogarithmic number of bits in the worst-case) can then be transformed into a small circuit separating\n10Here UA and VB are defined as above, by identifying the Ai(⃗p, ⃗q)’s and the Bi(⃗p,⃗r)’s with the sets of assignments\nthat satisfy them.\n27"},{"page":28,"text":"U from V (see Definition 7.2). This efficient transformation from protocols for Karchmer-Wigderson\ngames (described in a certain way) into circuits, was demonstrated by Razborov in [Razb95]. So\noverall, given a semantic refutation with proof-lines of low communication complexity, one can\nobtain a small circuit for separating the corresponding sets.\nFirst, we need to define the concept of communication complexity in a suitable way for the\ninterpolation theorem.\nDefinition 7.4 (Communication complexity). Let N = n + s + t and A ⊆{0, 1}N. Let u, v ∈\n{0, 1}n, qu ∈{0, 1}s, rv ∈{0, 1}t. Denote by ui, vi the ith bit of u, v, respectively, and let (u, qu, rv)\nand (v, qu, rv) denote the concatenation of strings u, qu, rv and v, qu, rv, respectively. Consider the\nfollowing three tasks:\n(1) Decide whether (u, qu, rv) ∈A;\n(2) Decide whether (v, qu, rv) ∈A;\n(3) If one of the following holds:\n(i) (u, qu, rv) ∈A and (v, qu, rv) ̸∈A; or\n(ii) (u, qu, rv) ̸∈A and (v, qu, rv) ∈A,\nthen find an i ∈[n], such that ui ̸= vi;\nConsider a game between two players, Player I and Player II, where Player I knows u ∈{0, 1}n , qu ∈\n{0, 1}s and Player II knows v ∈{0, 1}n , rv ∈{0, 1}t. The two players communicate by exchanging\nbits of information between them (following a protocol previously agreed on). The communication\ncomplexity of A, denoted CC(A), is the minimal (over all protocols) number of bits that players I\nand II need to exchange in the worst-case in solving each of Tasks 1, 2 and 3 above.11\nFor A ⊆{0, 1}n+s define\n ̇A :=\n \n(a, b, c)\n (a, b) ∈A and c ∈{0, 1}t \n,\nwhere a and b range over {0, 1}n and {0, 1}s, respectively. Similarly, for B ⊆{0, 1}n+t define\n ̇B :=\n \n(a, b, c)\n (a, c) ∈B and b ∈{0, 1}t \n,\nwhere a and c range over {0, 1}n and {0, 1}t, respectively.\nTheorem 23 ([Kra97]). Let A1, . . . , Ak ⊆{0, 1}n+s and B1, . . . , Bl⊆{0, 1}n+t. Let D1, . . . , Dm\nbe a semantic refutation from ̇A1, . . . , ̇Ak and ̇B1, . . . , ̇Bl. Assume that CC(Di) ≤ζ, for all i ∈[m].\nThen, the sets UA and VB (as defined above) can be separated by a Boolean circuit of size (m +\nn)2O(ζ).\nIn light of Theorem 23, to demonstrate that a certain propositional refutation system P possesses\na polynomial upper bound on interpolant circuits (see Definition 7.3) it suffices to show that any\nproof-line of P induces a set of assignments with at most a logarithmic (in the number of variables)\ncommunication complexity (Definition 7.4).\n7.2. Polynomial Upper Bounds on Interpolants for R0(lin). Here we apply Theorem 23 to\nshow that R0(lin) has polynomial upper bounds on its interpolant circuits. Again, in what follows\nwe sometime identify a disjunction of linear equations with the set of its satisfying assignments.\nTheorem 24. R0(lin) has a polynomial upper bounds on interpolant circuits (Definition 7.3).\nAccording to the paragraph after Theorem 23, all we need in order to establish Theorem 24 is\nthe following lemma:\n11In other words, CC(A) is the minimal number ζ, for which there exists a protocol, such that for every input\n(u, qu to Player I and v, rv to Player II) and every task (from Tasks 1, 2 and 3), the players need to exchange at most\nζ bits in order to solve the task.\n28"},{"page":29,"text":"Lemma 25. Let D be an R0(lin)-line with N variables and let eD be the set of assignments that\nsatisfy D.12Then, CC( eD) ≤O(log N).\nProof: Let N = n + s + t (and so eD ∈{0, 1}n+s+t). For the sake of convenience we shall assume\nthat the N variables in D are partitioned into (pairwise disjoint) three groups ⃗p := (p1 . . . , pn),\n⃗q := (q1, . . . , qs) and ⃗r := (r1, . . . , rt). Let u, v ∈{0, 1}n, qu ∈{0, 1}s, rv ∈{0, 1}t. Assume that\nPlayer I knows u, qu and Player II knows v, rv.\nBy the definition of an R0(lin)-line (see Definition 3.2) we can partition the disjunction D into\na constant number of disjuncts, where one disjunct is a (possibly empty, translation of a) clause in\nthe ⃗p, ⃗q,⃗r variables (see Section 3.1), and all other disjuncts have the following form:\n_\ni∈I\n \n⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r = li\n \n,\n(40)\nwhere I is (an unbounded) set of indices, li are integer numbers, for all i ∈I, and ⃗a,⃗b,⃗c denote\nvectors of n, s and t constant coefficients, respectively.\nLet us denote the (translation of the) clause from D in the ⃗p, ⃗q,⃗r variables by\nP ∨Q ∨R ,\nwhere P, Q and R denote the (translated) sub-clauses consisting of the ⃗p, ⃗q and ⃗r variables,\nrespectively.\nWe need to show that by exchanging O(log N) bits, the players can solve each of Tasks 1, 2 and\n3 from Definition 7.4, correctly.\nTask 1: The players need to decide whether (u, qu, rv) ∈eD. Player II, who knows rv, computes\nthe numbers ⃗c · rv, for every ⃗c pertaining to every disjunct of the form shown in Equation (40)\nabove. Then, Player II sends the (binary representation of) these numbers to Player I. Since there\nare only a constantly many such numbers and the coefficients in every ⃗c are also constants, this\namounts to O(log t) ≤O(log N) bits that Player II sends to Player I. Player II also computes the\ntruth value of the sub-clause R, and sends this (single-bit) value to Player I.\nNow, it is easy to see that Player I has sufficient data to compute by herself/himself whether\n(u, qu, rv) ∈eD (Player I can then send a single bit informing Player II whether (u, qu, rv) ∈eD).\nTask 2: This is analogous to Task 1.\nTask 3: Assume that (u, qu, rv) ∈eD and (v, qu, rv) ̸∈eD (the case (u, qu, rv) ̸∈eD and (v, qu, rv) ∈eD\nis analogous).\nThe first rounds of the protocol are completely similar to that described in Task 1 above: Player\nII, who knows rv, computes the numbers ⃗c · rv, for every ⃗c pertaining to every disjunct of the form\nshown in Equation (40) above. Then, Player II sends the (binary representation of) these numbers\nto Player I. Player II also computes the truth value of the sub-clause R, and sends this (single-bit)\nvalue to Player I. Again, this amounts to O(log N) bits that Player II sends to Player I.\nBy assumption (that (u, qu, rv) ∈eD and (v, qu, rv) ̸∈eD) the players need to deal only with the\nfollowing two cases:\nCase 1: The assignment (u, qu, rv) satisfies the clause P ∨Q∨R while (v, qu, rv) falsifies P ∨Q∨R.\nThus, it must be that ⃗u satisfies the sub-clause P while ⃗v falsifies P. This means that for any i ∈[n]\nsuch that ui sets to 1 a literal in P (there ought to exist at least one such i), it must be that ui ̸= vi.\nTherefore, all that Player I needs to do is to send the (binary representation of) index i to Player\nII. (This amounts to O(log N) bits that Player I sends to Player II.)\n12The notation eD has nothing to do with the same notation used in Section 3.\n29"},{"page":30,"text":"Case 2: There is some linear equation\n⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r = l\n(41)\nin D, such that ⃗a · u +⃗b · qu + ⃗c · rv = l. Note that (by assumption that (v, qu, rv) ̸∈eD) it must\nalso hold that: ⃗a · v +⃗b · qu + ⃗c · rv ̸= l(and so there is an i ∈[n], such that ui ̸= vi). Player I can\nfind linear equation (41), as he/she already received from Player II all the possible values of ⃗c · ⃗r\n(for all possible ⃗c ’s in D).\nRecall that the left hand side of a linear equation ⃗d · ⃗x = lis called the linear form of the\nequation. By the definition of an R0(lin)-line there are only constant many distinct linear forms\nin D. Since both players know these linear forms, we can assume that each linear form has some\nindex associated to it by both players. Player I sends to Player II the index of the linear form\n⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r from (41) in D. Since there are only constantly many such linear forms in D, it\ntakes only constant number of bits to send this index.\nNow both players need to apply a protocol for finding an i ∈[n] such that ui ̸= vi, where\n⃗a · ⃗u +⃗b · qu + ⃗c · rv = land ⃗a · ⃗v +⃗b · qu + ⃗c · rv ̸= l. Thus, it remains only to prove the following\nclaim:\nClaim 4. There is a communication protocol in which Player I and Player II need at most O(log N)\nbits of communication in order to find an i ∈[n] such that ui ̸= vi (under the above conditions).\nProof of claim: We invoke the well-known connection between Boolean circuit-depth and com-\nmunication complexity. Let f : {0, 1}N →{0, 1} be a Boolean function. Denote by dp(f) the\nminimal depth of a Boolean circuit computing f. Consider a game between two players: Player I\nknows some ⃗x ∈{0, 1}N and Player II knows some other ⃗y ∈{0, 1}N, such that f(⃗x) = 1 while\nf(⃗y) = 0. The goal of the game is to find an i ∈[N] such that xi ̸= yi. Denote by CC′(f) the\nminimal number of bits needed for the two players to communicate (in the worst case13) in order\nto solve this game.14 Then, for any function f it is known that dp(f) = CC′(f) (see [KW88]).\nTherefore, to conclude the proof of the claim it is enough to establish that the function f :\n{0, 1}N →{0, 1} that receives the input variables ⃗p, ⃗q,⃗r and computes the truth value of ⃗a · ⃗p +⃗b ·\n⃗q + ⃗c · ⃗r = lhas Boolean circuit of depth O(log N). In case all the coefficients in ⃗a,⃗b,⃗c are 1, it is\neasy to show15 that there is a Boolean circuit of depth O(log N) that computes the function f. In\nthe case that the coefficients in ⃗a,⃗b,⃗c are all constants, it is easy to show, by a reduction to the\ncase where all coefficients are 1,16 that there is a Boolean circuit of depth O(log N) that computes\nthe function f. We omit the details.\n8. Size Lower Bounds\nIn this section we establish an exponential-size lower bound on R0(lin) refutations of the clique-\ncoloring formulas. We shall employ the theorem of Bonet, Pitassi & Raz in [BPR97] that provides\nexponential-size lower bounds for any semantic refutation of the clique-coloring formulas, having\nlow communication complexity in each refutation-line.\n13Over all inputs ⃗x, ⃗y such that f(⃗x) = 1 and f(⃗y) = 0.\n14The measure CC′ is basically the same as CC defined earlier.\n15Using the known O(log N)-depth Boolean circuits for the threshold functions.\n16For instance, consider the simple case where we have only a single variable. That is, let c be a constant and\nassume that we wish to construct a circuit that computes c · x = l, for some integer l. Then, we take a circuit that\ncomputes the function f : {0, 1}c →{0, 1} that outputs the truth value of y1 + . . . + yc = l(thus, in f all coefficients\nare 1’s); and to compute c · x = lwe only have to substitute each yi in the circuit with the variable x.\n30"},{"page":31,"text":"First we recall the strong lower bound obtained by Alon & Boppana [AB87] (improving over\n[Razb85]; see also [And85]) for the (monotone) clique separator functions, defined as follows (a\nfunction f : {0, 1}n →{0, 1} is called monotone if for all α ∈{0, 1}n, α′ ≥α implies f(α′) ≥f(α)):\nDefinition 8.1 (Clique separator). A monotone boolean function Qn\nk,k′ is called a clique separator\nif it interprets its inputs as the edges of a graph on n vertices, and outputs 1 on every input\nrepresenting a k-clique, and 0 on every input representing a complete k′-partite graph (see Section\n6.3).\nRecall that a monotone Boolean circuit is a circuit that uses only monotone Boolean gates (for\ninstance, only the fan-in two gates ∧, ∨).\nTheorem 26 ([AB87]). Let k, k′ be integers such that 3 ≤k′ < k and k\n√\nk′ ≤n/(8 log n), then\nevery monotone Boolean circuit that computes a clique separator function Qn\nk,k′ requires size at least\n1\n8\n \nn\n4k\n√\nk′ log n\n (\n√\nk′+1)/2\n.\nFor the next theorem, we need a slightly different (and weaker) version of communication com-\nplexity, than that in Definition 7.4.\nDefinition 8.2 (Communication complexity (second definition)). Let X denote n Boolean variables\nx1, . . . , xn, and let S1, S2 be a partition of X into two disjoint sets of variables. The communication\ncomplexity of a Boolean function f : {0, 1}n →{0, 1} is the number of bits needed to be exchanged\nby two players, one knowing the values given to the S1 variables and the other knowing the values\ngiven to S2 variables, in the worst-case, over all possible partitions S1 and S2.\nTheorem 27 ([BPR97]). Every semantic refutation of ¬cliquen\nk,k′ (for k′ < k) with m refutation-\nlines and where each refutation-line (considered as a the characteristic function of the line) has\ncommunication complexity (as in Definition 8.2) ζ, can be transformed into a monotone circuit of\nsize m · 23ζ+1 that computes a separating function Qn\nk,k′.\nIn light of Theorem 26, in order to be able to apply Theorem 27 to R0(lin), and arrive at an\nexponential-size lower bound for R0(lin) refutations of the clique-coloring formulas, it suffices to\nshow that R0(lin) proof-lines have logarithmic communication complexity:\nLemma 28. Let D be an R0(lin)-line with N variables. Then, the communication complexity (as\nin Definition 8.2) of D is at most O(log N) (where D is identified here with the characteristic\nfunction of D).\nProof: The proof is similar to the proof of Lemma 25 for solving Task 1 (and the analogous Task\n2) in Definition 7.4.\nBy direct calculations we obtain the following lower bound from Theorems 26, 27 and Lemma\n28:\nCorollary 29. Let k be an integer such that 3 ≤k′ = k −1 and assume that 1\n2 · n/(8 log n) ≤\nk\n√\nk ≤n/(8 log n). Then, for all ε < 1/3, every R0(lin) refutation of ¬cliquen\nk,k′ is of size at least\n2Ω(nε).\nWhen considering the parameters of Theorem 20, we obtain a super-polynomial separation be-\ntween R0(lin) refutations and R(lin) refutations, as described below.\nFrom Theorems 26,27 and Lemma 28 we have (by direct calculations):\n31"},{"page":32,"text":"Corollary 30. Let k = √n and k′ = (log n)2/8 log log n.\nThen, every R0(lin) refutation of\n¬cliquen\nk,k′ has size at least nΩ\n“\nlog n\n√log log n\n”\n.\nBy Corollary 21, R(lin) admits polynomial-size in n refutations of ¬cliquen\nk,k′ under the param-\neters in Corollary 30. Thus we obtain the following separation result:\nCorollary 31. R(lin) is super-polynomially stronger than R0(lin).