paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006227 | We have to check the non-degeneracy axiom. Let us denote by MATH the matrix whose MATH entry is equal to the value of the MATH-framed NAME link with linking -REF and coloring of the components MATH. Then we have that MATH . We deduce that the MATH entry of the matrix MATH is equal to the invariant of the colored link d... |
math/0006227 | Let MATH be a maximal left ideal of MATH. Suppose that MATH does not contains MATH, then, using maximality of MATH, we get that the left ideal MATH is equal to MATH. We further have that MATH, MATH, MATH, and so MATH is in the ideal MATH, which contradicts the hypothesis. |
math/0006227 | We have to prove the dominating property. The proof is the same in all cases, so we will use the notation MATH, MATH for MATH, MATH where MATH is one of the categories mentioned in the claim. It is enough to show that the identity morphism of the object MATH decomposes using the simple objects in MATH. This is done by ... |
math/0006227 | An object MATH is transparent if and only if for any (non-negligible) MATH in the branching formula for MATH, the braiding coefficient is equal to one. Indeed, if all braiding coefficients are equal to one, by summing over MATH the left hand sides and right hand sides of REF and applying the branching formula we have M... |
math/0006227 | In the semi-simple category MATH we can decompose the identity of the object MATH as we did in REF . MATH . Here all simple subobjects MATH are transparent and hence have dimension MATH. By comparing the dimensions we see that there is only one such MATH with multiplicity MATH. It should be trivial, because the duality... |
math/0006227 | Let MATH. We show that MATH is a set of dominating simple objects. As in the proof of REF , we decompose the tensor products MATH, for MATH. The sublte point here is that some idempotent in the branching formula for the partition MATH (that is, MATH) is missing. We will avoid this difficulty by using the isomorphism in... |
math/0006227 | It is sufficient to show that the transparent simple objects have order REF, that is, any non-trivial transparent simple object MATH satisfies the equation: MATH. Clearly, MATH contains the trivial object and decomposes into a sum of transparent simple ones. Comparing the quantum dimensions on the left and right hand s... |
math/0006227 | We substitute REF and calculations of REF into REF . Let us consider the MATH case in details. Here MATH. By REF and the calculations of REF we have MATH . Furthermore, MATH . Here we used the bijection MATH sending MATH to MATH. Substituting the last two formulas into REF we get the result. For the third formula we us... |
math/0006227 | Let us call the graded sliding property the equality drawn in REF by replacing MATH on the left-hand side by MATH and MATH on the right-hand side by MATH with MATH. The proof of this identity can be adapted from the one of this proposition. Using twice the graded sliding property, we can see that the morphism drawn bel... |
math/0006227 | A colored MATH-component link invariant of a pre-modular category MATH with MATH as a representative set of simple objects can be considered as a multilinear function from MATH to MATH. Here we supply MATH with a ring structure by considering direct sums and tensor products. It is easy to see from the previous discussi... |
math/0006233 | Using REF , an evident inequality introducing an auxiliary object MATH, and twice REF again: MATH . |
math/0006233 | Rewrite the expectation MATH . Define MATH and MATH to obtain MATH . Given the program that computes MATH, we can approximate MATH by a MATH, and similarly for MATH. That is, the distributions MATH REF are lower semicomputable, and by REF , therefore, they are computable. It is known that for every computable probabili... |
math/0006233 | The lemma follows from the definition of conditional algorithic mutual information, REF , if we show that MATH, where the MATH term implicit in the MATH sign is independent of MATH. Equip the reference universal prefix machine, with a MATH length program to compute a NAME code from the auxiliary table of probabilities.... |
math/0006233 | By the triangle inequality, MATH . Thus, MATH . |
math/0006233 | It suffices to prove the case MATH and apply it twice. The proof is by replacing the program MATH that computes a particular string MATH from a particular MATH in REF . There, MATH possibly depends on MATH and MATH. Replace it by a program MATH that first computes MATH from MATH, followed by computing a recursive funct... |
math/0006233 | We have MATH . The negative logarithm of the left-hand side in the theorem is therefore MATH . Using REF , and the conditional additivity REF , this is MATH . |
