paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/9910010 | MATH is obvious. Using the techniques of REF we have MATH, for all MATH and some fixed MATH, hence MATH. |
cs/9910010 | MATH: Let MATH, MATH be the matrix defined by MATH, MATH be the row vector whose MATH-th entry is MATH. Note that the MATH-th row of MATH is MATH times MATH. Thus all rows of MATH are scalar multiples of each other, hence MATH has rank REF. Since MATH and MATH, we have MATH. MATH: Suppose MATH. Then there are MATH columns MATH in MATH which span the column space of MATH. Let MATH be the MATH matrix that has these MATH as columns. Let MATH be the MATH matrix whose MATH-th column is formed by the MATH coefficients of the MATH-th column of MATH when written out as a linear combination of MATH. Then MATH, hence MATH . Defining functions MATH by MATH and MATH, we have MATH. |
cs/9910010 | Clearly MATH. To prove the converse, consider MATH, the unique polynomial for the disjointness function. Note that this polynomial contains all and only even monomials (with coefficients MATH). Since DISJ has rank MATH, it follows from REF that DISJ cannot be decomposed in fewer then MATH terms. We will show how a decomposition of MATH with MATH would give rise to a decomposition of DISJ with fewer than MATH terms. Suppose we can write MATH . Let MATH be some even monomial in MATH and suppose the monomial MATH in DISJ has coefficient MATH. Now whenever MATH occurs in some MATH, replace that MATH by MATH. Using the fact that MATH contains only even monomials, it is not hard to see that the new polynomial obtained in this way is the same as MATH, except that the monomial MATH is replaced by MATH. Doing this sequentially for all monomials in MATH, we end up with a polynomial MATH (with MATH monomials and MATH) which is a subpolynomial of DISJ, in the sense that each monomial in MATH also occurs with the same coefficient in DISJ. Notice that by adding all MATH missing NAME to MATH, we obtain a decomposition of DISJ with MATH terms. But any such decomposition needs at least MATH terms, hence MATH, which implies MATH. |
cs/9910010 | It is known (and easy to see) that MATH, where MATH is the number of different rows of MATH (this equals the number of different columns in our case, because MATH). We count MATH. Firstly, if MATH then the MATH-row contains only zeroes. Secondly, if MATH and both MATH and MATH then it is easy to see that there exists a MATH such that MATH and MATH (or vice versa), hence MATH so the MATH-row and MATH-row are different. Accordingly, MATH equals the number of different MATH with MATH, MATH for REF-row, which gives the lemma. |
cs/9910010 | The upper bound follows from the fact that MATH cannot have monomials of degree MATH. For the lower bound we distinguish two cases. CASE: MATH. It is known that every symmetric MATH has degree MATH CITE. That is, an interval MATH such that MATH has no monomials of any degree MATH has length at most MATH. This implies that every interval MATH REF such that MATH has no monomials of any degree MATH has length at most MATH (by setting MATH variables to REF, we can reduce to a function on MATH variables where MATH occurs ``at the end"). Since MATH must have monomials of degree MATH, MATH must contain a monomial of degree MATH for some MATH. But because MATH is symmetric, it must then contain all MATH monomials of degree MATH. Hence by NAME 's approximation MATH, which implies the lemma. CASE: MATH. It is easy to see that MATH must contain all MATH monomials of degree MATH. Now MATH . Hence MATH. |
cs/9910010 | Let MATH be the smallest number such that MATH, then MATH. If MATH then MATH, so MATH. If MATH then MATH has no monomials of degree MATH, hence MATH . |
cs/9910010 | By REF we have MATH. REF implies MATH. Using NAME 's lower bound technique for disjointness CITE (see also CITE) we can easily show MATH, which implies the theorem. |
cs/9910010 | Let MATH be all the minimal monomials in MATH. Each MATH induces a rectangle MATH, where MATH and MATH. Because MATH is monotone increasing, MATH iff MATH makes at least one MATH true. Hence MATH iff there is a MATH such that MATH. Accordingly, the set of MATH is a REF-cover for MATH and MATH by REF . Plugging into REF gives the theorem. |
cs/9910010 | Suppose there exists a REF-round qubit protocol with MATH qubits: NAME sends a message MATH of MATH qubits to NAME, and NAME then has sufficient information to establish whether NAME 's MATH and NAME 's MATH are disjoint. Note that MATH is independent of MATH. If NAME 's input is MATH, then MATH is the negation of NAME 's MATH-th bit. But then the message is a MATH quantum random access code CITE: by choosing input MATH and continuing the protocol, NAME can extract from MATH the MATH-th bit of NAME (with probability MATH), for any MATH of his choice. For this the lower bound MATH is known CITE. |
cs/9910010 | By REF , it suffices to show that a MATH-approximation of the MATH identity matrix MATH requires full rank. Suppose that MATH approximates MATH entry-wise up to MATH but has rank MATH. Then MATH has some eigenvalue MATH. Gers̆gorin's Disc Theorem (see CITE) implies that all eigenvalues of MATH are in the set MATH where MATH. But if MATH is in this set, then for some MATH hence MATH, contradiction. |
cs/9910010 | The proof is by induction on MATH: Base step. For MATH the lemma is obvious. Induction step. Suppose after MATH qubits of communication the state can be written as MATH . We assume without loss of generality that it is NAME 's turn: she applies MATH to her part and the channel. Note that there exist complex numbers MATH and unit vectors MATH such that MATH . Thus every element of the superposition REF ``splits in two" when we apply MATH. Accordingly, we can write the state after MATH in the form required by the lemma. |
cs/9910010 | Consider a clean MATH-qubit protocol for MATH. By REF , we can write its final state as MATH . The protocol is clean, so the final state is MATH. Hence all parts of MATH and MATH other than MATH will cancel out, and we can assume without loss of generality that MATH for all MATH. Now the amplitude of the MATH-state is simply the sum of the amplitudes MATH of the MATH for which MATH. This sum is either REF or REF, and is the acceptance probability MATH of the protocol. Letting MATH (respectively, MATH) be the dimension-MATH vector whose entries are MATH (respectively, MATH) for the MATH with MATH: MATH . Since the protocol is exact, we must have MATH. Hence if we define MATH as the MATH matrix having the MATH as rows and MATH as the MATH matrix having the MATH as columns, then MATH. But now MATH and the theorem follows. |
cs/9910010 | We will first give a protocol based on a REF-cover. Let MATH and MATH be an optimal REF-cover. Let MATH. We will also use MATH to denote the MATH matrix of MATH-rows and MATH for the MATH matrix of MATH-columns. Call MATH type REF if MATH, and type REF otherwise. Note that MATH, hence at least one of MATH and MATH is MATH. The protocol is specified recursively as follows. NAME checks if her MATH occurs in some type REF MATH. If no, then she sends a REF to NAME; if yes, then she sends the index MATH and they continue with the reduced function MATH (obtained by shrinking NAME 's domain to MATH), which has MATH. If NAME receives a REF, he checks if his MATH occurs in some type REF MATH. If no, then he knows that MATH does not occur in any MATH, so MATH and he sends a REF to NAME to tell her; if yes, then he sends MATH and they continue with the reduced function MATH, which has MATH because MATH is type REF. Thus NAME and NAME either learn MATH or reduce to a function MATH with MATH, at a cost of at most MATH bits. It now follows by induction on the rank that MATH. Noting that MATH and MATH, we have MATH. |
cs/9910010 | By REF we can write the final state of a MATH-qubit bounded-error protocol for MATH as MATH . Let MATH be the part of the final state that corresponds to a REF-output of the protocol. For MATH, define functions MATH by MATH . Note that the acceptance probability is MATH . We have now decomposed MATH into MATH functions. However, we must have MATH for all MATH, hence MATH. It follows that MATH. |
cs/9910010 | Assume, by way of contradiction, that there exists a blocking set MATH of MATH with MATH. Obtain restrictions MATH and MATH of MATH and MATH, respectively, on MATH variables by fixing all MATH-variables to REF. Then MATH approximates MATH and all monomials of MATH have degree MATH (all MATH-monomials of higher degree have been set to REF because MATH is a blocking set for MATH). Since MATH approximates MATH we have MATH, MATH, and MATH for all other MATH. By standard symmetrization techniques CITE, we can turn MATH into a single-variate polynomial MATH of degree MATH, such that MATH, MATH, and MATH for MATH. Since MATH and MATH, we must have MATH for some real MATH. But then MATH by REF , contradiction. |