\nComment 1. Note that we do not need to assume that the coefficients in R0(lin)-lines are constants\nfor the lower bound argument. If the coefficients in R0(lin)-lines are only polynomially bounded\n(in the number of variables) then the same lower bound as in Corollary 30 also applies. This is\nbecause R0(lin)-lines in which coefficients are polynomially bounded integers, still have low (that\nis, logarithmic) communication complexity (as in Definition 8.2).\n9. Applications to Multilinear Proofs\nIn this section we arrive at one of the main benefits of the work we have done so far; Namely,\napplying results on resolution over linear equations in order to obtain new results for multilinear\nproof systems. Subsection 9.1 that follows, contains definitions, sufficient for the current paper,\nconcerning the notion of multilinear proofs introduced in [RT06].\n9.1. Background on Algebraic and Multilinear Proofs.\n9.1.1. Arithmetic and Multilinear Formulas.\nDefinition 9.1 (Arithmetic formula). Fix a field F. An arithmetic formula is a tree, with edges\ndirected from the leaves to the root, and with unbounded (finite) fan-in. Every leaf of the tree\n(namely, a node of fan-in 0) is labeled with either an input variable or a field element. A field\nelement can also label an edge of the tree. Every other node of the tree is labeled with either + or\n× (in the first case the node is a plus gate and in the second case a product gate). We assume that\nthere is only one node of out-degree zero, called the root. The size of an arithmetic formula F is\nthe total number of nodes in its graph and is denoted by |F|. An arithmetic formula computes a\npolynomial in the ring of polynomials F[x1, . . . , xn] in the following way. A leaf just computes the\ninput variable or field element that labels it. A field element that labels an edge means that the\npolynomial computed at its tail (namely, the node where the edge is directed from) is multiplied\nby this field element. A plus gate computes the sum of polynomials computed by the tails of all\nincoming edges. A product gate computes the product of the polynomials computed by the tails of\nall incoming edges. (Subtraction is obtained using the constant −1.) The output of the formula is\nthe polynomial computed by the root. The depth of a formula F is the maximal number of edges\nin a path from a leaf to the root of F.\nWe say that an arithmetic formula has a plus (resp., product) gate at the root if the root of the\nformula is labeled with a plus (resp., product) gate.\nA polynomial is multilinear if in each of its monomials the power of every input variable is at\nmost one.\nDefinition 9.2 (Multilinear formula). An arithmetic formula is a multilinear formula (or equiva-\nlently, multilinear arithmetic formula) if the polynomial computed by each gate of the formula is\nmultilinear (as a formal polynomial, that is, as an element of F[x1, . . . , xn]).\nAn additional definition we shall need is the following linear operator, called the multilinearization\noperator:\n32"},{"page":33,"text":"Definition 9.3 (Multilinearization operator). Given a field F and a polynomial q ∈F[x1, . . . , xn],\nwe denote by M[q] the unique multilinear polynomial equal to q modulo the ideal generated by all\nthe polynomials x2\ni −xi, for all variables xi.\nFor example, if q = x2\n1x2 + ax3\n4 (for some a ∈F) then M[q] = x1x2 + ax4 .\nThe simulation of R0(lin) by multilinear proofs will rely heavily on the fact that multilinear\nsymmetric polynomials have small depth-3 multilinear formulas over fields of characteristic 0 (see\n[SW01] for a proof of this fact). To this end we define precisely the concept of symmetric polyno-\nmials.\nA renaming of the variables x1, . . . , xn is a permutation σ ∈Sn (the symmetric group on [n])\nsuch that xi is mapped to xσ(i) for every 1 ≤i ≤n.\nDefinition 9.4 (Symmetric polynomial). Given a set of variables X = {x1, . . . , xn}, a symmetric\npolynomial f over X is a polynomial in (all the variables of) X such that renaming of variables\ndoes not change the polynomial (as a formal polynomial).\n9.1.2. Polynomial Calculus with Resolution. Here we define the PCR proof system, introduced by\nAlekhnovich et al. in [ABSRW02].\nDefinition 9.5 (Polynomial Calculus with Resolution (PCR)). Let F be some fixed field and\nlet Q := {Q1, . . . , Qm} be a collection of multivariate polynomials from the ring of polynomials\nF[x1, . . . , xn, ̄x1, . . . , ̄xn]. The variables ̄x1, . . . , ̄xn are treated as new formal variables. Call the\nset of polynomials x2 −x, for x ∈{x1, . . . , xn, ̄x1, . . . , ̄xn}, plus the polynomials xi + ̄xi −1, for\nall 1 ≤i ≤n, the set of Boolean axioms of PCR. A PCR proof from Q of a polynomial g is\na finite sequence π = (p1, ..., pl) of multivariate polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] (each\npolynomial pi is interpreted as the polynomial equation pi = 0), where pl= g and for each i ∈[l],\neither pi = Qj for some j ∈[m], or pi is a Boolean axiom, or pi was deduced from pj, pk , where\nj, k < i, by one of the following inference rules:\nProduct: From p deduce xi · p , for some variable xi ;\nFrom p deduce ̄xi · p , for some variable ̄xi ;\nAddition: From p and q deduce α · p + β · q, for some α, β ∈F.\nA PCR refutation of Q is a proof of 1 (which is interpreted as 1 = 0) from Q. The number of\nsteps in a PCR proof is the number of proof-lines in it (that is, lin the case of π above).\nNote that the Boolean axioms of PCR have only 0, 1 solutions, where ̄xi = 0 if xi = 1 and ̄xi = 1\nif xi = 0.\n9.1.3. Multilinear Proof Systems. In [RT06] the authors introduced a natural (semantic) algebraic\nproof system that operates with multilinear arithmetic formulas denoted fMC (which stands for\nformula multilinear calculus), defined as follows:\nDefinition 9.6 (Formula Multilinear Calculus (fMC)). Fix a field F and let Q := {Q1, . . . , Qm} be\na collection of multilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] (the variables ̄x1, . . . , ̄xn are\ntreated as formal variables). Call the set of polynomials consisting of xi + ̄xi −1 and xi · ̄xi for\n1 ≤i ≤n , the Boolean axioms of fMC. An fMC proof from Q of a polynomial g is a finite sequence\nπ = (p1, ..., pl) of multilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] , such that pl= g and for\neach i ∈[l], either pi = Qj for some j ∈[m], or pi is a Boolean axiom of fMC, or pi was deduced\nby one of the following inference rules using pj, pk for j, k < i:\nProduct: from p deduce q · p , for some polynomial q ∈F[x1, . . . , xn, ̄x1, . . . , ̄xn] such that\np · q is multilinear;\nAddition: from p, q deduce α · p + β · q, for some α, β ∈F.\n33"},{"page":34,"text":"All the polynomials in an fMC proof are represented as multilinear formulas. (A polynomial pi in\nan fMC proof is interpreted as the polynomial equation pi = 0.) An fMC refutation of Q is a proof\nof 1 (which is interpreted as 1 = 0) from Q. The size of an fMC proof π is defined as the total\nsum of all the formula sizes in π and is denoted by |π|.\nNote that the Boolean axioms have only 0, 1 solutions, where ̄xi = 0 if xi = 1 and ̄xi = 1 if\nxi = 0, for each 1 ≤i ≤n .\nDefinition 9.7 (Depth-k Formula Multilinear Calculus (depth-k fMC)). For a natural number k,\ndepth-k fMC denotes a restriction of the fMC proof system, in which proofs consist of multilinear\npolynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] represented as multilinear formulas of depth at most k.\n9.2. From R(lin) Proofs to PCR Proofs. We now demonstrate a general and straightforward\ntranslation from R(lin) proofs into PCR proofs over fields of characteristic 0. We use the term\n“translation” in order to distinguish it from a simulation; since here we are not interested in the\nsize of PCR proofs. In fact we have not defined the size of PCR proofs at all. We shall be interested\nonly in the number of steps in PCR proofs.\nFrom now on, all polynomials and arithmetic formulas are considered over some fix field F of\ncharacteristic 0.\nRecall that any field of characteristic 0 contains (an isomorphic copy of) the\ninteger numbers, and so we can use integer coefficients in the field.\nDefinition 9.8 (Polynomial translation of R(lin) proof-lines). Let D be a disjunction of linear\nequations:\n \na(1)\n1 x1 + . . . + a(1)\nn xn = a(1)\n0\n \n∨· · · ∨\n \na(t)\n1 x1 + . . . + a(t)\nn xn = a(t)\n0\n \n.\n(42)\nWe denote by bD its translation into the following polynomial:17\n \na(1)\n1 x1 + . . . + a(1)\nn xn −a(1)\n0\n \n· · ·\n \na(t)\n1 x1 + . . . + a(t)\nn xn −a(t)\n0\n \n.\n(43)\nIf D is the empty disjunction, we define bD to be the polynomial 1.\nIt is clear that every 0, 1 assignment to the variables in D, satisfies D, if and only if bD evaluates\nto 0 under the assignment.\nProposition 3. Let π = (D1, . . . , Dl) be an R(lin) proof sequence of Dl, from some collection\nof initial disjunctions of linear equations Q1, . . . , Qm. Then, there exists a PCR proof of bDlfrom\nbQ1, . . . , bQm with at most a polynomial in |π| number of steps.\nProof: We proceed by induction on the number of lines in π.\nThe base case is the translation of the axioms of R(lin) via the translation scheme in Definition\n9.8. An R(lin) Boolean axiom (xi = 0) ∨(xi = 1) is translated into xi · (xi −1) which is already a\nBoolean axiom of PCR.\nFor the induction step, we translate every R(lin) inference rule application into a polynomial-size\nPCR proof sequence as follows. We use the following simple claim:\nClaim 5. Let p and q be two polynomials and let s be the minimal size of an arithmetic formula\ncomputing q. Then one can derive in PCR, with only a polynomial in s number of steps, from p\nthe product q · p.18\nProof of claim: By induction on s.\n17This notation should not be confused with the same notation in Section 6.3.\n18Again, note that we only require that the number of steps in the proof is polynomial. We do not consider here\nthe size of the PCR proof.\n34"},{"page":35,"text":"Assume that Di = Dj ∨L was derived from Dj using the Weakening inference rule of R(lin),\nwhere j < i ≤land L is some linear equation. Then, by Claim 5, bDi = bDj · bL can be derived from\nbDj with a derivation of at most polynomial in |Dj ∨L| many steps.\nAssume that Di was derived from Dj where Dj is Di ∨(0 = k), using the Simplification inference\nrule of R(lin), where j < i ≤land k is a non-zero integer.\nThen, bDi can be derived from\nbDj = bDi · −k by multiplying with −k−1 (via the Addition rule of PCR).\nThus, it remains to simulate the resolution rule application of R(lin). Let A, B be two disjunc-\ntions of linear equations and assume that A ∨B ∨((⃗a + ⃗b) · ⃗x = a0 + b0) was derived in π from\nA ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0) (the case where A ∨B ∨((⃗a −⃗b) · ⃗x = a0 −b0) was derived from\nA ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0), is similar).\nWe need to derive bA · bB · ((⃗a +⃗b) · ⃗x −a0 −b0) from bA · (⃗a · ⃗x −a0) and bB · (⃗b · ⃗x −b0). This is\ndone by multiplying bA · (⃗a · ⃗x −a0) with bB and multiplying bB · (⃗b · ⃗x −b0) with bA (using Claim 5),\nand then adding the resulted polynomials together.\nRemark 4. When translating R(lin) proofs into PCR proofs we actually do not make any use of\nthe “negative” variables ̄x1, . . . , ̄xn. Nevertheless, the multilinear proof systems make use of these\nvariables in order to polynomially simulate PCR proofs (see Theorem 33 and its proof in [RT06]).\nWe shall need the following corollary in the sequel:\nCorollary 32. Let π = D1, . . . , Dlbe an R0(lin) proof of Dl, and let s be the maximal size of an\nR0(lin)-line in π. Then there is a PCR proof π′ of bDlwith polynomial-size in |π| number of steps\nand such that every line of π′ is a translation (via Definition 9.8) of an R0(lin)-line (Definition\n3.2), where the size of the R0(lin)-line is polynomial in s.\nProof: The simulation of R(lin) by PCR shown above, can be thought of as, first, considering\nbD1, . . . , bDlas the “skeleton” of a PCR proof of bDl. And second, for each Di that was deduced by\none of R(lin)’s inference rules from previous lines, one inserts the corresponding PCR proof sequence\nthat simulates the appropriate inference rule application (as described in the proof of Proposition\n3). By definition, those PCR proof-lines that correspond to lines in the skeleton bD1, . . . , bDlare\ntranslations of R0(lin)-lines (with size at most polynomial in s). Thus, to conclude the proof of\nthe corollary, one needs only to check that for any R0(lin)-line Di that was deduced by one of\nR(lin)’s inference rules from previous R0(lin)-lines (as demonstrated in the proof of Proposition 3),\nthe inserted corresponding PCR proof sequence uses only translations of R0(lin)-lines (with size\npolynomial in s). This can be verified by a straightforward inspection.\n9.3. From PCR Proofs to Multilinear Proofs. We now recall the general simulation result\nproved in [RT06] stating the following: Let π be a PCR refutation of some initial collection of\nmultilinear polynomials Q over some fixed field. Assume that π has polynomially many steps (that\nis, the number of proof lines in the PCR proof sequence is polynomial). If the ‘multilinearization’\n(namely, the result of applying the M[·] operator – see Definition 9.3) of each of the polynomials\nin π has a polynomial-size depth d multilinear formula (with a plus gate at the root), then there is\na polynomial-size depth-d fMC refutation of Q. More formally, we have:\nTheorem 33 ([RT06]). Fix a field F (not necessarily of characteristic 0) and let Q be a set of\nmultilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn]. Let π = (p1, . . . , pm) be a PCR refutation\nof Q. For each pi ∈π, let Φi be a multilinear formula for the polynomial M[pi]. Let s be the total\nsize of all formulas Φi, that is, s = Σm\ni=1|Φi|, and let d ≥2 be the maximal depth of all formulas\nΦi. Assume that the depth of all the formulas Φi that have a product gate at the root is at most\nd −1. Then there is a depth-d fMC refutation of Q of size polynomial in s.\n35"},{"page":36,"text":"9.3.1. Depth-3 Multilinear Proofs. Here we show that multilinear proofs operating with depth-3\nmultilinear formulas (that is, depth-3 fMC) over fields of characteristic 0 polynomially simulate\nR0(lin) proofs. In light of Proposition 32 and Theorem 33, to this end it suffices to show that any\nR0(lin)-line D translates into a corresponding polynomial p (via the translation in Definition 9.8)\nsuch that M[p] has a multilinear formula of size polynomial (in the number of variables) and depth\nat most 3 (with a plus gate at the root) over fields of characteristic 0.