math/0006233 | By REF and by typicality MATH. |
math/0006233 | Writing MATH, since MATH by REF , we have MATH. Hence, it suffices to show MATH. Now, from an implicit description MATH we can find the value MATH. To recover MATH we only require an extra MATH bits apart from MATH. Therefore, MATH. This reduces what we have to show to MATH which is asserted by REF . |
math/0006233 | A set MATH is optimal iff REF holds with equalities. Rewriting MATH the first inequality becomes an equality iff MATH, and the second inequality becomes an equality iff MATH (that is, MATH is a typical set). |
math/0006233 | If MATH is optimal for MATH, then MATH. From MATH we can find both MATH and MATH and hence MATH, that is, MATH. We have MATH by REF , respectively. This proves the first property. Substitution of MATH in the expression of REF proves the second property. |
math/0006233 | The lower bound is easiest. Denote by MATH of length MATH a shortest program for MATH. Every string MATH of length MATH can be described in a self-delimiting manner by prefixing it with MATH, hence MATH. For a large enough constant MATH, we have MATH and hence there are MATH strings that are in MATH. For the upper boun... |
math/0006233 | From MATH we can compute both MATH and MATH and recursively enumerate MATH. Since also MATH REF , the string MATH plus a fixed program is an implicit description of MATH so that MATH. Hence, MATH and since MATH is the shortest description by definition equality REF holds. That is, MATH is optimal for MATH. By REF MATH ... |
math/0006233 | The set MATH is a sufficient statistic for MATH since MATH. It is minimal since by REF we must have MATH for implicit, and hence for explicit sufficient statistics. It is evident that MATH is explicit: MATH. |
math/0006233 | We can describe MATH by MATH where MATH is the index of MATH in the enumeration of MATH. Moreover, MATH explicitly describes the set MATH. Namely, using MATH we can recursively enumerate MATH. At some point the first string MATH is enumerated (index MATH). By REF and MATH. Therefore, in the enumeration of MATH eventual... |
math/0006233 | If MATH is explicitly optimal for MATH, then we can find MATH from MATH (as in the proof of REF ), and given MATH and MATH we find MATH as in REF . Hence, given MATH, we can enumerate MATH and determine the maximal index MATH of a MATH. Since also MATH, the numbers MATH have a maximal common prefix MATH. Write MATH wit... |
math/0006233 | For every MATH in the set defined by the left-hand side of the inequality, we have MATH, and the length of continuation of MATH to the total padded index of MATH is MATH. Moreover, all these indices share the same first MATH bits. This proves the lemma. |
math/0006233 | Let us prove first MATH . By the NAME inequality, we have, with MATH, MATH since MATH is in REF correspondence with the prefix programs of length MATH. Hence MATH . For the statement of the lemma, we have MATH where in the last inequality we used REF . |
math/0006233 | When we talk about complexity with MATH in the condition, we use a NAME machine with MATH as an ``oracle". With the help of MATH, we can compute MATH, and so we can define the following new semicomputable (relative to MATH) function with MATH: MATH . We have, using REF and defining MATH so that MATH for MATH: MATH . Su... |
math/0006233 | Let MATH be a set with MATH and assume MATH. Tacitly understanding MATH in the conditions, and using the additivity REF , MATH . Therefore MATH . |
math/0006233 | Denote the conditional universal probability as MATH. We write MATH to indicate sets MATH that satisfy MATH. For every MATH, let us define a function over all strings MATH of length MATH as follows: MATH . The following lemma shows that this function of MATH is a semimeasure. We have MATH . We have MATH . There are con... |
math/0006233 | : Let MATH be one of the non-stochastic objects of which the existence is established by REF . Choose MATH with MATH so that the set MATH has complexity MATH and MATH has randomness deficiency MATH with respect to MATH. Because MATH is non-stochastic, this yields MATH. For every MATH we have MATH. Together it follows t... |
math/0006233 | By REF and by typicality MATH. |
math/0006233 | Writing MATH, since MATH by REF , we have MATH. Hence, it suffices to show MATH. Now, from an implicit description MATH we can find the value MATH. To recover MATH from MATH, we at most require an extra MATH bits. That is, MATH. This reduces what we have to show to MATH which is asserted by REF . This shows the first s... |
math/0006233 | A distribution MATH is optimal iff REF holds with equalities. Rewriting MATH the first inequality becomes an equality iff MATH, and the second inequality becomes an equality iff MATH (that is, MATH is a typical distribution). |