cs/9910010 | We use the probabilistic method to show the existence of a blocking set MATH. Randomly choose a set MATH of MATH elements. The probability that MATH does not hit some specific MATH is MATH . Then the probability that there is some edge MATH which is not hit by MATH is MATH . Thus with positive probability, MATH hits all MATH, which proves the existence of a blocking set. |
cs/9910010 | Let MATH be a polynomial which approximates MATH with MATH monomials. Let MATH, and MATH and MATH be the input and sets which achieve REF-block sensitivity of MATH. We assume without loss of generality that MATH. We derive a MATH-variable Boolean function MATH from MATH as follows: if MATH then we replace MATH in MATH by MATH, and if MATH for any MATH, then we fix MATH in MATH to the value MATH. Note that MATH satisfies CASE: MATH CASE: MATH for all unit MATH CASE: MATH, because we can easily derive an approximating polynomial for MATH from MATH, without increasing the number of monomials in MATH. It follows easily from combining the previous lemmas that any approximating polynomial for MATH requires at least MATH monomials, which concludes the proof. |
cs/9910012 | Let MATH and let MATH be any order preserving bijection. Suppose that MATH is MATH-satisfiable. Say that MATH, MATH and MATH. Let MATH where: An easy induction on the construction of formulas in MATH shows that MATH and so MATH is the right mosaic. Suppose mosaic MATH from structure MATH is a MATH-relativized MATH-mosaic. Thus MATH and MATH. Define MATH via MATH iff MATH for any atom MATH (including MATH). As MATH there is some MATH such that MATH and MATH. It is easy to show that MATH. |
cs/9910012 | Consider the forward direction of the proof. Suppose MATH is fully decomposed by the tactic shuffle MATH. By REF, we have a decomposition for MATH including each MATH and mosaics with each MATH in their starts or ends. MATH follows. We now establish REF . Each MATH is an MPC by the definition of a shuffle. To show the forward MATH property for MATH, suppose that MATH. We consider the case when MATH: the case with MATH is similar. We know REF that MATH is fully decomposed by MATH. If MATH is a type REF defect in MATH then it is cured in this decomposition and we can conclude that MATH is in the cover MATH of the first mosaic MATH. If MATH is not a type REF defect in MATH then MATH as well. Thus in any case MATH. I claim that MATH. If not then MATH and coherency CREF of MATH implies that MATH. This is a contradiction to the consistency of MATH. So MATH is either a type REF defect in MATH or not. In the former case it is cured in the full decomposition REF for MATH and so MATH appears in the start or end of a mosaic in some MATH or in some MATH. Thus MATH. This is KREF. If MATH is not a type REF defect in MATH then KREF or KREF holds by definition. We now show the converse part of the forward MATH property for MATH. Suppose that both MATH and KREF holds: the cases of REF or KREF holding are straightforward. Thus MATH, MATH is a type REF defect in MATH and so a cure is witnessed in some MATH or in the start or end of some mosaic in some MATH. Now look in the decomposition FREF (or FREF) for MATH in which we have MATH holding in all starts, ends and covers and MATH appearing somewhere. A simple induction shows that we must have MATH in the very start MATH as required. To show the backwards MATH property is the mirror image. Very similar arguments establish REF - REF . To show REF we just use the full decomposition REF for MATH and reason about type REF defects. To show REF , use the full decomposition REF for MATH and reason about type REF defects. REF follow from using the full decomposition REF for MATH and reasoning about type REF defects respectively. Now consider the converse: suppose that the seven REF - REF hold for mosaic MATH. First we must show that each of MATH, MATH, each MATH and any MATH (as defined from MATH, the MATH and the MATH in the definition of a shuffle) are mosaics. This follows from If the MPC MATH satisfies the forward MATH property, the MPC MATH satisfies the backwards MATH property and MATH then MATH is a mosaic. We must check the first two coherency conditions. The mirror images are mirror images. CASE: Suppose MATH and MATH. First we establish that we must have MATH. Suppose not for contradiction. Since MATH, KREF does not hold and so MATH, KREF does not hold and so MATH and KREF does not hold and so either MATH or MATH. We have a contradiction to the coherency REF of MATH. CASE: First, we show MATH. Otherwise, MATH. Thus, by REF, MATH and we have our contradiction. Next we show that either MATH or MATH. Suppose instead that MATH and MATH. Thus MATH. By coherency CREF of MATH, we must have MATH which is a contradiction. CASE: We show that MATH and MATH. We can not have MATH, as then KREF implies MATH. We can not have MATH as then coherency REF of MATH implies MATH, a contradiction. CASE: Assume MATH and MATH. By the forward MATH property for MATH, MATH and, since MATH, either KREF or KREF holds (for MATH). By the coherency CREF of MATH we can conclude that we can not have MATH. Thus MATH and CREF of MATH implies that MATH as required. Next we must check the fullness of the decompositions. This follows by Suppose MATH and MATH are as in the previous claim. Furthermore, suppose the sequence MATH of mosaics composes to MATH such that for each MATH with MATH, there is a mosaic in MATH other than the very first which includes MATH in its start. Then MATH is fully decomposed by MATH. Suppose MATH is a type REF defect of MATH. By the forward MATH property for MATH, MATH. As MATH is a type REF defect MATH and either MATH or MATH. By coherency CREF of MATH, MATH. So MATH and MATH must appear in a non-first mosaic in MATH and we have our cure. Type REF defects: mirror image. Type REF defects: Suppose MATH but MATH. Thus MATH appears in the start of a non-first mosaic in MATH. We have our witness. Thus MATH is fully decomposed by the tactic shuffle MATH as required. |
cs/9910012 | Given MATH from MATH choose any MATH with MATH. Let MATH realize MATH on MATH and MATH realize MATH on MATH. Define MATH via: MATH . It is straightforward to check that MATH realizes MATH on MATH. Use the facts that MATH and MATH. |
cs/9910012 | Say MATH. Choose a sequence MATH converging to MATH. For each MATH, and each MATH, let MATH realize MATH on MATH. Define MATH via MATH and if MATH, then put MATH. I claim that MATH realizes MATH on MATH. Consider the six realization conditions. The harder cases are REF . There are several subcases and their converses and they all involve similar sorts of reasoning so we will just present a few for illustration purposes. To show the forward direction of REF assume that MATH and MATH. The subcases concern whether MATH, MATH equals some MATH or MATH is in some MATH. We must show that RREF or RREF holds. Suppose MATH and MATH is a type REF defect in MATH. As MATH is a full decomposition of MATH, a cure to this defect is witnessed in this sequence. We can conclude that MATH is in the cover of MATH and so in the starts, covers and ends of each of the MATH. We can also conclude that MATH is in the start of some MATH and in the end of the preceding one MATH (with MATH if MATH). RREF follows easily with MATH. Suppose MATH and MATH is not a type REF defect in MATH. So MATH. If MATH then RREF holds. Otherwise RREF holds. Suppose MATH and MATH. So MATH. Now MATH is realized by MATH on MATH and so by REF (for MATH) either RREF holds and we are almost immediately done or RREF holds. In this latter case MATH, MATH, we may suppose MATH and for all MATH, if MATH then MATH. Possibly there are some MATH and MATH such that MATH ( so MATH) and either one of the following five holds: MATH; MATH; MATH; MATH; or MATH. If there is no such MATH then it is straightforward to show that RREF holds and we are done. If there are such MATH then we can suppose that they are chosen so that MATH is greatest possible. It follows that MATH. If RREF holds of MATH then it is easy to finish. So suppose not. Thus RREF holds of MATH and we can conclude via RREF that MATH, MATH and MATH. Because RREF does not hold of MATH we can also conclude via RREF that MATH and MATH. This contradicts our choice of MATH and MATH and we are done. For the converse direction of REF, we assume that MATH and either RREF or RREF holds. The subcases concern whether RREF or RREF holds and whether MATH, MATH equals some MATH or MATH is in some MATH. We must show that MATH. Suppose RREF holds with MATH in some MATH. So MATH, MATH and for all MATH, if MATH then MATH. A straightforward induction on MATH, using RREF for each MATH shows that for all such numbers with MATH, we have MATH. That MATH follows immediately by using RREF on MATH. For the forward direction of REF , suppose MATH is in the cover of MATH. Thus MATH is in the start, end and cover of each MATH as they compose (with MATH itself) to MATH. Also note that the end of MATH is the same as the start of MATH. By REF for each MATH, MATH for each MATH as required. For the converse direction of REF , suppose, for all MATH, MATH. It is clear that MATH is in the cover, start and end of each MATH. If MATH was not in the cover of MATH then the fact that MATH is a full decomposition of MATH would imply that MATH would be in the start of some MATH. Hence, by contradiction, MATH is in the cover of MATH as required. |