\nWe need the following proposition from [RT06]:\nProposition 4 ([RT06]). Let F be a field of characteristic 0. For a constant c, let X1, . . . , Xc\nbe c finite sets of variables (not necessarily disjoint), where Σc\ni=1|Xi| = n . Let f1, . . . , fc be c\nsymmetric polynomials over X1, . . . , Xc (over the field F), respectively. Then, there is a depth-3\nmultilinear formula for M[f1 · · · fc] of size polynomial (in n), with a plus gate at the root.\nThe following is the key lemma of the simulation:\nLemma 34. Let D be an R0(lin)-line with n variables and let p = bD (see Definition 9.8). Then,\nM[p] has a depth-3 multilinear formula over fields of characteristic 0, with a plus gate at the root\nand size at most polynomial in the size of D.\nProof: Assume that the underlying variables of D are ⃗x = x1 . . . , xn. By the definition of an\nR0(lin)-line (see Definition 3.2) we can partition the disjunction D into a constant number of\ndisjuncts, where one disjunct is a (possibly empty, translation of a) clause C,19 and all other\ndisjuncts have the following form:\nm\n_\ni=1\n(⃗a · ⃗x = li) ,\n(44)\nwhere the li’s are integers, m is not necessarily bounded and ⃗a denotes a vector of n constant\ninteger coefficients.\nLet us denote by q the polynomial representing the clause C.20\nConsider a disjunct as shown in (44). Since the coefficients ⃗a are constants, ⃗a · ⃗x can be written\nas a sum of constant number of linear forms, each with the same constant coefficient. In other\nwords, ⃗a · ⃗x can be written as z1 + . . . + zd, for some constant d, where for all i ∈[d]:\nzi := b ·\nX\nj∈J\nxj ,\n(45)\nfor some J ⊆[n] and some constant integer b. We shall assume without loss of generality that d is\nthe same constant for every disjunct of the form (44) inside D (otherwise, take d to be the maximal\nsuch d).\nThus, (44) is translated (via the translation scheme in Definition 9.8) into:\nm\nY\ni=1\n(z1 + ... + zd −li) .\n(46)\nBy fully expanding the product in (46), we arrive at:\nX\nr1+...+rd+1=m\n \nαrd+1 ·\nd\nY\nk=1\nzrk\nk\n!\n,\n(47)\n19If there is more than one clause in D, we simply combine all the clauses into a single clause.\n20C is a translation of a clause (that is, disjunction of literals) into a disjunction of linear equations, as defined\nin Section 3.1. The polynomial q is then the polynomial translation of this disjunction of linear equations, as in\nDefinition 9.8.\n36"},{"page":37,"text":"where the ri’s are non-negative integers, and where the αr’s, for every 0 ≤r ≤m are just integer\ncoefficients, formally defined as follows (this definition is not essential; we present it only for the\nsake of concreteness):\nαr :=\nX\nU⊆[m]\n| U|=r\nY\nj∈U\n(−lj) .\n(48)\nClaim 6. The polynomial bD (the polynomial translation of D) is a linear combination (over F) of\npolynomially (in |D|) many terms, such that each term can be written as\nq ·\nY\nk∈K\nzrk\nk ,\nwhere K is a collection of a constant number of indices, rk’s are non-negative integers, and the zk’s\nand q are as above (that is, the zk’s are linear forms, where each zk has a single coefficient for all\nvariables in it, as in (45), and q is a polynomial translation of a clause).\nProof of claim: Denote the total number of disjuncts of the form (44) in D by h. By definition\n(of R0(lin)-line), h is a constant. Consider the polynomial (47) above. In bD, we actually need to\nmultiply h many polynomials of the form shown in (47) and the polynomial q.\nFor every j ∈[h] we write the (single) linear form in the jth disjunct as a sum of constantly\nmany linear forms zj,1 + . . . + zj,d, where each linear form zj,k has the same coefficient for every\nvariable in it. Thus, bD can be written as:\nq ·\nh\nY\nj=1\n \n \n \n \n \n \n \nX\nr1+...+rd+1=mj\n \nα(j)\nrd+1 ·\nd\nY\nk=1\nzrk\nj,k\n!\n|\n{z\n}\n(⋆)\n \n \n \n \n \n \n \n,\n(49)\n(where the mj’s are not bounded, and the coefficients α(j)\nrd+1 are as defined in (48) except that here\nwe add the index (j) to denote that they depend on the jth disjunct in D). Denote the maximal\nmj, for all j ∈[h], by m0. The size of D, denoted |D|, is at least m0. Note that since d is a\nconstant, the number of summands in each (middle) sum in (49) is polynomial in m0, which is at\nmost polynomial in |D|. Thus, by expanding the outermost product in (49), we arrive at a sum of\npolynomially in |D| many summands. Each summand in this sum is a product of h terms of the\nform (⋆) multiplied by q.\nIt remains to apply the multilinearization operator (Definition 9.3) on bD, and verify that the re-\nsulting polynomial has a depth-3 multilinear formula with a plus gate at the root and of polynomial-\nsize (in |D|). Since M[·] is a linear operator, it suffices to show that when applying M[·] on each\nsummand in bD, as described in Claim 6, one obtains a (multilinear) polynomial that has a depth-3\nmultilinear formula with a plus gate at the root, and of polynomial-size in the number of variables\nn (note that clearly n ≤|D|). This is established in the following claim:\nClaim 7. The polynomial M\n \nq · Q\nk∈K zrk\nk\n \nhas a depth-3 multilinear formula of polynomial-size\nin n (the overall number of variables) and with a plus gate at the root (over fields of characteristic\n0), under the same notation as in Claim 6.\nProof of claim: Recall that a power of a symmetric polynomial is a symmetric polynomial in itself.\nSince each zk (for all k ∈K) is a symmetric polynomial, then its power zrk\nk is also symmetric. The\npolynomial q is a translation of a clause, hence it is a product of two symmetric polynomials: the\n37"},{"page":38,"text":"symmetric polynomial that is the translation of the disjunction of literals with positive signs, and\nthe symmetric polynomial that is the translation of the disjunction of literals with negative signs.\nTherefore, q ·Q\nk∈K zrk\nk is a product of constant number of symmetric polynomials. By Proposition\n4, M\n \nq · Q\nk∈K zrk\nk\n \n(where here the M[·] operator operates on the ⃗x variables in the zk’s and q) is\na polynomial for which there is a polynomial-size (in n) depth-3 multilinear formula with a plus\ngate at the root (over fields of characteristic 0).\nWe now come to the main corollary of this section.\nCorollary 35. Multilinear proofs operating with depth-3 multilinear formulas (that is, depth-3 fMC\nproofs) polynomially-simulate R0(lin) proofs.\nProof: Immediate from Corollary 32, Theorem 33 and Proposition 34.\nFor the sake of clarity we repeat the chain of transformations needed to prove the simulation.\nGiven an R0(lin) proof π, we first use Corollary 32 to transform π into a PCR proof π′, with number\nof steps that is at most polynomial in |π|, and where each line in π′ is a polynomial translation of\nsome R0(lin)-line with size at most polynomial in the maximal line in π (which is clearly at most\npolynomial in |π|). Thus, by Proposition 34 each polynomial in π′ has a corresponding multilinear\npolynomial with a polynomial-size in |π| depth-3 multilinear formula (and a plus gate at the root).\nTherefore, by Theorem 33, we can transform π′ into a depth-3 fMC proof with only a polynomial\n(in |π|) increase in size.\n9.4. Small Depth-3 Multilinear Proofs. Since R0(lin) admits polynomial-size (in n) refutations\nof the m to n pigeonhole principle (for any m > n) (as defined in 6.1), Corollary 35 and Theorem\n15 yield:\nTheorem 36. For any m > n there are polynomial-size (in n) depth-3 fMC refutations of the m\nto n pigeonhole principle PHPm\nn (over fields of characteristic 0).\nThis improves over the result in [RT06] that demonstrated a polynomial-size (in n) depth-3 fMC\nrefutations of a weaker principle, namely the m to n functional pigeonhole principle.\nFurthermore, corollary 35 and Theorem 19 yield:\nTheorem 37. Let G be an r-regular graph with n vertices, where r is a constant, and fix some\nmodulus p. Then there are polynomial-size (in n) depth-3 fMC refutations of Tseitin mod p formulas\n¬TseitinG,p (over fields of characteristic 0).\nThe polynomial-size refutations of Tseitin graph tautologies here are different than those demon-\nstrated in [RT06]. Theorem 37 establishes polynomial-size refutations over any field of characteristic\n0 of Tseitin mod p formulas, whereas [RT06] required the field to contain a primitive pth root of\nunity. On the other hand, the refutations in [RT06] of Tseitin mod p formulas do not make any use\nof the semantic nature of the fMC proof system, in the sense that they do not utilize the fact that\nthe base field is of characteristic 0 (which in turn enables one to efficiently represent any symmetric\n[multilinear] polynomial by a depth-3 multilinear formula).\n10. Relations with Extensions of Cutting Planes\nIn this section we tie some loose ends by showing that, in full generality, R(lin) polynomially\nsimulates R(CP) with polynomially bounded coefficients, denoted R(CP*). First we define the\nR(CP*) proof system – introduced in [Kra98] – which is a common extension of resolution and\nCP* (the latter is cutting planes with polynomially bounded coefficients). The system R(CP*),\nthus, is essentially resolution operating with disjunctions of linear inequalities (with polynomially\nbounded integral coefficients) augmented with the cutting planes inference rules.\n38"},{"page":39,"text":"A linear inequality is written as\n⃗a · ⃗x ≥a0 ,\n(50)\nwhere ⃗a is a vector of integral coefficients a1, . . . , an, ⃗x is a vector of variables x1, . . . , xn, and a0\nis an integer. The size of the linear inequality (50) is the sum of all a0, . . . , an written in unary\nnotation (this is similar to the size of linear equations in R(lin)). A disjunction of linear inequalities\nis just a disjunction of inequalities of the form in (50). The semantics of a disjunction of inequalities\nis the natural one, that is, a disjunction is true under an assignment of integral values to ⃗x if and\nonly if at least one of the inequalities is true under the assignment. The size of a disjunction of\nlinear inequalities is the total size of all linear inequalities in it. We can also add in the obvious\nway linear inequalities, that is, if L1 is the linear inequality ⃗a·⃗x ≥a0 and L2 is the linear inequality\n⃗b · ⃗x ≥b0, then L1 + L2 is the linear inequality (⃗a +⃗b) · ⃗x ≥a0 + b0.\nThe proof system R(CP*) operates with disjunctions of linear inequalities with integral coeffi-\ncients (written in unary representation), and is defined as follows (our formulation is similar to\nthat in [Koj07]):21\nDefinition 10.1 (R(CP*)). Let K := {K1, . . . , Km} be a collection of disjunctions of linear in-\nequalities (whose coefficients are written in unary representation). An R(CP*)-proof from K of a\ndisjunction of linear inequalities D is a finite sequence π = (D1, ..., Dl) of disjunctions of linear\ninequalities, such that Dl= D and for each i ∈[l]: either Di = Kj for some j ∈[m]; or Di is one\nof the following R(CP*)-axioms:\n(1) xi ≥0, for any variable xi;\n(2) −xi ≥−1, for any variable xi;\n(3) (⃗a · ⃗x ≥a0) ∨(−⃗a · ⃗x ≥1 −a0), where all coefficients (including a0) are integers;\nor Di was deduced from previous lines by one of the following R(CP*)-inference rules:\n(1) Let A, B be two disjunctions of linear inequalities and let L1, L2 be two linear inequalities.22\nFrom A ∨L1 and B ∨L2 derive A ∨B ∨(L1 + L2).\n(2) Let L be some linear equation.\nFrom a disjunction of linear equations A derive A ∨L.\n(3) Let A be a disjunction of linear equations\nFrom A ∨(0 ≥1) derive A.\n(4) Let c be a non-negative integer.\nFrom (⃗a · ⃗x ≥a0) ∨A derive (c⃗a · ⃗x ≥ca0) ∨A.\n(5) Let A be a disjunction of linear inequalities, and let c ≥1 be an integer.\nFrom (c⃗a · ⃗x ≥a0) ∨A derive (a · ⃗x ≥⌈a0/c⌉) ∨A.\nAn R(CP*) refutation of a collection of disjunctions of linear inequalities K is a proof of the empty\ndisjunction from K. The size of a proof π in R(CP*) is the total size of all the disjunctions of\nlinear inequalities in π, denoted |π|.\nIn order for R(lin) to simulate R(CP*) proofs, we need to fix the following translation scheme.\nEvery linear inequality L of the form ⃗a·⃗x ≥a0 is translated into the following disjunction, denoted\nbL:\n(⃗a · ⃗x = a0) ∨(⃗a · ⃗x = a0 + 1) ∨· · · ∨(⃗a · ⃗x = a0 + k) ,\n(51)\nwhere k is such that a0 + k equals the sum of all positive coefficients in ⃗a, that is, a0 + k =\nmax\n⃗x∈{0,1}n (⃗a · ⃗x) (in case the sum of all positive coefficients in ⃗a is less than a0, then we put k = 0).\nAn inequality with no variables of the form 0 ≥a0 is translated into 0 = a0 in case it is false (that\n21When we allow coefficients to be written in binary representation, instead of unary representation, the resulting\nproof system is denoted R(CP).\n22In all R(CP*)-inference rules, A, B are possibly the empty disjunctions.\n39"},{"page":40,"text":"is, in case 0 < a0), and into 0 = 0 in case it is true (that is, in case 0 ≥a0). Note that since the\ncoefficients of linear inequalities (and linear equations) are written in unary representation, any\nlinear inequality of size s translates into a disjunction of linear equations of size O(s2). Clearly,\nevery 0, 1 assignment to the variables ⃗x satisfies L if and only if it satisfies its translation bL. A\ndisjunction of linear inequalities D is translated into the disjunction of the translations of all the\nlinear inequalities in it, denoted bD.\nA collection K := {K1, . . . , Km} of disjunctions of linear\ninequalities, is translated into the collection\nn\nbK1, . . . , bKm\no\n.\nTheorem 38. R(lin) polynomially-simulates R(CP*). In other words, if π is an R(CP*) proof of\na linear inequality D from a collection of disjunctions of linear inequalities K1, . . . , Kt, then there\nis an R(lin) proof of bD from bK1, . . . , bKt whose size is polynomial in |π|.\nProof: By induction on the number of proof-lines in π.\nBase case: Here we only need to show that the axioms of R(CP*) translates into axioms of\nR(lin), or can be derived with polynomial-size (in the size of the original R(CP*) axiom) R(lin)\nderivations (from R(lin)’s axioms).\nR(CP*) axiom number (1): xi ≥0 translates into the R(lin) axiom (xi = 0) ∨(xi = 1).\nR(CP*) axiom number (2): −xi ≥−1, translates into (−xi = −1) ∨(−xi = 0).\nFrom the\nBoolean axiom (xi = 1) ∨(xi = 0) of R(lin), one can derive with a constant-size R(lin) proof the\nline (−xi = −1)∨(−xi = 0) (for instance, by subtracting twice each equation in (xi = 1)∨(xi = 0)\nfrom itself).