math/0006233 | If MATH is optimal for MATH, then MATH. From MATH we can find both MATH and MATH, and hence MATH, that is, MATH. We have MATH by REF , respectively. This proves the first property. Substitution of MATH in the expression of REF proves the second property. |
math/0006233 | Clearly, REF implies REF . We show that both REF implies REF implies REF : By REF we have MATH where we absorb a MATH additive term in the MATH sign. Together with REF implies MATH and vice versa REF together with REF implies REF . |
math/0006233 | (If) By assumption, MATH. Rearrange and add MATH (by typicality) to the right-hand side to obtain MATH. Substitute according to MATH (by sufficiency) in the right-hand side, and subsequently subtract MATH from both sides, to obtain REF . (Only If) Reverse the proof of the (If) case. |
math/0006233 | If MATH is a probabilistic sufficient statistic, then, by REF , equality of MATH-expectations REF holds. However, it is still consistent with this to have large positive and negative differences MATH for different MATH arguments, such that these differences cancel each other. This problem is resolved by appeal to the a... |
math/0006233 | By assumption, using REF , there is a positive constant MATH, such that, MATH . Therefore, MATH . On the other hand, since MATH we obtain MATH . Altogether, this implies REF , and by REF , the theorem. |
math/0006234 | First consider distinct fixed vertices connected by a path of nonfixed vertices. Since fixed vertices are the points where paths move between squares of parity MATH, such a path must be contained in a single unit square of parity MATH; moreover, for a unit square to contain two fixed vertices, it must be contained in M... |
nlin/0006024 | It suffices to show that there exists some time MATH such that MATH . Then it follows from the maximum principle that MATH for all MATH, and hence MATH satisfies the linear advection-diffusion equation MATH for MATH, which implies REF . We actually show that REF holds at MATH for sufficiently large MATH. Recall that MA... |
nlin/0006024 | Let us define the set MATH . As before, it suffices to show that solution of REF satisfies MATH and that is what we will do. First, we split the initial data for REF into two parts: one supported on a strip MATH, containing the flat part of MATH and another supported outside it. We will choose MATH such that any soluti... |
nlin/0006024 | The set of all profiles MATH satisfying NAME is dense in MATH so by REF the set of strongly quenching profiles is dense. To complete the proof, we will show that if MATH satisfies NAME, then there exists MATH such that if MATH then MATH is strongly quenching. From the proof of REF , we know that there exists a constant... |
nlin/0006024 | The proof of REF proceeds in several steps. We consider the initial data satisfying REF . First we find MATH such that there exists a MATH function MATH such that MATH, and MATH and MATH vanishes on the boundary of the disc of radius MATH centered at the point MATH: MATH . Then in the system of coordinates that moves w... |
nlin/0006024 | Let MATH be an initial data as in REF such that MATH with the function MATH constructed in the proof of REF and given explicitly by REF . Then we have MATH and therefore REF implies that MATH because MATH may not go to zero as MATH. |
nlin/0006024 | Let MATH be solutions of MATH and MATH with the same initial data MATH . The function MATH is assumed to be front-like and monotonic. That is, MATH for MATH, MATH for MATH and MATH . Then by maximum principle applied to the equation for MATH we have MATH and hence MATH . Therefore by maximum principle we have MATH . Ho... |
nlin/0006024 | In order to finish the proof of REF we apply REF to MATH given by REF , and observe that MATH satisfies the estimate in REF according to REF . |
nlin/0006024 | The first statement is a direct corollary of REF . The second statement is the content of REF . |
nlin/0006024 | We will show that solution of REF with MATH replaced by a KPP nonlinearity MATH and the same initial data drops below MATH before time MATH. That will imply REF . Choose MATH such that MATH (such MATH exists since MATH is NAME continuous) and let MATH be the solution of MATH with initial data MATH and periodic boundary... |
quant-ph/0006076 | First, apply NAME 's orthonormalization to the horizontal lift MATH of MATH, to obtain the orthonormal basis MATH such that MATH is a subset of the real span of MATH. We immerse NAME space MATH into MATH as MATH where MATH is an orthonormal basis in MATH. Then, letting MATH be a projection valued measure which is obtai... |