cs/9910012 | We can proceed in a two stage construction as follows. Stage one is the construction of the MATH. If MATH skip this stage. Stage one proceeds in MATH rounds starting with round MATH. Start with all the MATH empty. Before each round MATH will contain finitely many closed intervals within MATH. So there will be finitely many open maximal intervals partitioning the complement of MATH within MATH. Call these the spaces left before that round. In round REF put MATH in MATH. In general, for each space MATH left before round MATH (for integers MATH and MATH with MATH), put MATH in MATH. Notice that we leave spaces on each side of the new intervals and these spaces are one third as wide as the original space. After MATH rounds we have our final MATH-s. We will now use the separability property of MATH to show that there are still plenty of points of MATH not in any interval in any MATH. Let MATH be the set of these points. I claim that between every pair of intervals from MATH there are some elements of MATH. To show this by contradiction suppose that no element of MATH lies between MATH and MATH (where MATH). So, for every MATH there is some interval MATH with MATH. Let MATH be the ordering of intervals from MATH which lie within MATH inherited from their elements. This is isomorphic to the rationals order as it is countable, dense and without endpoints. Thus the order MATH has an uncountable order MATH of gaps. Define a map MATH as follows: given a gap MATH in MATH, let MATH. Let MATH which exists as MATH. Clearly MATH is order preserving and one-to-one. Furthermore, if MATH are gaps of MATH then there is MATH between them. Thus MATH and MATH must also be strictly between the interval MATH from MATH containing MATH and the interval MATH from MATH containing MATH. These two intervals must be disjoint. Thus MATH is an uncountable set of pairwise disjoint non-singleton intervals of MATH. This clearly contradicts separability. Thus MATH is a set of points densely located between the intervals in the MATH or, in the case that MATH, MATH. It is straightforward to partition MATH densely into the pairwise disjoint MATH as required. |
cs/9910012 | Let MATH and MATH be as constructed for MATH in REF . Suppose MATH and MATH. For each interval MATH choose a sequence MATH and for each MATH, let MATH be the interval MATH. Let MATH realize MATH on MATH. Define MATH via MATH, MATH, for each MATH in MATH within an interval MATH from MATH, MATH and for each MATH, MATH. Note that MATH may lie at the end of some MATH and the beginning of MATH. In that case, the fact that the mosaics in each MATH compose will guarantee that MATH is well-defined. I claim that MATH realizes MATH on MATH. Consider the six conditions. The harder cases are REF . It is useful to consider REF first. For the forward direction of REF , suppose MATH is in the cover of MATH. By REF , MATH is in each MATH and in the start, cover and end of each mosaic in each MATH. So MATH is in MATH for each MATH in each MATH and in each MATH in each MATH in each MATH. Thus MATH for each MATH as required. For the converse direction of REF , suppose, for all MATH, MATH. For contradiction suppose that MATH is not in the cover of MATH. Thus MATH is a type REF defect in MATH and this is cured in the full decomposition FREF. Thus MATH appears in the start of a mosaic in MATH or in MATH or in the start of MATH. Thus MATH appears in the start of a mosaic in one of the MATH or appears in one of the MATH, MATH or MATH. Thus MATH for MATH being the start of some MATH for some MATH in some MATH or for MATH where some MATH is in some MATH or for MATH in some MATH. Thus MATH can not be in MATH and we have our contradiction. Hence, MATH is in the cover of MATH as required. To show the forward direction of REF assume that MATH and MATH. The subcases concern whether MATH, MATH is in some MATH or MATH is in some MATH for some MATH in some MATH. We must show that RREF or RREF holds. Again there are several subcases and their converses using similar sorts of arguments. We give a selection for illustration purposes. First consider MATH. So MATH. Now MATH is fully decomposed by MATH so, by definition of a full decomposition, either REF MATH is a type REF defect cured in the decomposition or REF MATH and either ( MATH and MATH) or MATH. These latter REF give us the desired result immediately. If MATH is cured in the full decomposition of MATH then it is clear that MATH is in the cover of the first mosaic, MATH. But this cover is MATH itself so MATH and MATH for all MATH. Now MATH appears in the end of a mosaic in the full decomposition and so in MATH for some MATH. Thus we are done. Now consider the case of MATH with MATH. The case of MATH is a slightly special case of what follows and can be proved with slight modifications so we will omit that case. Assume MATH. Thus MATH is fully decomposed by MATH and we must have MATH. This is whether or not MATH is a type REF defect in MATH or not. By the argument above for the RREF case, MATH for all MATH. If MATH then MATH is a type REF defect in MATH and thus is cured in the full decomposition. Thus MATH appears in the start of a mosaic in MATH or in MATH or in the start of MATH. Thus MATH appears in the start of a mosaic in one of the MATH or appears in one of the MATH, MATH or MATH. Thus MATH for MATH being the start of some MATH for some MATH in some MATH or for MATH where some MATH is in some MATH or for MATH in some MATH. Combined with the observation about MATH this gives us RREF Otherwise, MATH and so coherency CREF along with the fact that MATH gives us MATH. By the fullness of the decomposition of MATH, either MATH appears in the start of a mosaic in MATH or in MATH or in the start of MATH (and we proceed as above), MATH (and RREF holds) or MATH and MATH (and RREF holds). We are done. The case of MATH in some MATH for MATH is similar but a little more complex. For the converse direction of REF, we assume that MATH and either RREF or RREF holds. The subcases concern whether MATH, MATH is in some MATH or MATH is in some MATH for some MATH in some MATH. We must show that MATH. Consider just the case of REF holding for MATH in some MATH for some MATH in some MATH. Let MATH be such that each MATH. Thus MATH. We have MATH and MATH and for all MATH, MATH. There are three possibilities for MATH. Suppose MATH. Thus RREF holds for MATH and so MATH. Suppose MATH. By REF or RREF, MATH. An easy induction using RREF establishes that MATH. Then RREF gives us MATH as required. Suppose MATH. For each MATH, choose MATH with MATH. So MATH. For each MATH, choose MATH with MATH. Say MATH and for each MATH, MATH. Now MATH, MATH and MATH. We can conclude MATH for all MATH and so by REF that MATH and so in the cover of all mosaics in each MATH and each MATH and each MATH. There are two cases now: either MATH or not. Suppose MATH so that MATH for all MATH. So MATH for any such MATH. We know that MATH so it follows that MATH. Thus MATH. Coherency CREF implies that MATH and CREF gives us MATH. By REF, MATH as required. The other case is that MATH so that MATH is a type REF defect in MATH and so appears in the start of a mosaic in some MATH, at the end of some MATH or in some MATH. If MATH then MATH is fully decomposed by MATH. If MATH then MATH is fully decomposed by MATH. In either case MATH is in the starts, covers and ends of all the mosaics and MATH is in the start of one of the mosaics. A simple induction using coherency CREF tells us that MATH. Now consider MATH say with MATH and MATH in all the starts, ends and covers. A simple induction using coherency CREF gives us MATH in the start and end of each MATH. So we have MATH and MATH in MATH. If MATH we are done. Otherwise, MATH. However, MATH realizes MATH on MATH and so REF gives us MATH as required. |
cs/9910012 | let MATH. Given the real mosaic system MATH of MATH-mosaics, we can easily proceed by induction on MATH to show that, for any MATH from MATH, for any level MATH member MATH there is MATH which realises MATH on MATH. Each step of the induction is just a use of one or two of the preceding REF , its mirror image and REF. So we have MATH which realizes MATH on MATH. Define MATH by MATH iff MATH and let MATH. I claim, for all MATH, for all MATH, MATH . This is a straightforward proof by induction on the construction of MATH. The case of MATH is as follows. Note that if MATH then MATH and so both MATH and MATH are also in MATH by REF . First suppose MATH. Thus there is MATH with MATH, MATH and for all MATH, if MATH then MATH. By the inductive hypothesis, MATH and for all MATH, if MATH then MATH. By REF, MATH as required. Now suppose MATH. Note that MATH as no MATH is in MATH as MATH is relativized. By REF, either RREF or RREF holds and it can not be the latter as that entails MATH amongst other things. So RREF holds and there is MATH with MATH, MATH and for all MATH, if MATH then MATH. It follows via the inductive hypothesis that MATH as required. From the claim and REF on realization, it follows that MATH, MATH and the cover of MATH contains exactly those MATH which hold at all points in between MATH and MATH. Thus MATH as required. |