\nR(CP*) axiom number (3): (⃗a ·⃗x ≥a0) ∨(−⃗a ·⃗x ≥1−a0). The inequality (⃗a ·⃗x ≥a0) translates\ninto\nh_\nb=a0\n(⃗a · ⃗x = b) ,\nwhere h is the maximal value of ⃗a · ⃗x over 0, 1 assignments to ⃗x (that is, h is just the sum of all\npositive coefficients in ⃗a). The inequality (−⃗a · ⃗x ≥1 −a0) translates into\nf_\nb=1−a0\n(−⃗a · ⃗x = b) ,\nwhere f is the maximal value of −⃗a · ⃗x over 0, 1 assignments to ⃗x (that is, f is just the sum of\nall negative coefficients in ⃗a). Note that one can always flip the sign of any equation ⃗a · ⃗x = b in\nR(lin). This is done, for instance, by subtracting twice ⃗a · ⃗x = b from itself. So overall R(CP*)\naxiom number (3) translates into\nh_\nb=a0\n(⃗a · ⃗x = b) ∨\nf_\nb=1−a0\n(−⃗a · ⃗x = b) ,\nthat can be converted inside R(lin) into\na0−1\n_\nb=−f\n(⃗a · ⃗x = b) ∨\nh_\nb=a0\n(⃗a · ⃗x = b) .\n(52)\nLet A′ := {−f, −f + 1, . . . , a0 −1, a0, a0 + 1, . . . , h} and let A be the set of all possible values that\n⃗a · ⃗x can get over all possible Boolean assignments to ⃗x. Notice that A ⊆A′. By Lemma 8, for any\n⃗a · ⃗x, there is a polynomial-size (in the size of the linear form ⃗a · ⃗x) derivation of W\nα∈A(⃗a · ⃗x = α).\nBy using the R(lin) Weakening rule we can then derive W\nα∈A′(⃗a · ⃗x = α) which is equal to (52).\nInduction step: Here we simply need to show how to polynomially simulate inside R(lin) every\ninference rule application of R(CP*).\n40"},{"page":41,"text":"Rule (1): Let A, B be two disjunctions of linear inequalities and let L1, L2 be two linear inequalities.\nAssume we already have a R(lin) proofs of bA ∨bL1 and bB ∨bL2. We need to derive bA ∨bB ∨\\\nL1 + L2.\nCorollary 7 shows that there is a polynomial-size (in the size of bL1 and bL2; which is polynomial\nin the size of L1 and L2) derivation of\n\\\nL1 + L2 from bL1 and bL2, from which the desired derivation\nimmediately follows.\nRule (2): The simulation of this rule in R(lin) is done using the R(lin) Weakening rule.\nRule (3): The simulation of this rule in R(lin) is done using the R(lin) Simplification rule (remem-\nber that 0 ≥1 translates into 0 = 1 under our translation scheme).\nRule (4): Let c be a non-negative integer. We need to derive\n\\\n(c⃗a · ⃗x ≥ca0)∨bA from\n\\\n(⃗a · ⃗x ≥a0)∨bA\nin R(lin).\nThis amounts only to “adding together” c times the disjunction\n\\\n(⃗a · ⃗x ≥a0) in\n\\\n(⃗a · ⃗x ≥a0) ∨bA. This can be achieved by c many applications of Corollary 7. We omit the details.\nRule (5): We need to derive\n\\\n(⃗a · ⃗x ≥⌈a0/c⌉) ∨bA, from\n\\\n(c⃗a · ⃗x ≥a0) ∨bA. Consider the disjunction\nof linear equations\n\\\n(c⃗a · ⃗x ≥a0), which can be written as:\n(c⃗a · ⃗x = a0) ∨(c⃗a · ⃗x = a0 + 1) ∨. . . ∨(c⃗a · ⃗x = a0 + r) ,\n(53)\nwhere a0 + r is the maximal value c⃗a · ⃗x can get over 0, 1 assignments to ⃗x. By Lemma 8 there is a\npolynomial-size (in the size of ⃗a · ⃗x) R(lin) proof of\n_\nα∈A\n(⃗a · ⃗x = α) ,\n(54)\nwhere A is the set of all possible values of ⃗a · ⃗x over 0, 1 assignments to ⃗x.\nWe now use (53) to cut-offfrom (54) all equations (⃗a · ⃗x = β) for all β < ⌈a0/c⌉(this will give\nus the desired disjunction of linear equations). Consider the equation (⃗a · ⃗x = β) in (54) for some\nfixed β < ⌈a0/c⌉. Use the resolution rule of R(lin) to add this equation to itself c times inside (54).\nWe thus obtain\n(c⃗a · ⃗x = cβ) ∨\n_\nα∈A\\{β}\n(⃗a · ⃗x = α) .\n(55)\nSince β is an integer and β < ⌈a0/c⌉, we have cβ < a0. Thus, the equation (c⃗a · ⃗x = cβ) does\nnot appear in (53). We can then successively resolve (c⃗a · ⃗x = cβ) in (55) with each equation\n(c⃗a · ⃗x = a0), . . . , (c⃗a · ⃗x = a0 + r) in (53). Hence, we arrive at W\nα∈A\\{β} (⃗a · ⃗x = α). Overall, we\ncan cut-offall equations (⃗a · ⃗x = β), for β < ⌈a0/c⌉, from (54). We then get the disjunction\n_\nα∈A′\n(⃗a · ⃗x = α) ,\nwhere A′ is the set of all elements of A greater or equal to ⌈a0/c⌉(in other words, all values greater\nor equal to ⌈a0/c⌉that ⃗a·⃗x can get over 0, 1 assignments to ⃗x). Using the Weakening rule of R(lin)\n(if necessary) we can arrive finally at the desired disjunction\n\\\n(⃗a · ⃗x ≥⌈a0/c⌉), which concludes the\nR(lin) simulation of R(CP*)’s inference Rule (5).\nAppendix A. Feasible Monotone Interpolation\nHere we formally define the feasible monotone interpolation property. The definition is taken\nmainly from [Kra97].\nRecall that for two binary strings of length n (or equivalently, Boolean assignments for n propo-\nsitional variables) α, α′, we denote by α′ ≥α that α′ is bitwise greater than α, that is, that for all\ni ∈[n], α′\ni ≥αi (where α′\ni and αi are the ith bits of α′ and α, respectively). Let A(⃗p, ⃗q), B(⃗p,⃗r)\nbe two collections of formulas in the displayed variables only, where ⃗p, ⃗q,⃗r are pairwise disjoint\n41"},{"page":42,"text":"sequences of distinct variables (similar to the notation at the beginning of Section 7). Assume\nthat there is no assignment that satisfies both A(⃗p, ⃗q) and B(⃗p,⃗r). We say that A(⃗p, ⃗q), B(⃗p,⃗r) are\nmonotone if one of the following conditions hold:\n(1) If ⃗α is an assignment to ⃗p and ⃗β is an assignment to ⃗q such that A(⃗α, ⃗β) = 1, then for any\nassignment ⃗α′ ≥⃗α it holds that A(⃗α′, ⃗β) = 1.\n(2) If ⃗α is an assignment to ⃗p and ⃗β is an assignment to ⃗r such that B(⃗α, ⃗β) = 1, then for any\nassignment ⃗α′ ≤⃗α it holds that B(⃗α′, ⃗β) = 1.\nFix a certain proof system P. Recall the definition of the interpolant function (corresponding\nto a given unsatisfiable A(⃗p, ⃗q) ∧B(⃗p,⃗r); that is, functions for which (39) in Section 7 hold).\nAssume that for every monotone A(⃗p, ⃗q), B(⃗p,⃗r) there is a transformation from every P-refutation\nof A(⃗p, ⃗q)∧B(⃗p,⃗r) into the corresponding interpolant monotone Boolean circuit C(⃗p) (that is, C(⃗p)\nuses only monotone gates23) and whose size is polynomial in the size of the refutation (note that\nfor every monotone A(⃗p, ⃗q), B(⃗p,⃗r) the corresponding interpolant circuit must compute a monotone\nfunction;24 the interpolant circuit itself, however, might not be monotone, namely, it may use non-\nmonotone gates). In such a case, we say that P has the feasible monotone interpolation property.\nThis means that, if a proof system P has the feasible monotone interpolation property, then an\nexponential lower bound on monotone circuits that compute the interpolant function corresponding\nto A(⃗p, ⃗q) ∧B(⃗p,⃗r) implies an exponential-size lower bound on P-refutations of A(⃗p, ⃗q) ∧B(⃗p,⃗r).\nDefinition A.1 (Feasible monotone interpolation property). Let P be a propositional refu-\ntation system. Let A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r) be two collections of formulas\nwith the displayed variables only (where ⃗p has n variables, ⃗q has s variables and ⃗r has t variables),\nsuch that either (the set of satisfying assignments of) A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) meet condition 1 above\nor (the set of satisfying assignments of) B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r) meet condition 2 above. Assume\nthat for any such A1(⃗p, ⃗q), . . . , Ak(⃗p, ⃗q) and B1(⃗p,⃗r), . . . , Bl(⃗p,⃗r), if there exists a P-refutation for\nA1(⃗p, ⃗q) ∧· · · ∧Ak(⃗p, ⃗q) ∧B1(⃗p,⃗r) ∧. . . ∧Bl(⃗p,⃗r) of size S then there exists a monotone Boolean\ncircuit separating UA from VB (as defined in Section 7.1) of size polynomial in S. In this case we\nsay that P possesses the feasible monotone interpolation property.\nAcknowledgments\nWe wish to thank Arist Kojevnikov for useful correspondence on his paper.\nThis work was\ncarried out in partial fulfillment of the requirements for the Ph.D. degree of the second author.\nReferences\n[AB87]\nNoga Alon and Ravi B. Boppana. The monotone circuit complexity of boolean functions. Combinatorica,\n7(1):1–22, 1987. 8, 26\n[ABE02]\nAlbert Atserias, Maria L. Bonet, and Juan L. Esteban. Lower bounds for the weak pigeonhole principle\nand random formulas beyond resolution. Information and Computation, 176:152–136, August 2002. 1.2,\n3, 6.3, 6.3, 20, 3\n[ABSRW02] Michael Alekhnovich, Eli Ben-Sasson, Alexander A. Razborov, and Avi Wigderson. Space complexity\nin propositional calculus. SIAM J. Comput., 31(4):1184–1211 (electronic), 2002. 9.1.2\n[And85]\nA. E. Andreev. On a method for obtaining lower bounds for the complexity of individual monotone\nfunctions. Dokl. Akad. Nauk SSSR (in Russian), 282(5):1033–1037, 1985. [Engl. Transl. Soviet Math.\nDokl., vol. 31 (1985), pp. 530-534]. 8\n23For instance, a monotone Boolean circuit is a circuit that uses only ∧, ∨gates of fan-in two (see also Section\n8). In certain cases, the monotone interpolation technique is also applicable for a larger class of circuits, that is,\ncircuits that compute with real numbers and that can use any nondecreasing real functions as gates (this was proved\nby Pudl ́ak in [Pud97]).\n24That is, if α′ ≥α then C(α′) ≥C(α).\n42"},{"page":43,"text":"[BGIP01]\nSamuel Buss, Dima Grigoriev, Russell Impagliazzo, and Toniann Pitassi. Linear gaps between degrees for\nthe polynomial calculus modulo distinct primes. J. Comput. System Sci., 62(2):267–289, 2001. Special\nissue on the 14th Annual IEEE Conference on Computational Complexity (Atlanta, GA, 1999). 6.2\n[BP97]\nSamuel Buss and Toniann Pitassi. Resolution and the weak pigeonhole principle. In Computer science\nlogic (Aarhus, 1997), volume 1414 of Lecture Notes in Comput. Sci., pages 149–156. Springer, Berlin,\n1997. 3\n[BPR97]\nMaria Bonet, Toniann Pitassi, and Ran Raz. Lower bounds for cutting planes proofs with small coeffi-\ncients. The Journal of Symbolic Logic, 62(3):708–728, 1997. 1.1, 1.2, 1.2, 6.3, 2, 7.1.1, 8, 27\n[CR79]\nStephen A. Cook and Robert A. Reckhow. The relative efficiency of propositional proof systems. The\nJournal of Symbolic Logic, 44(1):36–50, 1979. 2\n[Hak85]\nArmin Haken. The intractability of resolution. Theoret. Comput. Sci., 39(2-3):297–308, 1985. 1\n[HK06]\nEdward Hirsch and Arist Kojevnikov. Several notes on the power of Gomory-Chv ́atal cuts. Annals of\nPure and Applied Logic, 141:429–436, 2006. 1.1\n[IPU94]\nRussel Impagliazzo, Toniann Pitassi, and Alasdair Urquhart. Upper and lower bounds for tree-like\ncutting planes proofs. In Ninth Annual Symposium on Logic in Computer Science, pages 220–228. IEEE\nComput. Soc. Press, 1994. 7.1.1\n[Koj07]\nArist Kojevnikov. Improved lower bounds for tree-like resolution over linear inequalities. In In Proceed-\nings of the 10th International Conference on Theory and Applications of Satisfiability Testing (SAT),\n2007. Preliminary version in Electronic Colloquium on Computational Complexity, ECCC, January 2007.\nReport No. TR07-010. 1.1, 10\n[Kra94]\nJan Kraj ́ıˇcek. Lower bounds to the size of constant-depth propositional proofs. The Journal of Symbolic\nLogic, 59(1):73–86, 1994. 7\n[Kra97]\nJan Kraj ́ıˇcek. Interpolation theorems, lower bounds for proof systems, and independence results for\nbounded arithmetic. The Journal of Symbolic Logic, 62(2):457–486, 1997. 1.1, 1.2, 1.2, 6.3, 7.1, 7.1.1,\n23, A\n[Kra98]\nJan Kraj ́ıˇcek. Discretely ordered modules as a first-order extension of the cutting planes proof system.\nThe Journal of Symbolic Logic, 63(4):1582–1596, 1998. 1.1, 1.2, 6.3, 10\n[Kra01]\nJan Kraj ́ıˇcek. On the weak pigeonhole principle. Fund. Math., 170(1-2):123–140, 2001. Dedicated to the\nmemory of Jerzy Lo ́s. 6.3\n[Kra07]\nJan Kraj ́ıˇcek. An exponential lower bound for a constraint propagation proof system based on ordered\nbinary decision diagrams. To appear in The Journal of Symbolic Logic. Preliminary version available in\nElectronic Colloquium on Computational Complexity, ECCC, January 2007. Report No. TR07-007. 6.3\n[KW88]\nMauricio Karchmer and Avi Wigderson. Monotone circuits for connectivity require super-logarithmic\ndepth. In Proceedings of the 20th Annual ACM Symposium on Theory of Computing, pages 539–550.\nACM, 1988. 7.1.1, 7.2\n[Pud97]\nPavel Pudl ́ak. Lower bounds for resolution and cutting plane proofs and monotone computations. The\nJournal of Symbolic Logic, 62(3):981–998, Sept. 1997. 6.3, 23\n[Razb85]\nAlexander A. Razborov. Lower bounds on the monotone complexity of some Boolean functions. Dokl.\nAkad. Nauk SSSR (in Russian), 281(4):798–801, 1985. [English translation in Sov. Math. Dokl., vol . 31\n(1985), pp. 354-357.]. 8\n[Razb95]\nAlexander A. Razborov. Unprovability of lower bounds on circuit size in certain fragments of bounded\narithmetic. Izv. Ross. Akad. Nauk Ser. Mat., 59(1):201–224, 1995. 7.1.1\n[Razb02]\nAlexander A. Razborov. Proof complexity of pigeonhole principles. In Developments in language theory\n(Vienna, 2001), volume 2295 of Lecture Notes in Comput. Sci., pages 110–116. Springer, Berlin, 2002. 1\n[Raz04]\nRan Raz. Multi-linear formulas for permanent and determinant are of super-polynomial size. In Pro-\nceedings of the 36th Annual ACM Symposium on the Theory of Computing, pages 633–641, Chicago, IL,\n2004. ACM. 1\n[Raz06]\nRan Raz. Separation of multilinear circuit and formula size. Theory of Computing, Vol. 2, article 6,\n2006. 1\n[RT06]\nRan Raz and Iddo Tzameret. The strength of multilinear proofs. Comput. Complexity (to appear). Pre-\nliminary version in Electronic Colloquium on Computational Complexity, ECCC, January 2006. Report\nNo. TR06-001. (document), 1, 1.1, 1.2, 1.2, 1, 2, 6.2, 9, 9.1.3, 4, 9.3, 33, 9.3.1, 4, 9.4, 9.4\n[SW01]\nAmir Shpilka and Avi Wigderson. Depth-3 arithmetic circuits over fields of characteristic zero. Comput.\nComplexity, 10:1–27, 2001. 9.1.1\n[Tse68]\nG. C. Tseitin. On the complexity of derivations in propositional calculus. Studies in constructive math-\nematics and mathematical logic Part II. Consultants Bureau, New-York-London, 1968. 6.2\n43"},{"page":44,"text":"Department of Applied Mathematics and Computer Science, Weizmann Institute, Rehovot 76100,\nIsrael\nE-mail address: ranraz@wisdom.weizmann.ac.il\nSchool of Computer Science, Tel Aviv University, Tel Aviv 69978, Israel\nE-mail address: tzameret@tau.ac.il\n44"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"disjunctions of linear equations with integral coefficients over the variables ⃗x = x1, . . . , xn.