quant-ph/0006076 | Suppose that any member of the manifold MATH in MATH is invariant by the antiunitary operator MATH, and let MATH. Then, we have MATH . Conversely, if MATH is real for any MATH, by NAME 's orthonormalization, we can obtain the orthonormal basis MATH such that MATH is subset of the real span of MATH, which means any memb... |
quant-ph/0006125 | Consider the real-valued function MATH defined for unit vectors MATH lying in the unit sphere MATH in MATH . As MATH varies over the compact space MATH this function attains a maximum at some point MATH . Let MATH be any orthonormal basis of MATH containing MATH and expand MATH as in the statement of the theorem. Since... |
quant-ph/0006125 | Consider the real-valued function MATH defined for unit vectors MATH lying in the unit sphere MATH in MATH. As MATH varies over the compact space MATH, this function attains a maximum at some point MATH. Let MATH be any orthonormal basis of MATH containing MATH, and expand MATH as in the statement of the theorem. Since... |
quant-ph/0006125 | REF for a stationary value of MATH become MATH where MATH, MATH and MATH. Clearly there is a solution MATH. We have to show that any other solution has MATH. Using REF to eliminate MATH and MATH, REF become MATH . For MATH, it follows that MATH since MATH. Now MATH which is positive if MATH (unless MATH), and the gradi... |
cond-mat/0007425 | We calculate MATH where we have set MATH. Note that MATH and that MATH satisfies all the assumptions in REF. We then conclude from REF that the NAME transform of the function MATH is non-negative, where MATH is a function such that MATH as MATH. [The detailed bounds from CITE show that we may in fact choose MATH, since... |
cond-mat/0007425 | If MATH is a symmetric function. Then MATH where MATH . The lemma follows since it is clear that MATH . |
cond-mat/0007425 | Simply note that MATH and now use the same arguments as before. |
cond-mat/0007425 | Choose MATH and MATH. Then we may assume that MATH since MATH is large. From REF we see immediately that MATH . The corollary follows since MATH and with the given choice of MATH and MATH it is easy to see that MATH. |
cond-mat/0007425 | We simply choose MATH in REF . This is allowed since MATH is ensured from the assumption that MATH is large. We then obtain MATH . The bound on MATH follows since the bound on the gap REF implies that MATH. |
cond-mat/0007425 | The terms containing MATH are MATH using the commutation relation MATH. |
cond-mat/0007425 | The terms containing MATH or MATH are MATH . Note that MATH commutes with MATH. We have that MATH . Hence MATH . Since MATH we obtain the operator inequality MATH and the lemma follows. |
cond-mat/0007425 | The operator MATH with integral kernel MATH is a non-negative NAME operator on MATH with norm less than MATH. Denote the eigenvalues of MATH by MATH, MATH and corresponding orthonormal eigenfunctions by MATH. We may assume that these functions are real. The eigenvalues satisfy MATH. We then have MATH . The identity REF... |
cond-mat/0007425 | The terms containing MATH, MATH,MATH, or MATH are MATH . In the last term we have used the representation REF and the commutation relation MATH. For all MATH we get that the above expression is bounded below by MATH . The bound REF follows from REF . The second bound REF follows in the same way if we notice that the te... |
cond-mat/0007425 | The terms containing MATH, MATH,MATH, or MATH are MATH . Using that MATH we may write this as MATH . The lemma now follows from REF . |
cond-mat/0007425 | It is clear that for all MATH and hence MATH . Now simply use this with MATH. |
cond-mat/0007425 | Let MATH, with MATH, be a parameter to be chosen below. Recall that MATH is the parameter used in the definition of MATH in REF. Then since MATH we have MATH where the last inequality follows from REF. We can now repeat this calculation to get MATH . If we therefore use REF and recall that MATH we arrive at MATH . Note... |
cond-mat/0007425 | We may complete the square MATH if MATH . We choose the solution MATH . Hence MATH . |
cond-mat/0007425 | We consider a state with MATH. Then MATH. We shall use REF . Note first that MATH by REF and the fact that MATH. We may of course rewrite MATH. By REF we have MATH where MATH . Since MATH we have that MATH . Note that MATH . The lemma now follows from REF by a simple change of variables in the MATH integral. |
cond-mat/0007425 | REF holds for all MATH hence also if we had replaced MATH by MATH in this case we get REF . The integral MATH satisfies the bound MATH . By REF we may assume that MATH. Hence MATH is bounded by a constant as long as MATH and MATH are sufficiently large and MATH is sufficiently small (which also ensures that MATH is clo... |