cs/9910012 | We will choose a finite set of points from MATH at which we will decompose MATH. For each defect MATH in MATH choose some MATH or MATH witnessing its cure between MATH and MATH as follows. If MATH is a type REF defect then it is clear that there must be MATH with MATH and for all MATH, MATH. Similarly find MATH witnessing a cure for type REF defects. If MATH is a type REF defect in MATH then it is clear that there is MATH with MATH. Collect all the MATH-s and MATH-s so defined into a finite set and add the midpoint MATH of MATH and MATH. Order these points between MATH and MATH as MATH. Note that some points might be in this list for two or more reasons. It is clear that because of our choice of witnesses, the sequence of MATH is a full decomposition. |
cs/9910012 | Say MATH for MATH from MATH. Construct a decomposition tree with root labelled by MATH by repeated use of REF . |
cs/9910012 | To find the first branch, start at the node and recursively move to the first child of the current node which lies on an infinite MATH-branch. |
cs/9910012 | Suppose that the mosaic MATH appears infinitely often as a mosaic label along MATH in the tree MATH. Thus the cover of MATH is MATH. For each MATH such that MATH choose MATH such that MATH and for all MATH, MATH. We say that MATH is constantly true for a while before MATH iff there is some MATH such that if MATH then MATH. For each MATH such that MATH is true for a while before MATH choose some MATH such that if MATH then MATH. Now choose any node MATH strictly below MATH and labelled by MATH such that MATH is strictly greater than each MATH and each MATH. This can be done as MATH halves in each generation but always MATH. Now say that MATH appears as MATH for some node MATH below MATH and labelled by MATH. Let MATH be the sequence of parents of leaf nodes of MATH before MATH in order. By the mirror image of REF , there is a sequence MATH such that the label of each MATH is MATH. Note that we apply REF to the subtree of MATH without the leaf nodes of MATH. In the statement of the lemma we required that all siblings of leaves are themselves leaves in order to guarantee that this subtree is a decomposition tree. If MATH then we are done. Assume MATH. We say that a formula MATH is true (in MATH) arbitrarily recently before a point MATH iff for every MATH there is some MATH with MATH and MATH. For each MATH which is true arbitrarily soon before MATH, choose some MATH such that MATH and MATH. Now find a node MATH below MATH and labelled by MATH such that MATH and MATH is greater than each MATH. Let MATH be the sequence of parents of leaf nodes of MATH below MATH and before MATH in order. Let MATH be the sequence of children in MATH of the MATH in order. By the mirror image of REF (used in the subtree rooted at MATH), there is a sequence MATH such that the label of each MATH is MATH. If MATH then let MATH be the parents of leaf nodes of MATH below MATH and before MATH in order. By REF (applied to the subtree of MATH rooted at MATH and not including the leaf nodes from MATH), there is a sequence MATH such that each MATH is the label of MATH. The long sequence MATH is as required and we are done. Now assume MATH. Let MATH. I claim that MATH too. The start of both is the start of MATH. The end of both is just MATH. Finally the cover of both is just the set of MATH such that MATH holds constantly for a while before MATH. We will now show that MATH fully decomposes MATH. The composition is MATH by REF . Before we show that the decomposition is full consider the nodes below MATH. Say that the children of node MATH are MATH labelled by MATH respectively. Thus MATH is a full decomposition of MATH. Also MATH and there is some MATH such that MATH and MATH lies on MATH. Because MATH is the lexically first infinite branch on which MATH lies there are only a finite number of nodes (in MATH) below any MATH with MATH. Also, we can start with the sequence MATH and repeatedly replace a node from MATH by the sequence of its children in order and end up with a prefix sequence of MATH and a corresponding sequence MATH. Note that MATH will be a subsequence of this. Now let us return to consider defects in MATH. Type REF defects: Suppose MATH is a type REF defect of MATH. If it happens that MATH then MATH is also a type REF defect in MATH and thus cured in the full decomposition MATH. Thus the cure of MATH is witnessed in this sequence. Say MATH and MATH. As MATH labelled by MATH lies on MATH, MATH does not contain MATH. Thus MATH. However, we have seen that then MATH appears as one of the MATH and thus we can find a witness to the cure of the type REF defect MATH of MATH in the decomposition MATH as required. Now assume MATH. For MATH to be a type REF defect in MATH we thus must have MATH and either MATH or MATH. Since MATH holds at MATH we must have MATH true somewhere between MATH and MATH. So MATH is a type REF defect of MATH and so cured in MATH. Thus MATH must appear in the end of MATH say. It can not appear after MATH as MATH is not true at MATH so MATH. As above this implies MATH appears in the end of a mosaic in MATH in MATH. Type REF defects: No type REF defects are possible in MATH. Suppose MATH. It is not possible that MATH as MATH. It is not possible that MATH and MATH as MATH. It is not possible that MATH and MATH as MATH. Type REF defects: Suppose MATH but MATH. So MATH does not hold constantly for a while before MATH and so is true arbitrarily recently before MATH. So MATH and MATH. Say that MATH. So MATH is within the label of MATH whose parent is MATH say. Maybe MATH is in the start or end of MATH where MATH is the label of MATH. Otherwise MATH and so MATH appears in the end of a mosaic in the full decomposition of MATH. So MATH is witnessed in MATH as required. |
cs/9910012 | Say MATH, MATH and MATH. Let MATH be the set of all MATH for MATH from MATH. Clearly MATH. I claim that MATH is a real mosaic system of depth MATH. In fact, I show that for all MATH, for all MATH, if MATH has cover containing at least MATH formulas then MATH is a level MATH member of MATH. We proceed by induction on MATH. Suppose that we have shown this for every MATH and mosaic MATH has cover containing at least MATH formulas. All full trees will be MATH-full trees. lightening There is a tapering full decomposition tree MATH with root with mosaic label MATH such that: Choose any MATH such that MATH. Use REF to find a complete and tapering full decomposition tree MATH with root with label MATH. Let MATH be the sub-tree of MATH containing only the nodes with mosaic label with cover MATH and all their children and grandchildren. Thus, any leaf node in MATH and any parent of a leaf node in MATH will have cover strictly including MATH and, by the inductive hypothesis will be a level MATH member of MATH. Also, any sibling of a leaf node of MATH will also be a leaf node of MATH. Enumerate the MATH-sticks in MATH. This can be done as for each stick MATH we can choose some node MATH which lies on it and on no other infinite MATH-branches. We can use a step by step process of gradually constructing MATH from MATH. Each step removes one stick MATH by only changing the subtree of MATH rooted at MATH. The step introduces no other infinite MATH-branches. So it suffices to just show how to so remove one MATH-stick MATH from MATH to make a tree MATH. Choose any node MATH below MATH (so MATH lies on no other infinite MATH-branches apart from MATH). Say that MATH are the children of MATH in order and MATH lies on MATH. Say that MATH is labelled by MATH and that the limit of MATH is MATH (so MATH). To make MATH we will just replace MATH from MATH and all its descendents by a sequence of new children of MATH who will be parents of leaf nodes in MATH. The new children of MATH will lie later than MATH and earlier than MATH. In fact, we may replace MATH by one sequence of children with labels partitioning the interval MATH and another later sequence with labels partitioning the interval MATH. Such a change can be seen to be as required in effecting a removal of MATH without any other infinite branches being introduced or even affected. Note that as the start of the mosaic label of the first new child of MATH will just be the same as the end of MATH (or the start of MATH in case that MATH), and similarly for the end of the last new mosaic label, the children of MATH in MATH will still carry a