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"i∈I(xi = 1)∨W","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"j∈J(xj = 0). Let A and B be two disjunctions of linear equations, and let ⃗a·⃗x = a0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"and ⃗b · ⃗x = b0 be two linear equations (where ⃗a,⃗b are two vectors of n integral coefficients, and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"i=1 aixi; and similarly for ⃗b · ⃗x). The rules of inference belonging to","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"R(lin) allow to derive A ∨B ∨((⃗a +⃗b) · ⃗x = a0 + b0) from A ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0) (or","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"similarly, to derive A ∨B ∨((⃗a −⃗b) ·⃗x = a0 −b0) from A ∨(⃗a ·⃗x = a0) and B ∨(⃗b ·⃗x = b0)). We can","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"also simplify disjunctions by discarding (unsatisfiable) equations of the form (0 = k), for k ̸= 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"In addition, for every variable xi, we shall add an axiom (xi = 0) ∨(xi = 1), which forces xi to","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"(⃗a · ⃗x = a0) ∨(⃗a · ⃗x = a0 + 1) ∨· · · ∨(⃗a · ⃗x = a0 + k) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"where a0 + k equals the sum of all positive coefficients in ⃗a (that is, a0 + k =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"equation a1x1 + . . . + anxn = a0 is the sum of all a0, . . . , an written in unary notation. The size of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"i=1 aixi. If ⃗b is another","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"L1 : ⃗a · ⃗x = a0 and L2 : ⃗b · ⃗x = b0, their addition (⃗a +⃗b) · ⃗x = a0 + b0 is denoted L1 + L2 (and their","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"subtraction (⃗a −⃗b) · ⃗x = a0 −b0 is denoted L1 −L2). For two Boolean assignments (identified as","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"clause in the sequence; (2) the last clause Dl= D. The size of a resolution proof is the sum of all","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"finite alphabet (“the (proposed) refutation” of F), such that there exists a π with A(F, π) = 1 if","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"of F”) so that A(F, π) = 1. The soundness of a (Cook-Reckhow) proof system stands for the fact","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"that every formula F so that A(F, π) = 1 for some string π is unsatisfiable (in other words, no","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"X := {x1, . . . , xn}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"3.1. Disjunctions of Linear Equations. For L a linear equation a1x1 + . . . + anxn = a0, the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"n xn = a(1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"n xn = a(t)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"n xn = a(j)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"The symbol |= denotes the semantic implication relation, that is, for every collection D1, . . . , Dm","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"D1, . . . , Dm |= D0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"The size of a linear equation a1x1 + . . . + anxn = a0 is Pn","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"i=0 |ai|, i.e., the sum of the bit sizes","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"i=1 |ai|. The size of a disjunction of linear equations is the total size of all linear equations in it.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"i∈I(xi = 1) ∨W","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"j∈J(xj = 0). For a clause D we denote by eD its translation into a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"about Boolean variables we augment the system with the axioms (xi = 0)∨(xi = 1), for all i ∈[n],","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"Definition 3.1 (R(lin)). Let K := {K1, . . . , Km} be a collection of disjunctions of linear equations.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"An R(lin)-proof from K of a disjunction of linear equations D is a finite sequence π = (D1, ..., Dl)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"of disjunctions of linear equations, such that Dl= D and for every i ∈[l], either Di = Kj for","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"some j ∈[m], or Di is a Boolean axiom (xh = 0) ∨(xh = 1) for some h ∈[n], or Di was deduced","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"Simplification: From A ∨(0 = k) derive A, where A is a disjunction of linear equations","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"and k ̸= 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"in D1, . . . , Dm, stipulating that the collection D1, . . . , Dm contains all disjunctions of the form (xj = 0) ∨(xj = 1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"and B ∨¬xi, then in R(lin) we subtract (xi = 0) from (xi = 1) in eB ∨(xi = 0) and eA ∨(xi = 1),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"respectively, to obtain eA ∨eB ∨(0 = 1). Then, using the Simplification rule, we can cut-off(0 = 1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"from eA ∨eB ∨(0 = 1), and arrive at eA ∨eB.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"(x1 + . . . + xl= 1) ∨· · · ∨(x1 + . . . + xl= l) ∨(xl+1 = 1) ∨· · · ∨(xn = 1),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"⃗a(1) · ⃗x = l(1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"⃗a(k) · ⃗x = l(k)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"(xj = bj) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"all i ∈[l] there is an R(lin) derivation of Ei from z = ai and K with size at most s where a1, . . . , al","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"i=1 Ei from K and (z = a1)∨· · ·∨(z = al),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"Proof: Denote by D the disjunction (z = a1) ∨· · · ∨(z = al) and by πi the R(lin) proof of Ei from","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"K and z = ai (with size at most s), for all i ∈[l]. It is easy to verify that for all i ∈[l] the sequence","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"j∈[l]\\{i}(z = aj) is an R(lin) proof of Ei ∨W","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"j∈[l]\\{i}(z = aj) from K and D. So overall, given","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"(z = aj), . . . , El∨","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"(z = aj) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"We now use the Resolution rule to cut-offall the equations (z = ai) inside all the disjunctions","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"(z = aj) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"and so putting k = l, will conclude the proof of the lemma.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"The base case for k = 1 is immediate (from (3)).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"(z = aj) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"j∈[l]\\[k](z = aj) and W","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"j∈[l]\\{k+1}(z = aj) from (4) and (5),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"Resolve (4) with (5) over (z = ak+1) and (z = a1), respectively, to obtain","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"(0 = a1 −ak+1) ∨E1 ∨· · · ∨Ek ∨Ek+1 ∨","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"(z = aj) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"Since a1 ̸= ak+1, we can use the Simplification rule to cut-off(0 = a1 −ak+1) from (6), and we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"(z = aj) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"Now, similarly, resolve (4) with (7) over (z = ak+1) and (z = a2), respectively, and use Simplification","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"(z = aj) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"(z = aj) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"all i ∈[l] there is an R0(lin) derivation of Ei from z = ai and K with size at most s, where the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"i=1 Ei is an R0(lin)-line, there is an R0(lin) proof of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"i=1 Ei from K and (z = a1) ∨· · · ∨(z = al), with size polynomial in s and l.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"by saying that “if z = ai implies Ei (with a polynomial-size proof) for all i ∈[l], then Wl","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"i=1(z = ai)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"i=1 Ei (with a polynomial-size proof)”.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq76","equation_number":null,"raw_text":"i=1(z = ai) above is just the Boolean axiom (xi = 0) ∨(xi = 1), for some i ∈[n], and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq77","equation_number":null,"raw_text":"xi = 0 implies E0 and xi = 1 implies E1 (both with polynomial-size proofs), then to simplify the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq78","equation_number":null,"raw_text":"can be depicted by saying that “if xi = 0 implies E0 with a polynomial-size proof and xi = 1 implies","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq79","equation_number":null,"raw_text":"α∈A(z1 = α) and let D2 be","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq80","equation_number":null,"raw_text":"β∈B (z2 = β), where A, B are two (finite) sets of integers. Then there is a polynomial-size (in the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq81","equation_number":null,"raw_text":"(z1 + z2 = α + β) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq82","equation_number":null,"raw_text":"Proof: Denote the elements of A by α1, . . . , αk. In case z1 = αi, for some i ∈[k] then we can add","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq83","equation_number":null,"raw_text":"z1 = αi to every equation in W","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq84","equation_number":null,"raw_text":"β∈B (z2 = β) to get W","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq85","equation_number":null,"raw_text":"β∈B(z1 + z2 = αi + β). Therefore, there exist","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq86","equation_number":null,"raw_text":"(z1 + z2 = α1 + β) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq87","equation_number":null,"raw_text":"(z1 + z2 = α2 + β) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq88","equation_number":null,"raw_text":"(z1 + z2 = αk + β)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq89","equation_number":null,"raw_text":"5Note that when the proofs of Ei from z = ai, for all i ∈[l], are all done inside R0(lin), then the linear form z","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq90","equation_number":null,"raw_text":"from z1 = α1, z1 = α2 ,. . . ,z1 = αk, respectively.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq91","equation_number":null,"raw_text":"(z1 + z2 = α + β)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq92","equation_number":null,"raw_text":"Corollary 7. Let z1 abbreviate ⃗a·⃗x and z2 abbreviate ⃗b·⃗x. Let D1 be (z1 = a0)∨(z1 = a0 +1) . . .∨","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq93","equation_number":null,"raw_text":"(z1 = a1), and let D2 be (z2 = b0) ∨(z2 = b0 + 1) . . . ∨(z2 = b1). Then there is a polynomial-size","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq94","equation_number":null,"raw_text":"(z1 + z2 = a0 + b0) ∨(z1 + z2 = a0 + b0 + 1) ∨. . . ∨(z1 + z2 = a1 + b1) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq95","equation_number":null,"raw_text":"Lemma 8. Let ⃗a · ⃗x be a linear form with n variables, and let A := {⃗a · ⃗x | ⃗x ∈{0, 1}n} be the set","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq96","equation_number":null,"raw_text":"(⃗a · ⃗x = α) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq97","equation_number":null,"raw_text":"Boolean axiom (x1 = 0) ∨(x1 = 1) and the (first) coefficient a1 from ⃗a. Assume that a1 ≥1. Add","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq98","equation_number":null,"raw_text":"(x1 = 0) to itself a1 times, and arrive at (a1x1 = 0) ∨(x1 = 1). Then, in the resulted line, add","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq99","equation_number":null,"raw_text":"(x1 = 1) to itself a1 times, until the following is reached:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq100","equation_number":null,"raw_text":"(a1x1 = 0) ∨(a1x1 = a1) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq101","equation_number":null,"raw_text":"Similarly, in case a1 ≤−1 we can subtract (|a1| + 1 many times) (x1 = 0) from itself in (x1 =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq102","equation_number":null,"raw_text":"0) ∨(x1 = 1), and then subtract (|a1| + 1 many times) (x1 = 1) from itself in the resulted line.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq103","equation_number":null,"raw_text":"In the same manner, we can derive the disjunctions: (a2x2 = 0) ∨(a2x2 = a2), . . . , (anxn =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq104","equation_number":null,"raw_text":"0) ∨(anxn = an).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq105","equation_number":null,"raw_text":"Consider (a1x1 = 0) ∨(a1x1 = a1) and (a2x2 = 0) ∨(a2x2 = a2). From these two lines, by","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq106","equation_number":null,"raw_text":"(a1x1 + a2x2 = 0) ∨(a1x1 + a2x2 = a1) ∨(a1x1 + a2x2 = a2) ∨(a1x1 + a2x2 = a1 + a2) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq107","equation_number":null,"raw_text":"In a similar fashion, now consider (a3x3 = 0) ∨(a3x3 = a3) and apply again Lemma 6, to obtain","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq108","equation_number":null,"raw_text":"(a1x1 + a2x2 + a3x3 = α) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq109","equation_number":null,"raw_text":"i=1 |ai|, that is, the size of the unary representation of ⃗a.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq110","equation_number":null,"raw_text":"Continue to consider, successively, all other lines (a4x4 = 0) ∨(a4x4 = a4), . . . , (anxn = 0) ∨","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq111","equation_number":null,"raw_text":"(anxn = an), and apply the same reasoning. Each step uses a derivation of size at most polynomial","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq112","equation_number":null,"raw_text":"i=1 |ai|. And so overall we reach the desired line (11), with a derivation of size polynomial in","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq113","equation_number":null,"raw_text":"(x1 = 1) ∨· · · ∨(xn = 1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq114","equation_number":null,"raw_text":"(x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq115","equation_number":null,"raw_text":"Proof: We show that for every i ∈[n], there is a polynomial-size (in n) R0(lin) proof from (xi = 1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq116","equation_number":null,"raw_text":"of (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n). This concludes the proof since, by Lemma 5,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq117","equation_number":null,"raw_text":"which each (xi = 1) (for all i ∈[n]) is replace by (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq118","equation_number":null,"raw_text":"Claim 2. For every i ∈[n], there is a a polynomial-size (in n) R0(lin) proof from (xi = 1) of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq119","equation_number":null,"raw_text":"(x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq120","equation_number":null,"raw_text":"(x1 + . . . + xi−1 + xi+1 + . . . + xn = 0) ∨· · · ∨(x1 + . . . + xi−1 + xi+1 + . . . + xn = n −1) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq121","equation_number":null,"raw_text":"Now add successively (xi = 1) to every equation in (16) (note that this can be done in R0(lin)).