cond-mat/0007425 | The lower bound follows from REF . To prove the upper bound on MATH we choose MATH (the maximally allowed value) and MATH, where we shall choose MATH sufficiently small, in particular MATH. We then have that MATH is large. Moreover MATH for some constant MATH and we get from REF that MATH where we have again used that ... |
cond-mat/0007425 | Inserting the bound MATH into REF gives MATH . We now choose MATH and MATH, where we shall choose MATH below, independently of MATH, MATH, and MATH. Note that since MATH is small we may assume that MATH as required and since MATH is small we may assume that MATH. Moreover MATH and MATH. Hence, since MATH (see REF), we ... |
cond-mat/0007425 | As explained above we choose MATH to be of order MATH. We then choose MATH as explained above. Then REF holds. We also know that the possible MATH values of MATH range in an interval of length MATH. We do not know however, where this interval is located. REF will allow us to say more about the location of the interval.... |
cond-mat/0007425 | If we replace all the operators MATH and MATH with MATH in the Foldy Hamiltonian by MATH and MATH we get a unitarily equivalent operator. This operator however differs from the Hamiltonian MATH only by a change of sign on the part that we denoted MATH. Since both operators satisfy the bound in REF we conclude that MATH... |
cond-mat/0007425 | We assume MATH. We proceed as in the beginning of REF, but we now use REF instead of REF . We then get MATH . If we now use REF and the facts that MATH, MATH, and MATH we see with appropriate choices of MATH and MATH that MATH . If we finally insert the choices of MATH and MATH and use REF we arrive at the bound in the... |
cond-mat/0007425 | According to REF we may assume that MATH where MATH is at least as big as the constant in REF . We still assume that MATH. We begin by bounding MATH and MATH using REF . We have from REF that MATH . In order to bound MATH we shall use REF . Together with REF this gives (choosing MATH say) MATH . Inserting the choices f... |
cond-mat/0007425 | We have accumulated various errors and we want to show that they can all be made small. There are basically two parameters that can be adjusted, MATH and MATH. Instead of MATH it is convenient to use MATH. We shall choose MATH as a function of MATH such that MATH as MATH. From REF we know that for some fixed MATH. Henc... |
cond-mat/0007425 | It is convenient to extend the matrix MATH to all MATH by defining MATH unless MATH. Similarly, we extend the vector MATH and we define the numbers MATH and the matrix MATH to be zero when MATH. We shall give the construction for MATH odd, the MATH even case being similar. For MATH set MATH if MATH and MATH otherwise. ... |
cond-mat/0007425 | We calculate the commutator MATH . Likewise we calculate the double commutator MATH . Note that MATH and thus the first term above is positive. We claim that MATH . To see this we simply calculate MATH . The last two terms are bounded by MATH. For the first term we have by the NAME inequality for operators, MATH, for a... |
cs/0007002 | Given MATH, MATH, and MATH three subsets of MATH, with MATH, let MATH be an element of MATH and MATH. CASE: Given any real MATH, we have by definition of Inner that MATH. By definition of Outer, we have that MATH, which leads to MATH. Consequently, MATH; CASE: For any MATH, MATH implies MATH, by definition of Inner; si... |
cs/0007002 | Immediate consequence of Inner and Outer definitions. |
cs/0007002 | Consider Line MATH of REF. By completeness of MATH, MATH. Consequently, MATH. Since the set MATH returned eventually is the union of all the sets computed on Line MATH, we have that MATH. We can use the dual reasoning to prove that MATH. |
cs/0007002 | Immediate. |
cs/0007002 | Termination of REF depends on a reasonable choice for the functions NAME and MATH. More specifically, termination is ensured whenever MATH creates boxes MATH that are all strictly smaller than MATH, and when NAME checks for the size of all dimensions of MATH but the one of MATH being smaller than some threshold that is... |
cs/0007002 | For each constraint MATH, we consider only the boxes that were put in MATH by the preceding constraint (Lines MATH and MATH). By soundness of REF respectively, REF, when considering constraint MATH, MATH then contains on line MATH only the boxes MATH verifying MATH (respectively, MATH). Once again, by soundness of REF ... |
cs/0007015 | The invariance of MATH is verified by observing that none of REF - SREF change any variable value at a timer-final state. Convergence is demonstrated by a sequence of claims about an arbitrary computation MATH. Let MATH be a suffix of MATH beginning following the second round of MATH; by definition, MATH is a based com... |