full decomposition of the mosaic label of MATH. If MATH or MATH then we do not add any new children in the first or second sequence respectively. Note that there will be some new children to add in one or other or both sequences as we can not have MATH. Here we just show how to construct the first sequence of new children with labels partitioning MATH in the case that MATH. Constructing the later second sequence is via a mirror image argument. So suppose MATH. REF applied to the subtree MATH of MATH consisting of MATH and all its descendents in MATH tells us we have two cases. Possibly there is a sequence MATH such that each MATH is the label of a parent, MATH say, of a leaf node in MATH. In this case the earlier sequence of new children of MATH in MATH will be new nodes MATH with each MATH labelled by MATH. We also give each MATH leaf node children with exactly the same labels as the leaf node children of MATH. We are done. In the other case there is a sequence MATH such that each MATH with MATH is the label of a parent MATH of a leaf node in MATH and a sequence MATH such that each MATH with MATH is the label of a leaf node in MATH and MATH is fully decomposed by MATH. In this case the earlier sequence of new children of MATH in MATH will be new nodes MATH with each MATH labelled by MATH. For MATH, we give each MATH leaf node children with exactly the same labels as the leaf node children of MATH. Thus for MATH each MATH is a level MATH member of MATH. For the node MATH labelled by MATH we give it MATH children, MATH in that order. We label each MATH by MATH. Now MATH is fully decomposed by tactic MATH and each of these mosaics are mosaic labels of leaf nodes in MATH and so are level MATH members of MATH. Thus MATH, the mosaic label of both MATH and MATH is a level MATH member of MATH. Again we are done. Construct such a MATH with root MATH labelled by MATH. If MATH has no infinite branches then it is clear that the mosaic labels on the leaf nodes taken in lexical order form a decomposition of MATH. They are all in MATH and so we have our required decomposition. Thus MATH is a level MATH member of MATH and hence trivially a level MATH member of MATH and we are done. We can thus assume that MATH has two or more infinite branches: a lone one would be a stick. Say that MATH is the lexically first one and MATH is the lexically last one. Possibly MATH. If not, that is, if MATH, then we can use REF to find either a sequence of level MATH members of MATH which compose to MATH or sequences MATH and MATH of level MATH members of MATH and a mosaic MATH, such that MATH is fully decomposed by MATH and MATH composes to MATH. It follows that MATH is the composition of level MATH members of MATH. Similarly, with MATH, and MATH. We are done when we show that MATH is fully decomposed by a shuffle of level MATH members of MATH. Then it follows that MATH is a level MATH member of MATH as required. Note that if MATH then we would already be done, so we are assuming MATH. Let MATH, the set of type REF defects in MATH. Let MATH be the deepest node on both MATH and MATH. Let MATH be any descendent of MATH on MATH which has two children which each lie on an infinite MATH-branch. Such a node exists as MATH is not a stick. Let MATH be a child of MATH which does not lie on MATH but does lie on another infinite MATH-branch. Clearly MATH has mosaic label with cover MATH and MATH is lexically after every node on MATH. Say that the children of MATH are MATH in order. Because the children are labelled with a full decomposition, each MATH appears in the start of some MATH for MATH. befaft For each MATH, if MATH is labelled by MATH, there are two infinite MATH-branches MATH and MATH such that MATH and, if MATH then, MATH can be decomposed as a non-empty sequence of level MATH member of MATH which includes some mosaic with start or end equal to MATH. We find MATH and, if MATH, a sequence MATH of level MATH members of MATH which composes to MATH. Finding MATH and a similar sequence MATH which composes to MATH is (almost) a mirror image. The sequence MATH will be as required. Note that as MATH, MATH will have a next earlier sibling MATH which will be labelled with MATH for some MATH. Use REF applied to MATH to find a node MATH labelled by MATH lying on an infinite branch of MATH and a sequence MATH with each MATH the label of a leaf node of MATH. Note that there is an infinite branch of MATH which lies after MATH as MATH does. Possible MATH in which case let MATH be the empty sequence of mosaics. Otherwise, if MATH then let MATH be the sequence of MATH-s in order. These are each level MATH members of MATH and the composition of the sequence is MATH. Let MATH be the lexically first infinite branch on which MATH lies. If MATH then we are done as MATH will do. So suppose that MATH. We will find a sequence MATH of level MATH members of MATH which compose to MATH and then we can put MATH and we will be done. By REF we have two cases. Possibly there is a sequence of mosaic labels of parents of leaf nodes in MATH which compose to MATH and we can use that as our MATH. The other possibility is that we have a sequence MATH of mosaic labels of parents of leaf nodes in MATH followed by one final mosaic MATH such that MATH composes to MATH and MATH is fully decomposed by the tactic MATH where MATH is a sequence of mosaic labels of leaf nodes of MATH. Again the mosaics in MATH are level MATH members of MATH and MATH is a level MATH member of MATH and so we are done. Let MATH and MATH be the sets of all the pairs of infinite MATH-branches got using REF on each MATH for MATH, such that each MATH and MATH have equal limits but each MATH and MATH do not. For each MATH, let MATH. For each MATH let MATH be a sequence of level MATH member of MATH which compose to MATH. By the claim and our original choice of MATH, we can do this and ensure that for each MATH, either there is a MATH with MATH or a MATH with MATH in the start or end of some mosaic in MATH. Now choose any infinite branch MATH of MATH as follows. Start at MATH and proceed recursively. Choose a child of the current node which lies on an infinite branch. When there is a choice of such children (as there will be infinitely often) infinitely often choose an earlier child, infinitely often a later child. Let MATH be the limit of MATH. Let MATH. Also let MATH. Say that an infinite MATH-branch MATH is left dense if for all nodes MATH there is a descendent MATH of MATH in MATH such that MATH has at least two children MATH and MATH on infinite MATH-branches such that MATH is earlier than MATH but MATH lies on MATH. Define right dense as the mirror image. Note that due to the absence of sticks each infinite MATH-branch is either left dense or right dense or both. leftandright Each MATH and MATH is left dense and each MATH and MATH is right dense. Each MATH and MATH is found as the lexically last infinite MATH-branch on which some node lies. Thus it can not be right dense. As it is not a stick it must be left dense. Similarly with MATH and each MATH. MATH was chosen to be left dense and right dense by construction. atend If MATH is true at the limit of a right dense infinite branch of MATH then MATH is in the cover of MATH. Suppose MATH is a right dense infinite branch with limit MATH and MATH. If MATH then there is some MATH such that MATH and for all MATH, if MATH then MATH. Choose some node MATH labelled by MATH such that MATH which we can do as the width of labels halves with each generation. Since MATH is right dense we can choose some descendent MATH of MATH with a child MATH on an infinite MATH-branch and an earlier child MATH lying on MATH. Say that MATH is labelled with MATH. Clearly MATH. Thus MATH is in the cover of MATH which is just MATH. arby If MATH is a type REF defect in MATH and MATH is a left dense infinite branch then MATH is true arbitrarily recently before MATH. There is a mirror image using right dense infinite branches and arbitrarily soon afterwards. Let MATH and MATH. Choose a node MATH on MATH labelled with MATH such that MATH. Choose a node MATH on MATH below MATH with two children MATH before MATH with MATH on MATH and MATH on another infinite branch. So MATH. Now the cover of MATH is MATH and MATH is a type REF defect in MATH and so in MATH. Consider the full decomposition exhibited by the children of MATH. Thus there is some non-first child MATH of MATH with MATH. Thus MATH and MATH as required. coverob The cover of MATH is MATH. As MATH, the cover is contained in MATH. For each MATH, MATH is a type REF defect in MATH and so by REF , MATH is true arbitrarily soon before MATH. Thus MATH is also not in the cover of MATH. beforetheend If MATH is a right dense infinite branch of MATH then MATH. (And mirror image). It is clear that MATH. We must rule out the case of MATH. Choose any node MATH labelled by MATH say on MATH which has a later sibling MATH labelled by MATH on another infinite branch MATH say. Thus MATH. Now choose a