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq122","equation_number":null,"raw_text":"We obtain precisely (x1 + . . . + xn = 1) ∨· · · ∨(x1 + . . . + xn = n).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq123","equation_number":null,"raw_text":"Lemma 10. There is a polynomial-size (in n) R0(lin) proof of (x1+. . .+xn = 0)∨(x1+. . .+xn = 1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq124","equation_number":null,"raw_text":"from the collection of disjunctions consisting of (xi = 0) ∨(xj = 0), for all 1 ≤i < j ≤n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq125","equation_number":null,"raw_text":"Proof: We proceed by induction on n. The base case for n = 1 is immediate from the Boolean","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq126","equation_number":null,"raw_text":"axiom (x1 = 0) ∨(x1 = 1). Assume we already have a polynomial-size proof of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq127","equation_number":null,"raw_text":"(x1 + . . . + xn = 0) ∨(x1 + . . . + xn = 1).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq128","equation_number":null,"raw_text":"If xn+1 = 0 we add xn+1 = 0 to both of the equations in (17), and reach:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq129","equation_number":null,"raw_text":"(x1 + . . . + xn+1 = 0) ∨(x1 + . . . + xn+1 = 1).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq130","equation_number":null,"raw_text":"Otherwise, xn+1 = 1, and so we can cut-off(xn+1 = 0) in all the initial disjunctions (xi = 0) ∨","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq131","equation_number":null,"raw_text":"(xn+1 = 0), for all 1 ≤i ≤n. We thus obtain (x1 = 0), . . . , (xn = 0). Adding together (x1 =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq132","equation_number":null,"raw_text":"0), . . . , (xn = 0) and (xn+1 = 1) we arrive at","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq133","equation_number":null,"raw_text":"(x1 + . . . + xn+1 = 1) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq134","equation_number":null,"raw_text":"Recall the definition of the semantic implication relation |= from Section 3.1. Formally, we say","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq135","equation_number":null,"raw_text":"D0, D1, . . . , Dm, it holds that D1, . . . , Dm |= D0 implies that there is an R(lin) proof of D0 from","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq136","equation_number":null,"raw_text":"The base case n = 0. We need to show that D1, . . . , Dm |= D0 implies that there is an R(lin)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq137","equation_number":null,"raw_text":"This means that each Di is a disjunction of equations of the form (0 = a0) for some integer a0 (if","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq138","equation_number":null,"raw_text":"this means precisely that D0 is the equation (0 = 0). Thus, D0 can be derived easily from any","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq139","equation_number":null,"raw_text":"axiom in R(lin) (for instance, by subtracting each equation in (x1 = 0) ∨(x1 = 1) from itself, to","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq140","equation_number":null,"raw_text":"reach (0 = 0) ∨(0 = 0), which is equal to (0 = 0), since we discard duplicate equations inside","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq141","equation_number":null,"raw_text":"In the second case D0 is unsatisfiable. Thus, since D1, . . . , Dm |= D0, there is no assignment sat-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq142","equation_number":null,"raw_text":"is a disjunction of zero or more unsatisfiable equations of the form (0 = a0), for some integer a0 ̸= 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq143","equation_number":null,"raw_text":"D1, . . . , Dm |= D0 .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq144","equation_number":null,"raw_text":"j=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq145","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq146","equation_number":null,"raw_text":"n+1xn+1 = a(j)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq147","equation_number":null,"raw_text":"denote by D↾xi=b the disjunction D, where in every equation in D the variable xi is substituted by","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq148","equation_number":null,"raw_text":"{0, 1}, and assume that xn+1 = b. Then, from D1, . . . , Dm we can derive (inside R(lin)):","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq149","equation_number":null,"raw_text":"D1↾xn+1=b, . . . , Dm↾xn+1=b .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq150","equation_number":null,"raw_text":"The only variables occurring in (22) are x1, . . . , xn. From assumption (20) we clearly have D1↾xn+1=b","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq151","equation_number":null,"raw_text":", . . . , Dm↾xn+1=b |= D0↾xn+1=b. And so by the induction hypothesis there is an R(lin) derivation of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq152","equation_number":null,"raw_text":"D0↾xn+1=b from D1↾xn+1=b, . . . , Dm↾xn+1=b. So overall, assuming that xn+1 = b, there is an R(lin)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq153","equation_number":null,"raw_text":"derivation of D0↾xn+1=b from D1, . . . , Dm.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq154","equation_number":null,"raw_text":"We now consider the two possible cases: xn+1 = 0 and xn+1 = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq155","equation_number":null,"raw_text":"In case xn+1 = 0, by the above discussion, we can derive D0↾xn+1=0 from D1, . . . , Dm. For every","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq156","equation_number":null,"raw_text":"n+1 times) the equation xn+1 = 0 to the jth equation in D0↾xn+1=0 (see","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq157","equation_number":null,"raw_text":"In case xn+1 = 1, again, by the above discussion, we can derive D0↾xn+1=1 from D1, . . . , Dm. For","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq158","equation_number":null,"raw_text":"n+1 times) the equation xn+1 = 1 to the jth equation in D0↾xn+1=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq159","equation_number":null,"raw_text":"by 1 is performed in D0↾xn+1=1). Again, we obtain precisely D0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq160","equation_number":null,"raw_text":"(xi,1 = 1) ∨· · · ∨(xi,n = 1), for all 1 ≤i ≤m;","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq161","equation_number":null,"raw_text":"(xi,k = 0) ∨(xj,k = 0),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq162","equation_number":null,"raw_text":"n , for any m > n). Thus, we fix m = n+1. In this subsection","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq163","equation_number":null,"raw_text":"(xi,1 + . . . + xi,n = 1) ∨· · · ∨(xi,1 + . . . + xi,n = n)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq164","equation_number":null,"raw_text":"(x1,j + . . . + xm,j = 0) ∨(x1,j + . . . + xm,j = 1),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq165","equation_number":null,"raw_text":"S :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq166","equation_number":null,"raw_text":"(S = m) ∨(S = m + 1) · · · ∨(S = m · n).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq167","equation_number":null,"raw_text":"Proof: For every i ∈[m] fix the abbreviation zi := xi,1 + . . . + xi,n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq168","equation_number":null,"raw_text":"(zi = 1) ∨· · · ∨(zi = n).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq169","equation_number":null,"raw_text":"Consider (z1 = 1) ∨· · · ∨(z1 = n) and (z2 = 1) ∨· · · ∨(z2 = n). By Corollary 7, we can derive","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq170","equation_number":null,"raw_text":"(z1 + z2 = 2) ∨(z1 + z2 = 3) ∨· · · ∨(z1 + z2 = 2n)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq171","equation_number":null,"raw_text":"Now, consider (z3 = 1) ∨· · · ∨(z3 = n) and (25). By Corollary 7 again, from these two lines we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq172","equation_number":null,"raw_text":"(z1 + z2 + z3 = 3) ∨(z1 + z2 + z3 = 4) ∨· · · ∨(z1 + z2 + z3 = 3n) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq173","equation_number":null,"raw_text":"(z1 + . . . + zm = m) ∨(z1 + . . . + zm = m + 1) ∨· · · ∨(z1 + . . . + zm = m · n) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq174","equation_number":null,"raw_text":"(S = 0) ∨· · · ∨(S = n).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq175","equation_number":null,"raw_text":"Proof: For all j ∈[n], fix the abbreviation yj := x1,j + . . . + xm,j.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq176","equation_number":null,"raw_text":"(yj = 0) ∨(yj = 1), for all j ∈[n]. Now the proof is similar to the proof of Lemma 8, except that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq177","equation_number":null,"raw_text":"If y1 = 0 then we can add y1 to the two sums in (y2 = 0) ∨(y2 = 1), and reach (y1 + y2 =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq178","equation_number":null,"raw_text":"0) ∨(y1 + y2 = 1) and if y1 = 1 we can do the same and reach (y1 + y2 = 1) ∨(y1 + y2 = 2). So, by","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq179","equation_number":null,"raw_text":"(y1 + y2 = 0) ∨(y1 + y2 = 1) ∨(y1 + y2 = 2) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq180","equation_number":null,"raw_text":"Now, we consider the three cases in (27): y1 +y2 = 0 or y1 +y2 = 1 or y1 +y2 = 2, and the clause","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq181","equation_number":null,"raw_text":"(y3 = 0) ∨(y3 = 1). We arrive in a similar manner at (y1 + y2 + y3 = 0) ∨· · · ∨(y1 + y2 + y3 = 3).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq182","equation_number":null,"raw_text":"We continue in the same way until we arrive at (S = 0) ∨· · · ∨(S = n).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq183","equation_number":null,"raw_text":"Proof: By Lemmas 13 and 14 above, all we need is to show a polynomial-size refutation of (S =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq184","equation_number":null,"raw_text":"m) ∨· · · ∨(S = m · n) and (S = 0) ∨· · · ∨(S = n).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq185","equation_number":null,"raw_text":"Since n < m, for all 0 ≤k ≤n, if S = k then using the Resolution and Simplification rules we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq186","equation_number":null,"raw_text":"can cut-offall the sums in (S = m) ∨· · · ∨(S = m · n) and arrive at the empty clause. Thus, by","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq187","equation_number":null,"raw_text":"Lemma 5, there is a polynomial-size R0(lin) proof of the empty clause from (S = 0) ∨· · · ∨(S = n)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq188","equation_number":null,"raw_text":"and (S = m) ∨· · · ∨(S = m · n).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq189","equation_number":null,"raw_text":"Let G = (V, E) be a connected","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq190","equation_number":null,"raw_text":"is no sub-graph G′ = (V, E′), where E′ ⊆E, so that for every vertex v ∈V , the number of edges","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq191","equation_number":null,"raw_text":"obtain the Tseitin mod p principle. Let p ≥2 be some fixed integer and let G = (V, E) be a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq192","equation_number":null,"raw_text":"connected undirected r-regular graph with n vertices and no double edges. Let G′ = (V, E′) be the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq193","equation_number":null,"raw_text":"let G = (V, E) be a connected undirected r-regular graph with n vertices and no double edges, and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq194","equation_number":null,"raw_text":"assume that n ≡1 (mod p). Let G′ = (V, E′) be the corresponding directed graph that results","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq195","equation_number":null,"raw_text":"MODp,1(v):=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq196","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq197","equation_number":null,"raw_text":"(xe[v,k],ik = 0)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq198","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq199","equation_number":null,"raw_text":"i=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq200","equation_number":null,"raw_text":"(xe,i = 1) , for all e ∈E′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq201","equation_number":null,"raw_text":"2. (xe,i = 0) ∨(xe,j = 0) , for all i ̸= j ∈{0, . . . , p −1} and all e ∈E′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq202","equation_number":null,"raw_text":"3. (xe,i = 1) ∨(x ̄e,p−i = 0) and (xe,i = 0) ∨(x ̄e,p−i = 1), 7","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq203","equation_number":null,"raw_text":"one i ∈{0, . . . , p −1} so that xe,i = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq204","equation_number":null,"raw_text":"graph G = (V, E) with n vertices and no double edges, such that n ≡1 mod p and r is a constant.