cs/0007015 | Provided MATH, arguments in the proof of REF show that for each perturbed region MATH, some process MATH executes SREF within the first round, where MATH satisfies either MATH or MATH for some MATH. REF then implies that within MATH additional rounds, each MATH satisfies MATH. Each unperturbed process clock variable re... |
cs/0007015 | REF directly shows that perturbed processes assign clock variables to at most MATH within the first MATH rounds. REF establishes that MATH increases its clock to at least MATH after MATH rounds following the execution of REF establishes that for each perturbed region, some process MATH either within or neighboring the ... |
cs/0007015 | Consider a MATH-faulty initial state. If MATH, then from MATH and REF , the system stabilizes to MATH and hence MATH in MATH rounds, which proves the conclusion. The remaining case is MATH for a MATH-faulty initial state. For this case, we first show that any faulty process MATH is time accurate within the first MATH r... |
cs/0007015 | The essence of the proof is that neither SREF nor SREF increment a clock to a value two greater than any neighbor. Since REF involves x variables, the effect of REF requires examination. Reading a register to assign an x variable only increases the accuracy of the image variable; in particular, given MATH as a precondi... |
cs/0007015 | By definition of a rising computation, each process has a lower bound on neighboring clock variables in its x variable, because clock values cannot decrease in a reset-free computation. Suppose MATH is the first of MATH to increment its clock. A precondition for this step is MATH, which implies MATH, which in turn impl... |
cs/0007015 | By REF , neighboring MATH establish MATH at or before MATH; by REF such processes continue to satisfy this property for the remainder of the reset-free computation segment. |
cs/0007015 | Because we consider a based computation, and not a rising computation in this lemma, the invariance of MATH stated in REF is not applicable. Note that MATH holds at a smooth state. The invariance of smoothness is therefore verified from the conditions of REF - SREF, since no gap exists at a smooth state and SREF preser... |
cs/0007015 | In the first round, MATH reads neighboring clock values and detects local minimality. If MATH increments in this round, the lemma holds; and if MATH does not increment, it writes its clock and detects cEcho in the next round, and local minimality implies MATH will increment MATH either by REF or SREF. |
cs/0007015 | Observe that MATH holds at least until some clock exceeds MATH so that SREF can execute. And MATH, so process MATH does not execute the assignment of REF. This implies that the computation is reset-free until some clock obtains the value exceeding MATH. REF implies that each minimal clock value increments in any pair o... |
cs/0007015 | REF shows that the computation contains a smooth state, so the obligation here is to show the MATH time bound. By REF each minimal clock value increments at least once in any two consecutive rounds, so within MATH rounds, some clock attains the value MATH, establishing a smooth state. |
cs/0007015 | The proof begins with a claim on the first MATH rounds of the based computation: within MATH rounds there is a state satisfying MATH . The claim is shown by induction. The first state of the computation satisfies the claim for MATH as the base case. Suppose the claim holds for MATH and consider two processes MATH and M... |
cs/0007015 | Let MATH be a state satisfying MATH. By REF such a state MATH occurs with MATH rounds of the based computation. So long as every clock is at most MATH, no step subsequent to MATH decreases a w variable; and if no w variable is reset by REF in a consecutive pair of rounds, then the minimum value of the set of w variable... |
cs/0007015 | REF by induction on MATH. For MATH let MATH to satisfy the base case. For MATH, we have MATH by hypothesis. By the Echo conditions of REF, SREF and SREF, the clock and w values of MATH remain at most MATH until all neighbors either REF complete cycles that observe these values and write corresponding images to output r... |
cs/0007015 | Note that the lemma holds trivially if the initial state is MATH-perturbed. For the case MATH we use induction on MATH and nested induction on MATH and suppose a based computation. For the base case MATH consider MATH. Since MATH is perturbed, there is a path MATH from MATH to some perturbed MATH (possibly through MATH... |
cs/0007015 | REF states that a process executes SREF at most once in a computation, so it suffices to show that MATH either does not execute SREF or executes SREF within the first MATH rounds. If MATH is unperturbed, REF implies the result. If MATH is perturbed, then for some perturbed region MATH containing MATH, there is an unper... |
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