descendent MATH of MATH labelled by MATH which lies on MATH and has a later sibling MATH lying on another infinite branch. Thus MATH as required. rdfwd If MATH is a right dense infinite branch of MATH then MATH satisfies the forward MATH property. There is a mirror image. Suppose MATH is a right dense infinite branch. First suppose MATH. By REF , MATH. Now either MATH is a type REF defect in MATH so MATH (so KREF) or MATH is not a type REF defect of MATH so MATH. However, MATH is true at MATH so there is MATH with MATH true at MATH and MATH true everywhere in MATH. Thus MATH and MATH is true everywhere in MATH. It follows that MATH and MATH hold everywhere in MATH. Now MATH so MATH holds at MATH if MATH or MATH and MATH hold at MATH if MATH. Thus KREF or KREF holds as required. To show the converse suppose that MATH and KREF, KREF or KREF holds with respect to MATH and the forward MATH property. Thus MATH holds everywhere between MATH and MATH. If KREF or KREF holds then it is clear that MATH is true at MATH. If KREF holds then REF tell us that MATH is true somewhere in between MATH and MATH. Again it follows that MATH is true at MATH as required. finally The mosaic MATH is fully decomposed by the tactic shuffle MATH. Let MATH and MATH be as in the definition of a shuffle. We use REF . All the necessary forward and backward MATH REF follow from REF above. SREF holds by virtue of REF holds by choice of the MATH and MATH. This gives us our result as MATH is fully decomposed by a shuffle in which each mosaic in each MATH is a level MATH member of MATH. |
cs/9910012 | To show REF implies REF is straightforward and it follows that MATH is a real mosaic system. It is then clear that REF implies REF. |
cs/9910012 | If MATH is satisfiable then REF implies there exists a MATH-relativized MATH-mosaic MATH which is fully MATH-satisfiable, and so is MATH-satisfiable. REF implies MATH appears in a real mosaic system of depth MATH. REF follows from REF . (MATH). If MATH-relativized MATH-mosaic MATH appears in a real mosaic system then REF implies that MATH is fully MATH-satisfiable. Thus REF tells us that MATH is MATH-satisfiable. |
cs/9910012 | For each MATH choose a mosaic from MATH to witness MATH. We can choose either a non-first mosaic which has MATH in its start, a non-last mosaic which has MATH in its end or any mosaic from MATH which does not have MATH in its cover. Call these MATH mosaics the important ones in MATH. Construct MATH by including the important mosaics and a composing subsequence of the mosaics between each consecutive pair of important mosaics which contains no repeated mosaics. A simple iterative procedure allows us to successively remove one copy of each repeat and the mosaics in between. Thus there will be at most MATH mosaics in MATH in between important mosaics. The maximum length of MATH will be MATH. It is straightforward to check that the composition of MATH is MATH: the cover is right because of the inclusion of the important mosaics. |
cs/9910012 | Use the idea of important mosaics as in the proof of the previous lemma but include, as important, a witness for the cure of each defect in MATH. |
cs/9910012 | By REF we need only enough MATH and MATH such that each element of MATH appears in some MATH or in the start or end of a mosaic in some MATH. Thus MATH and MATH can be chosen so that MATH. As in the proofs of the previous lemmas we can reduce each of the chosen MATH to be of length MATH by removing repeats in between important mosaics: in this case just the witnesses of elements of MATH. |
cs/9910012 | The algorithms are defined in terms of each other. First consider SH. Given MATH of length MATH, MATH and MATH, first check whether MATH is a mosaic and check that its start satisfies the forward MATH property and its end satisfies the backwards MATH property. Return the answer ``no" if any of these checks or subsequent checks fail. Also collect the set MATH of type REF defects in MATH and guess MATH. For each MATH to MATH, guess MATH and check that it is an MPC containing the cover of MATH and satisfying the forwards and backwards MATH property and remove any MATH from the set MATH. Guess MATH. For each MATH (if any), guess the start of the first mosaic in MATH and check that it satisfies the backwards MATH property and guess the end of the last mosaic in MATH and check that it satisfies the forwards MATH property. Also guess and check ``on the fly" a composing sequence MATH of up to MATH mosaics (with appropriate start and ends). Check (via MATH) that each MATH of these is a level MATH member of MATH and remove from MATH any formula which appears in the start or end of MATH. Check that the start, cover and end of each MATH contains the cover of MATH. Return ``yes" if MATH ends up empty. Otherwise return ``no". Now consider LV. To decide whether or not MATH is a level MATH member of MATH we need to guess a sequence of mosaics which compose to MATH and check that each of these, MATH say, is either a level MATH member of MATH (so use MATH) or is fully decomposed by a shuffle with each mosaic in each sequence in the shuffle being a level MATH member of MATH (so use MATH). LD is as follows. To decide whether or not there is some MATH such that MATH can be fully decomposed by tactic MATH with each mosaic in MATH being a level MATH member of MATH, we need to guess and check a sequence MATH which is a full decomposition of MATH and check that each mosaic in MATH returns yes from MATH. TR is similar to LD. CP is easy: we already know how to guess and check decompositions. LD' and TR' are very similar to LD and TR. CM uses LD' and TR' in the same way CP uses LD and TR. We already know how to guess and check decompositions. CASE: The algorithms use polynomial space and are correct. Fix MATH of length MATH. We proceed by induction on the number MATH used. Assume MATH and that we have shown that the algorithms work for any MATH and any MATH. By REF and the inductive hypothesis, MATH gives the correct result. By REF , the other algorithms are correct. The space bounds follow as each algorithm needs only a small constant amount of information about each mosaic and the composition so far in a possibly long composing sequence of mosaics. They may also need about MATH bits to represent, in binary, the value of a counter as we check that the sequence is not too long. Each call that they make to another algorithm also requires a polynomial amount of space but we know that the depth of nesting of such calls is just linear in MATH. |
cs/9910012 | An NPSPACE algorithm is as follows. Given MATH of length MATH, choose some atom MATH not appearing in MATH. Guess a MATH-relativized MATH-mosaic MATH (checking that it is is straightforward and uses polynomial space). Use LV from REF to check whether there is a real mosaic system of depth MATH including MATH. By REF , this approach gives ``yes" answers to satisfiable input and the approach does not give incorrect ``yes" answers. By a theorem in CITE the problem is also in PSPACE. |
cs/9910012 | The proof of REF (as one possible example amongst many in the literature) contains a formula which we can easily modify. The idea is to simulate the running of any polynomial space bounded NAME machine in a formula. Let MATH be a one-tape deterministic NAME Machine where MATH is the set of states, MATH is the alphabet including blank MATH, MATH, MATH is the set of accepting states, MATH is the set of rejecting states and MATH is the initial state. Suppose that MATH is MATH-space bounded, where MATH is bounded by a polynomial in MATH. Without loss of generality, we may assume that MATH is MATH-time bounded where MATH is also bounded by a polynomial in MATH. We also assume that once MATH enters a state in MATH (or MATH) then it stays in states in MATH (MATH respectively,). Let MATH be an input to MATH. We can represent runs of MATH via tape configurations in the usual way. These may be supposed to be sequences of MATH symbols each from MATH. We are going to effectively construct a MATH formula MATH which is of polynomial size in MATH such that the satisfiability of MATH is equivalent to the acceptance of MATH by MATH. The atoms we use for MATH are from MATH along with MATH new atoms MATH. The idea of the proof will be that MATH is MATH-satisfiable iff it is satisfiable in a certain structure MATH in this language. MATH will represent an accepting run of MATH on MATH in a straightforward way. MATH has an initial tick point MATH say. From then on, every tick point has a successor tick point so we can name the points MATH etc but there may be more tick points after those. At every MATH-th tick point, starting at MATH, the atom MATH will hold. The MATH tick points in between MATH points will represent the contents of MATH's tape configuration at a particular instant. The points MATH represent the tape configuration at the initial instant of MATH's run with input MATH. For MATH, the atom MATH from MATH will be true at the MATH-th