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq205","equation_number":null,"raw_text":"As in Definition 6.2, we let G′ = (V, E′) be the corresponding directed graph that results from G","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq206","equation_number":null,"raw_text":"αv :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq207","equation_number":null,"raw_text":"j=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq208","equation_number":null,"raw_text":"i=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq209","equation_number":null,"raw_text":"l=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq210","equation_number":null,"raw_text":"(αv = 1 + l· p) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq211","equation_number":null,"raw_text":"7If i = 0 then x ̄e,p−i denotes x ̄e,0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq212","equation_number":null,"raw_text":"Tv |=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq213","equation_number":null,"raw_text":"l=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq214","equation_number":null,"raw_text":"(αv = 1 + l· p) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq215","equation_number":null,"raw_text":"to the edges coming out of v, that sums up to 1 mod p. This means precisely that αv = 1 mod p","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq216","equation_number":null,"raw_text":"under the assignment σ. Thus, there exists a nonnegative integer k, such that αv = 1 + kp under","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq217","equation_number":null,"raw_text":"l=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq218","equation_number":null,"raw_text":"αv = n + l· p","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq219","equation_number":null,"raw_text":"Formally, we show that for every subset of vertices V ⊆V , with |V| = k, there is a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq220","equation_number":null,"raw_text":"l=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq221","equation_number":null,"raw_text":"αv = k + l· p","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq222","equation_number":null,"raw_text":"and so putting V = V , will conclude the proof.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq223","equation_number":null,"raw_text":"We proceed by induction on the size of V. The base case, |V| = 1, is immediate from Lemma 16.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq224","equation_number":null,"raw_text":"such that |V| = k < n. Let u ∈V \\ V. By Lemma 16, we can derive","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq225","equation_number":null,"raw_text":"l=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq226","equation_number":null,"raw_text":"(αu = 1 + l· p)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq227","equation_number":null,"raw_text":"l=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq228","equation_number":null,"raw_text":"αv = k + 1 + l· p","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq229","equation_number":null,"raw_text":"relation |=.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq230","equation_number":null,"raw_text":"(i · xe,i + (p −i) · x ̄e,p−i = 0) ∨(i · xe,i + (p −i) · x ̄e,p−i = p) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq231","equation_number":null,"raw_text":"(xe,i + x ̄e,p−i = 0) ∨(xe,i + x ̄e,p−i = 2) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq232","equation_number":null,"raw_text":"(xe,i + x ̄e,p−i = 0) ∨(xe,i + x ̄e,p−i = 2) |=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq233","equation_number":null,"raw_text":"(i · xe,i + (p −i) · x ̄e,p−i = 0) ∨(i · xe,i + (p −i) · x ̄e,p−i = p) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq234","equation_number":null,"raw_text":"αv =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq235","equation_number":null,"raw_text":"l=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq236","equation_number":null,"raw_text":"αv = n + l· p","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq237","equation_number":null,"raw_text":"v∈V αv = 1 mod p (since n = 1 mod p). On the other","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq238","equation_number":null,"raw_text":"(i · xe,i + (p −i) · x ̄e,p−i) = 0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq239","equation_number":null,"raw_text":"Theorem 19. Let G = (V, E) be an r-regular graph with n vertices, where r is a constant. Fix","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq240","equation_number":null,"raw_text":"l=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq241","equation_number":null,"raw_text":"αv = n + l· p","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq242","equation_number":null,"raw_text":"(i · xe,i + (p −i) · x ̄e,p−i = p) ∨(i · xe,i + (p −i) · x ̄e,p−i = 0) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq243","equation_number":null,"raw_text":"for every pair of opposite directed edges in G′ = (V, E′) (as in Definition 6.2) and every residue","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq244","equation_number":null,"raw_text":"{0, . . . , p −1}. If i · xe,i + (p −i) · x ̄e,p−i = 0, then subtract this equation successively from every","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq245","equation_number":null,"raw_text":"Otherwise, i·xe,i+(p−i)·x ̄e,p−i = p, then subtract this equation successively from every equation","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq246","equation_number":null,"raw_text":"So overall, in both cases (i · xe,i + (p −i) · x ̄e,p−i = 0 and i · xe,i + (p −i) · x ̄e,p−i = p) we obtained","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq247","equation_number":null,"raw_text":"Therefore, we arrive at a disjunction of equations of the form (0 = γ) for some γ = 1 mod p.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq248","equation_number":null,"raw_text":"the following basic combinatorial idea. Let G = (V, E) be an undirected graph with n vertices and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq249","equation_number":null,"raw_text":"for all i ̸= j ∈[k′], all the vertices in Gi are connected by edges (in E) to all the vertices in","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq250","equation_number":null,"raw_text":"Each variable pi,j, for all i ̸= j ∈[n], is an indicator variable for the fact that there is an edge in","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq251","equation_number":null,"raw_text":"(i) (ql,1 = 1) ∨· · · ∨(ql,n = 1), for all l∈[k]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq252","equation_number":null,"raw_text":"(ii) (ql,i = 0) ∨(ql,j = 0), for all i ̸= j ∈[n], l∈[k]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq253","equation_number":null,"raw_text":"(iii) (ql,i = 0) ∨(ql′,i = 0), for all i ∈[n], l̸= l′ ∈[k]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq254","equation_number":null,"raw_text":"(iv) (ql,i = 0) ∨(ql′,j = 0) ∨(pi,j = 1), for all l̸= l′ ∈[k], i ̸= j ∈[n]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq255","equation_number":null,"raw_text":"(v) (r1,i = 1) ∨· · · ∨(rk′,i = 1), for all i ∈[n]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq256","equation_number":null,"raw_text":"(vi) (rl,i = 0) ∨(rl′,i = 0), for all l̸= l∈[k′], i ∈[n]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq257","equation_number":null,"raw_text":"(vii) (pi,j = 0) ∨(rt,i = 0) ∨(rt,j = 0), for all i ̸= j ∈[n], t ∈[k′]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq258","equation_number":null,"raw_text":"k,k′, when k = √n and k′ = (log n)2/8 log log n. These are rather weak parameters, but","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq259","equation_number":null,"raw_text":"sequence of 2-DNFs D1, D2, . . . , Ds , such that Ds = D, and every Dj is either from K or was","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq260","equation_number":null,"raw_text":"i=1 li and B∨W2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq261","equation_number":null,"raw_text":"i=1 ¬li derive A∨B, where the li’s are (not necessarily distinct)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq262","equation_number":null,"raw_text":"i=1 li.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq263","equation_number":null,"raw_text":"i=1 li , where the li’s are (not necessarily dis-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq264","equation_number":null,"raw_text":"xi into xi, and ¬xi into 1−xi. A 2-term l1 ∧l2 is first transformed into the equation bl1 +bl2 = 2, and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq265","equation_number":null,"raw_text":"then moving the free-terms in the left hand side of bl1 + bl2 = 2 (in case there are such free-terms)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq266","equation_number":null,"raw_text":"right hand side. A disjunction of 2-terms (that is, a 2-DNF) D = W","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq267","equation_number":null,"raw_text":"Theorem 20 ([ABE02]). Let k = √n and k′ = (log n)2/8 log log n. Then ¬cliquen","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq268","equation_number":null,"raw_text":"C(⃗α) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq269","equation_number":null,"raw_text":"C(⃗α) = 0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq270","equation_number":null,"raw_text":"i=1 Ei = ∅. A semantic refutation from E1, . . . , Ek is a sequence D1, . . . , Dm ⊆","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq271","equation_number":null,"raw_text":"{0, 1}N with Dm = ∅and such that for every i ∈[m], Di is either one of the Ej’s or is deduced","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq272","equation_number":null,"raw_text":"Definition 7.2 (Separating circuit). Let U, V ⊆{0, 1}n, where U ∩V = ∅, be two disjoint sets. A","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq273","equation_number":null,"raw_text":"Boolean circuit C with n input variables is said to separate U from V if C(⃗x) = 1 for every ⃗x ∈U,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq274","equation_number":null,"raw_text":"and C(⃗x) = 0 for every ⃗x ∈V. In this case we also say that U and V are separated by C.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq275","equation_number":null,"raw_text":"Let N = n+s+t be fixed from now on. Let A1, . . . , Ak ⊆{0, 1}n+s and let B1, . . . , Bl⊆{0, 1}n+t.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq276","equation_number":null,"raw_text":"UA :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq277","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq278","equation_number":null,"raw_text":"VB :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq279","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq280","equation_number":null,"raw_text":"Definition 7.4 (Communication complexity). Let N = n + s + t and A ⊆{0, 1}N. Let u, v ∈","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq281","equation_number":null,"raw_text":"then find an i ∈[n], such that ui ̸= vi;","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq282","equation_number":null,"raw_text":"̇A :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq283","equation_number":null,"raw_text":"̇B :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq284","equation_number":null,"raw_text":"Proof: Let N = n + s + t (and so eD ∈{0, 1}n+s+t). For the sake of convenience we shall assume","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq285","equation_number":null,"raw_text":"that the N variables in D are partitioned into (pairwise disjoint) three groups ⃗p := (p1 . . . , pn),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq286","equation_number":null,"raw_text":"⃗q := (q1, . . . , qs) and ⃗r := (r1, . . . , rt). Let u, v ∈{0, 1}n, qu ∈{0, 1}s, rv ∈{0, 1}t. Assume that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq287","equation_number":null,"raw_text":"⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r = li","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq288","equation_number":null,"raw_text":"such that ui sets to 1 a literal in P (there ought to exist at least one such i), it must be that ui ̸= vi.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq289","equation_number":null,"raw_text":"⃗a · ⃗p +⃗b · ⃗q + ⃗c · ⃗r = l","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq290","equation_number":null,"raw_text":"in D, such that ⃗a · u +⃗b · qu + ⃗c · rv = l. Note that (by assumption that (v, qu, rv) ̸∈eD) it must","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq291","equation_number":null,"raw_text":"also hold that: ⃗a · v +⃗b · qu + ⃗c · rv ̸= l(and so there is an i ∈[n], such that ui ̸= vi). Player I can","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq292","equation_number":null,"raw_text":"Recall that the left hand side of a linear equation ⃗d · ⃗x = lis called the linear form of the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq293","equation_number":null,"raw_text":"Now both players need to apply a protocol for finding an i ∈[n] such that ui ̸= vi, where","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq294","equation_number":null,"raw_text":"⃗a · ⃗u +⃗b · qu + ⃗c · rv = land ⃗a · ⃗v +⃗b · qu + ⃗c · rv ̸= l. Thus, it remains only to prove the following","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq295","equation_number":null,"raw_text":"bits of communication in order to find an i ∈[n] such that ui ̸= vi (under the above conditions).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq296","equation_number":null,"raw_text":"knows some ⃗x ∈{0, 1}N and Player II knows some other ⃗y ∈{0, 1}N, such that f(⃗x) = 1 while","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq297","equation_number":null,"raw_text":"f(⃗y) = 0. The goal of the game is to find an i ∈[N] such that xi ̸= yi. Denote by CC′(f) the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq298","equation_number":null,"raw_text":"to solve this game.14 Then, for any function f it is known that dp(f) = CC′(f) (see [KW88]).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq299","equation_number":null,"raw_text":"⃗q + ⃗c · ⃗r = lhas Boolean circuit of depth O(log N). In case all the coefficients in ⃗a,⃗b,⃗c are 1, it is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq300","equation_number":null,"raw_text":"13Over all inputs ⃗x, ⃗y such that f(⃗x) = 1 and f(⃗y) = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq301","equation_number":null,"raw_text":"assume that we wish to construct a circuit that computes c · x = l, for some integer l. Then, we take a circuit that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq302","equation_number":null,"raw_text":"computes the function f : {0, 1}c →{0, 1} that outputs the truth value of y1 + . . . + yc = l(thus, in f all coefficients","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq303","equation_number":null,"raw_text":"are 1’s); and to compute c · x = lwe only have to substitute each yi in the circuit with the variable x.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq304","equation_number":null,"raw_text":"Corollary 29. Let k be an integer such that 3 ≤k′ = k −1 and assume that 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq305","equation_number":null,"raw_text":"Corollary 30. Let k = √n and k′ = (log n)2/8 log log n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq306","equation_number":null,"raw_text":"For example, if q = x2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq307","equation_number":null,"raw_text":"4 (for some a ∈F) then M[q] = x1x2 + ax4 .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq308","equation_number":null,"raw_text":"Definition 9.4 (Symmetric polynomial). Given a set of variables X = {x1, . . . , xn}, a symmetric","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq309","equation_number":null,"raw_text":"let Q := {Q1, . . . , Qm} be a collection of multivariate polynomials from the ring of polynomials","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq310","equation_number":null,"raw_text":"a finite sequence π = (p1, ..., pl) of multivariate polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] (each","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq311","equation_number":null,"raw_text":"polynomial pi is interpreted as the polynomial equation pi = 0), where pl= g and for each i ∈[l],","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq312","equation_number":null,"raw_text":"either pi = Qj for some j ∈[m], or pi is a Boolean axiom, or pi was deduced from pj, pk , where","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq313","equation_number":null,"raw_text":"A PCR refutation of Q is a proof of 1 (which is interpreted as 1 = 0) from Q. The number of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq314","equation_number":null,"raw_text":"Note that the Boolean axioms of PCR have only 0, 1 solutions, where ̄xi = 0 if xi = 1 and ̄xi = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq315","equation_number":null,"raw_text":"if xi = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq316","equation_number":null,"raw_text":"Definition 9.6 (Formula Multilinear Calculus (fMC)). Fix a field F and let Q := {Q1, . . . , Qm} be","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq317","equation_number":null,"raw_text":"π = (p1, ..., pl) of multilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn] , such that pl= g and for","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq318","equation_number":null,"raw_text":"each i ∈[l], either pi = Qj for some j ∈[m], or pi is a Boolean axiom of fMC, or pi was deduced","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq319","equation_number":null,"raw_text":"an fMC proof is interpreted as the polynomial equation pi = 0.) An fMC refutation of Q is a proof","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq320","equation_number":null,"raw_text":"of 1 (which is interpreted as 1 = 0) from Q. The size of an fMC proof π is defined as the total","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq321","equation_number":null,"raw_text":"Note that the Boolean axioms have only 0, 1 solutions, where ̄xi = 0 if xi = 1 and ̄xi = 1 if","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq322","equation_number":null,"raw_text":"xi = 0, for each 1 ≤i ≤n .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq323","equation_number":null,"raw_text":"n xn = a(1)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq324","equation_number":null,"raw_text":"n xn = a(t)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq325","equation_number":null,"raw_text":"Proposition 3. Let π = (D1, . . . , Dl) be an R(lin) proof sequence of Dl, from some collection","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq326","equation_number":null,"raw_text":"9.8. An R(lin) Boolean axiom (xi = 0) ∨(xi = 1) is translated into xi · (xi −1) which is already a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq327","equation_number":null,"raw_text":"Assume that Di = Dj ∨L was derived from Dj using the Weakening inference rule of R(lin),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq328","equation_number":null,"raw_text":"where j < i ≤land L is some linear equation. Then, by Claim 5, bDi = bDj · bL can be derived from","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq329","equation_number":null,"raw_text":"Assume that Di was derived from Dj where Dj is Di ∨(0 = k), using the Simplification inference","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq330","equation_number":null,"raw_text":"bDj = bDi · −k by multiplying with −k−1 (via the Addition rule of PCR).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq331","equation_number":null,"raw_text":"tions of linear equations and assume that A ∨B ∨((⃗a + ⃗b) · ⃗x = a0 + b0) was derived in π from","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq332","equation_number":null,"raw_text":"A ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0) (the case where A ∨B ∨((⃗a −⃗b) · ⃗x = a0 −b0) was derived from","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq333","equation_number":null,"raw_text":"A ∨(⃗a · ⃗x = a0) and B ∨(⃗b · ⃗x = b0), is similar).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq334","equation_number":null,"raw_text":"Corollary 32. Let π = D1, . . . , Dlbe an R0(lin) proof of Dl, and let s be the maximal size of an","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq335","equation_number":null,"raw_text":"multilinear polynomials from F[x1, . . . , xn, ̄x1, . . . , ̄xn]. Let π = (p1, . . . , pm) be a PCR refutation","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq336","equation_number":null,"raw_text":"size of all formulas Φi, that is, s = Σm","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq337","equation_number":null,"raw_text":"i=1|Φi|, and let d ≥2 be the maximal depth of all formulas","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq338","equation_number":null,"raw_text":"i=1|Xi| = n . Let f1, . . . , fc be c","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq339","equation_number":null,"raw_text":"Lemma 34. Let D be an R0(lin)-line with n variables and let p = bD (see Definition 9.8). Then,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq340","equation_number":null,"raw_text":"Proof: Assume that the underlying variables of D are ⃗x = x1 . . . , xn. By the definition of an","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq341","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq342","equation_number":null,"raw_text":"(⃗a · ⃗x = li) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq343","equation_number":null,"raw_text":"zi := b ·","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq344","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq345","equation_number":null,"raw_text":"r1+...+rd+1=m","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq346","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq347","equation_number":null,"raw_text":"αr :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq348","equation_number":null,"raw_text":"| U|=r","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq349","equation_number":null,"raw_text":"j=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq350","equation_number":null,"raw_text":"r1+...+rd+1=mj","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq351","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq352","equation_number":null,"raw_text":"Definition 10.1 (R(CP*)). Let K := {K1, . . . , Km} be a collection of disjunctions of linear in-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq353","equation_number":null,"raw_text":"disjunction of linear inequalities D is a finite sequence π = (D1, ..., Dl) of disjunctions of linear","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq354","equation_number":null,"raw_text":"inequalities, such that Dl= D and for each i ∈[l]: either Di = Kj for some j ∈[m]; or Di is one","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq355","equation_number":null,"raw_text":"(⃗a · ⃗x = a0) ∨(⃗a · ⃗x = a0 + 1) ∨· · · ∨(⃗a · ⃗x = a0 + k) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq356","equation_number":null,"raw_text":"where k is such that a0 + k equals the sum of all positive coefficients in ⃗a, that is, a0 + k =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq357","equation_number":null,"raw_text":"⃗x∈{0,1}n (⃗a · ⃗x) (in case the sum of all positive coefficients in ⃗a is less than a0, then we put k = 0).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq358","equation_number":null,"raw_text":"An inequality with no variables of the form 0 ≥a0 is translated into 0 = a0 in case it is false (that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq359","equation_number":null,"raw_text":"is, in case 0 < a0), and into 0 = 0 in case it is true (that is, in case 0 ≥a0). Note that since the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq360","equation_number":null,"raw_text":"A collection K := {K1, . . . , Km} of disjunctions of linear","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq361","equation_number":null,"raw_text":"R(CP*) axiom number (1): xi ≥0 translates into the R(lin) axiom (xi = 0) ∨(xi = 1).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq362","equation_number":null,"raw_text":"R(CP*) axiom number (2): −xi ≥−1, translates into (−xi = −1) ∨(−xi = 0).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq363","equation_number":null,"raw_text":"Boolean axiom (xi = 1) ∨(xi = 0) of R(lin), one can derive with a constant-size R(lin) proof the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq364","equation_number":null,"raw_text":"line (−xi = −1)∨(−xi = 0) (for instance, by subtracting twice each equation in (xi = 1)∨(xi = 0)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq365","equation_number":null,"raw_text":"b=a0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq366","equation_number":null,"raw_text":"(⃗a · ⃗x = b) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq367","equation_number":null,"raw_text":"b=1−a0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq368","equation_number":null,"raw_text":"(−⃗a · ⃗x = b) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq369","equation_number":null,"raw_text":"all negative coefficients in ⃗a). Note that one can always flip the sign of any equation ⃗a · ⃗x = b in","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq370","equation_number":null,"raw_text":"R(lin). This is done, for instance, by subtracting twice ⃗a · ⃗x = b from itself. So overall R(CP*)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq371","equation_number":null,"raw_text":"b=a0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq372","equation_number":null,"raw_text":"(⃗a · ⃗x = b) ∨","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq373","equation_number":null,"raw_text":"b=1−a0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq374","equation_number":null,"raw_text":"(−⃗a · ⃗x = b) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq375","equation_number":null,"raw_text":"b=−f","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq376","equation_number":null,"raw_text":"(⃗a · ⃗x = b) ∨","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq377","equation_number":null,"raw_text":"b=a0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq378","equation_number":null,"raw_text":"(⃗a · ⃗x = b) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq379","equation_number":null,"raw_text":"Let A′ := {−f, −f + 1, . . . , a0 −1, a0, a0 + 1, . . . , h} and let A be the set of all possible values that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq380","equation_number":null,"raw_text":"α∈A(⃗a · ⃗x = α).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq381","equation_number":null,"raw_text":"α∈A′(⃗a · ⃗x = α) which is equal to (52).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq382","equation_number":null,"raw_text":"ber that 0 ≥1 translates into 0 = 1 under our translation scheme).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq383","equation_number":null,"raw_text":"(c⃗a · ⃗x = a0) ∨(c⃗a · ⃗x = a0 + 1) ∨. . . ∨(c⃗a · ⃗x = a0 + r) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq384","equation_number":null,"raw_text":"(⃗a · ⃗x = α) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq385","equation_number":null,"raw_text":"We now use (53) to cut-offfrom (54) all equations (⃗a · ⃗x = β) for all β < ⌈a0/c⌉(this will give","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq386","equation_number":null,"raw_text":"us the desired disjunction of linear equations). Consider the equation (⃗a · ⃗x = β) in (54) for some","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq387","equation_number":null,"raw_text":"(c⃗a · ⃗x = cβ) ∨","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq388","equation_number":null,"raw_text":"(⃗a · ⃗x = α) .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq389","equation_number":null,"raw_text":"Since β is an integer and β < ⌈a0/c⌉, we have cβ < a0. Thus, the equation (c⃗a · ⃗x = cβ) does","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq390","equation_number":null,"raw_text":"not appear in (53). We can then successively resolve (c⃗a · ⃗x = cβ) in (55) with each equation","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq391","equation_number":null,"raw_text":"(c⃗a · ⃗x = a0), . . . , (c⃗a · ⃗x = a0 + r) in (53). Hence, we arrive at W","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq392","equation_number":null,"raw_text":"α∈A\\{β} (⃗a · ⃗x = α). Overall, we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq393","equation_number":null,"raw_text":"can cut-offall equations (⃗a · ⃗x = β), for β < ⌈a0/c⌉, from (54). We then get the disjunction","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq394","equation_number":null,"raw_text":"(⃗a · ⃗x = α) ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq395","equation_number":null,"raw_text":"(1) If ⃗α is an assignment to ⃗p and ⃗β is an assignment to ⃗q such that A(⃗α, ⃗β) = 1, then for any","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq396","equation_number":null,"raw_text":"assignment ⃗α′ ≥⃗α it holds that A(⃗α′, ⃗β) = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq397","equation_number":null,"raw_text":"(2) If ⃗α is an assignment to ⃗p and ⃗β is an assignment to ⃗r such that B(⃗α, ⃗β) = 1, then for any","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq398","equation_number":null,"raw_text":"assignment ⃗α′ ≤⃗α it holds that B(⃗α′, ⃗β) = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":148826,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}