point. The atom MATH will hold at point REF. The MATH points in between the MATH at point MATH and the MATH at point MATH will similarly contain the tape configuration at the second instant of MATH's run. And so on. We will use the MATH-s to count up to MATH in binary at MATH points because we are only interested in the first MATH steps in MATH's computation. The formula MATH will be the conjunction of MATH as defined below. It should be clear that MATH is satisfiable iff it is satisfiable in a model like MATH which represents a run of MATH (on input MATH) which is accepting, iff MATH accepts MATH. That will complete our proof. We use abbreviations MATH, MATH, MATH and MATH. We also write MATH for MATH and MATH. Note that MATH and MATH are thus strict. The discreteness of ticks is given by MATH. The distribution of MATH-s is given by MATH. MATH, which we will not write out in detail just prevents any two different configuration symbol atoms from MATH from holding at any one point and prevents any of these symbols holding at non-tick points. The initial configuration is given by MATH defined as follows. Let MATH for each MATH. Now define each MATH by recursion down from MATH to MATH. MATH, each MATH, and MATH. This is a formula of length MATH. The start of the second configuration is determined by MATH where MATH and MATH are such that MATH. (M must move right at first). The relationship between a consecutive sequence of three symbols in any configuration and the corresponding symbols at the next step is given in cases by MATH. MATH is the conjunction of all MATH for each MATH such that MATH. MATH is the conjunction of all MATH for each MATH such that MATH. MATH is the conjunction of all MATH for each MATH such that MATH. MATH is the conjunction of all MATH for each MATH such that MATH. MATH is the conjunction of all MATH for each MATH. MATH is the conjunction of all MATH for each MATH. MATH is the conjunction of all MATH for each MATH. It is straightforward to show that MATH along with MATH ensures the progress of configurations represented in any model of MATH matches those of a run of MATH. MATH says that of the MATH-s only MATH holds at time point MATH. MATH forces the MATH-s to count MATH points. This large conjunct of MATH is still of size polynomial in MATH. It is MATH . MATH says that when the MATH-s are next all false at a MATH point then from then on the only MATH atoms holding are those with MATH. This forces the structure to be representing an accepting run of MATH as described above. |
cs/9910013 | For the ``only if" part, suppose MATH is a map graph. Let MATH be the set of nations in a map of MATH; for convenience we identify the MATH vertices of MATH with the corresponding nations in MATH. Consider a single nation MATH. Clearly at most MATH boundary points will account for all the adjacencies of MATH with other nations, and so a finite collection MATH of boundary points witnesses all the adjacencies among the nations in MATH. In each nation MATH we choose a representative interior point, and connect it with edges through the interior of MATH to the points of MATH bounding MATH. In this way we construct a planar embedding of the bipartite graph MATH, such that any two nations MATH and MATH overlap if and only if they have distance two in MATH. In other words, MATH is the half-square MATH. For the ``if" part, given a bipartite planar graph MATH, we embed it in the plane. By drawing a sufficiently thin star-shaped nation around each MATH and its edges in MATH, we obtain a map for MATH. |
cs/9910013 | Construct MATH as above. A point MATH is redundant if all pairs of its neighbors are also connected through other points of MATH. Deleting a redundant point does not change the half-square; we repeat this until MATH has no redundant points. Consider a drawing of MATH. For each MATH, we choose a pair of nations MATH and MATH connected only by MATH. Remove each MATH and its arcs, and replace them by a single arc from MATH to MATH. In this way, we draw a simple planar subgraph MATH of MATH with edge set MATH and vertex set MATH. Hence MATH, and MATH has less than MATH vertices. Since MATH is simple and bipartite, by NAME 's formula it has at most MATH edges, which is less than MATH. |
cs/9910013 | By REF , the map graph has a witness MATH with less than MATH edges. So, some nation MATH has degree at most REF in MATH, and consequently MATH has degree less than MATH in the map graph. Now we delete MATH, and prove our edge bound by induction on MATH. Since MATH where MATH ranges over all subsets of MATH containing at least two vertices CITE, and each MATH is again a MATH-map graph, the edge bound implies the arboricity bound. |
cs/9910013 | Let MATH be the graph realized by the hole-free REF-map in REF , and let MATH be a witness of MATH. We may assume that MATH has no point of degree REF. Let MATH be the bipartite graph in REF , and let MATH be the graph obtained from MATH by deleting points MATH and MATH and their incident edges. Since each MATH has exactly two neighbors in MATH, the degree of MATH in MATH is either REF or REF; we may assume the former case because in the latter case, the two points adjacent to nation MATH in MATH can be identified. By this assumption, MATH is an induced subgraph of MATH. In turn, by the existence of edges MATH and MATH in MATH and the planarity of MATH, MATH must be a (not necessarily induced) subgraph of MATH. Now, by REF and planarity, point MATH must also be adjacent to nations REF in MATH. Our argument so far did not depend on edge MATH in MATH, but now this edge has been forced by considering other edges. So in other words, MATH is not a map graph. |
cs/9910013 | By REF , we have a bipartite planar witness graph MATH such that the half-square MATH is the clique MATH, where MATH. Let MATH be the maximum degree of all points MATH. If MATH, we have a pizza. If MATH, we have a pizza-with-crust. So we may assume MATH. If MATH, then the map graph is planar; since MATH is not planar, this forces MATH and MATH, the rice-ball. We now assume MATH. Pick point MATH of maximum degree MATH, and nations MATH and MATH not adjacent to MATH. Consider the set MATH of all points connecting MATH or MATH to the nations around MATH. We claim that there is a point MATH connecting MATH, MATH, and at least two nations adjacent to MATH; otherwise, by drawing arcs through the points of MATH, we could get a planar MATH with the MATH nations on a common face, which is impossible. Since MATH has maximum degree, there are also two nations adjacent to MATH but not MATH. In summary, the following three disjoint sets each contain at least two nations: MATH . We will choose six distinct nations MATH, MATH, and MATH; no matter how we choose, the graph MATH will contain the induced subgraph in REF , with the cycle MATH. The graph MATH is a complete bipartite graph, with a planar embedding inherited from MATH. Each face of MATH is a REF-cycle; furthermore all nations in MATH must lie inside one face, in order to be connected by other points. So, we choose MATH on this face. Then this face is bounded by MATH; we have an embedding with MATH inside MATH, and MATH outside MATH. By an appropriate choice of nations MATH and MATH, we arrive at REF , the embedding of MATH, MATH, and all their edges to adjacent nations. In this figure, the three occurrences of MATH locate any other nations in MATH. There must exist a third point MATH inside MATH connecting MATH and MATH. These edges now separate MATH from MATH, so all these nations are adjacent to MATH as well, yielding REF . This figure is not necessarily an induced subgraph, since the edges MATH and MATH may occur in MATH. But by the maximality of the degree of MATH, if exactly MATH of these edges exist, then there exist MATH other nations MATH, necessarily outside MATH. So, no matter whether these edges exist or not, the points MATH, MATH, and MATH support a hamantasch on MATH. Hence, we are done if MATH. For contradiction, suppose MATH contains some nation MATH not adjacent to MATH or MATH. We need to place MATH, and some new points and edges, in REF so that MATH has neighbor points connected to the other nations. However by planarity of MATH, if MATH or MATH is an edge in MATH, then MATH cannot be placed so that both MATH and MATH have neighbor points connected to MATH. Similarly, if neither MATH nor MATH is an edge in MATH, then MATH cannot be placed so that all of MATH, MATH, and MATH have neighbor points connected to MATH. |
cs/9910013 | We may assume that MATH is connected. As in REF , we choose a planar witness MATH for MATH where MATH is the set of nations, MATH is the set of at most MATH points, and MATH is the set of edges. Fix a plane embedding of MATH. If vertices MATH appear in that cyclic order as distinct neighbors of some vertex MATH in MATH, then we say that the pairs MATH and MATH cross at MATH. Each point can contribute to at most one maximal pizza, and so there are at most MATH maximal pizzas in MATH. Note that each MATH is a pizza. Next, let MATH be the maximal cliques in MATH that are either non-pizza MATH's or hamantaschen. For each hamantasch MATH, we may choose three points MATH and three nations MATH such that MATH is an induced cycle in MATH and MATH consists of all nations adjacent to at least two of the points MATH in MATH. For each pair MATH of vertices in MATH such that some non-pizza MATH contains both MATH and MATH, let MATH be a point in MATH that is adjacent to both MATH and MATH in MATH. For each non-pizza MATH, let MATH, MATH, MATH, MATH, and MATH be the induced cycle MATH in MATH. In either case, we define MATH and note that MATH consists of all nations adjacent to at least two points of MATH. This implies MATH for distinct MATH and MATH, because otherwise MATH would be a larger clique. Define MATH as the simple graph with vertex set MATH and edge set MATH for some MATH. We claim that MATH is planar. To see this, we embed MATH in MATH by drawing the edge MATH of MATH through their neighbor MATH in MATH, and similarly for the other two edges MATH and MATH. Towards a contradiction, assume that two edges of MATH cross in the embedding. Then, by cycle symmetries, we may assume that for some distinct MATH and MATH, pairs MATH and MATH cross at nation MATH (call it MATH) in MATH. Since the cycles MATH and MATH cross at MATH in MATH, they must cross again, sharing either another nation or a point. CASE: MATH and MATH share another nation but no point. By symmetry, it suffices to consider the case MATH. If MATH and MATH were both non-pizza MATH's, then we would have MATH, contradicting the crossing. So at least one of MATH and MATH is a hamantasch, we suppose MATH. Then MATH has another nation MATH also adjacent to MATH and MATH or else MATH would be a pizza-with-crust. Because of MATH, it must be MATH. Now since MATH is adjacent to MATH, in order for MATH to be a non-pizza, MATH must also be a hamantasch. Then MATH has another nation MATH adjacent to MATH and MATH, but planarity of MATH makes this impossible. CASE: MATH and MATH share a point. Since MATH and MATH are induced, MATH is adjacent to neither MATH nor MATH, so the only possible shared point is MATH. In turn, MATH is a clique of MATH. So, neither MATH nor MATH is a MATH, and both are hamantaschen. Now as in the previous case, we find it is impossible to add a nation MATH between MATH and MATH and a nation MATH between MATH and MATH. By this, both MATH and MATH are pizza-with-crusts, a contradiction. By the above case-analysis, the claim holds, and MATH is a planar graph with at most MATH vertices and at least MATH distinct triangles. An easy exercise shows that any simple planar graph with MATH vertices has at most MATH triangles. So, MATH. There are at most MATH rice-balls, since they all have different center nations. It remains to bound the number of maximal pizza-with-crusts of size REF or more. Fix a point MATH in MATH, let MATH denote the set of nations adjacent to MATH in MATH, and let MATH be the maximal pizza-with-crusts with center MATH and size REF or more. We claim that MATH. This claim implies that MATH has at most MATH maximal pizza-with-crusts of size REF or more. This sum equals MATH; since MATH and MATH, the sum is less than MATH. Now we prove the claim. The embedding gives a cyclic clockwise order on the nations of MATH around MATH; this order defines ``consecutive" nations and ``intervals" of nations around MATH. For nations MATH, let MATH denote the circular interval of nations starting at MATH, proceeding clockwise around MATH, and ending at MATH. For each clique MATH in MATH, let MATH be the crust of MATH. Since MATH is not a pizza, we can choose distinct nations MATH and distinct points MATH satisfying the following three conditions: CASE: MATH is a simple cycle in MATH. CASE: If a nation of MATH is adjacent to MATH or MATH, then it lies outside MATH, otherwise it lies inside. CASE: All nations of MATH lying inside the cycle MATH lie in the interval MATH. Denote the unordered pair MATH by MATH, and MATH by MATH. By considering such REF-cycles MATH and the planarity of MATH, there are no cliques MATH such that MATH and MATH cross at MATH; consequently the graph MATH is simple outerplanar, where we use the same cyclic order on MATH for the outerplanar embedding. Since MATH is simple outerplanar, it can have at most MATH edges. Thus, to prove MATH, it suffices to prove that each edge of MATH equals MATH for at most two MATH. For contradiction, assume that there exist three distinct cliques MATH with MATH. Say that two of these cliques are nested if the crust of one is inside the cycle of the other. We consider two cases. CASE: There are two non-nested cliques. We may suppose that they are MATH and MATH. By planarity, the interiors of MATH and MATH are disjoint, with MATH and MATH. Moreover, no matter whether MATH or not, no nation of MATH is adjacent to both MATH and at least one point of MATH in MATH. Thus, the set of nations lying inside MATH and the set of nations lying inside MATH form a partition of MATH. By this and the maximality of cliques in MATH, the crust of MATH must lie inside MATH or MATH; by symmetry we suppose it lies inside MATH. On the other hand, since MATH is a maximal clique of size REF or more, there exists a nation MATH lying inside MATH. Since MATH cannot be adjacent to MATH or MATH, there is another point MATH lying inside MATH that connects MATH with MATH. So, we have a path MATH sharing only its endpoints with MATH, and bisecting the interior of MATH. Now the crust MATH must lie inside MATH, to one side or the other of MATH. To achieve MATH, MATH must be adjacent to MATH in MATH. But then we would see that MATH, and so either MATH or MATH was chosen incorrectly. CASE: All three cliques nest. Again by planarity, the cycles cannot cross. So, their interiors nest in some order; we may assume that MATH lies inside MATH, and MATH lies inside MATH. We have MATH and MATH. Since MATH is a maximal clique, it contains some nation MATH not adjacent to MATH in MATH. By planarity, MATH must lie inside MATH, and there is some point MATH connecting MATH with MATH in MATH. We cannot have MATH, by the choice of MATH; so again we have a path MATH, sharing only its endpoints with MATH and bisecting its interior. Now the crust MATH must lie inside MATH, to one side or the other of MATH; the rest of the argument proceeds as in the last case. |
cs/9910013 | Two edges would imply two MATH's, sitting inside REF-clique MATH. |
cs/9910013 | We first prove REF . If MATH, then nations MATH and MATH are disjoint disc homeomorphs in MATH, and hence removing them from MATH leaves exactly one connected region. So we know MATH. Let MATH be the number of MATH-segments in MATH. Consider the following three cases. CASE: MATH. We erase the MATH-points in MATH. Then, MATH becomes an atlas of MATH and nations MATH and MATH are disjoint disc homeomorphs in it. So, removing MATH and MATH from MATH leaves exactly one connected region. CASE: MATH. We erase the MATH-points in MATH. MATH remains an atlas of MATH. Moreover, edge MATH becomes good in MATH. Since the union of nations MATH and MATH is a disc homeomorph in MATH, removing them from MATH leaves exactly one connected region. CASE: MATH. We erase the MATH-points in MATH. MATH remains an atlas of MATH. Moreover, there are exactly MATH disjoint holes in MATH. So, removing nations MATH and MATH from MATH leaves exactly MATH connected regions. Each of these regions forms a connected component of MATH. This completes the proof of REF . We next prove REF . For each MATH, let MATH. Each hole in MATH is a REF-hole and is touched only by MATH and MATH, and hence erasing the holes in MATH yields an atlas of MATH. On the other hand, given an atlas MATH of each MATH, we erase any MATH-points in MATH. MATH remains an atlas of MATH, because edge MATH is marked in MATH and so there exists a MATH-segment in MATH. Since MATH is connected, REF implies there is exactly one MATH-segment. Thus removing nations MATH and MATH from MATH leaves exactly one connected region, and the closure MATH of this region is a disc homeomorph. The boundary of MATH can be divided into two curve segments MATH and MATH such that MATH (respectively, MATH) is a portion of the boundary of nation MATH (respectively, MATH) in MATH. Now, we can obtain an atlas of MATH as follows. First, put MATH, , MATH on the sphere in such a way that no two of them touch and each MATH appears on the upper half of the sphere while each MATH appears on the lower half. Second, draw nation MATH (respectively, MATH) to occupy the area of the upper (respectively, lower) half of the sphere that is occupied by no MATH. This gives an atlas of MATH. |
cs/9910013 | By REF , the ``if" part is obvious. For the other direction, suppose MATH is REF-connected. Let MATH be an atlas of MATH. If no edge of MATH is bad in MATH, then MATH is well-formed and we are done. So, suppose that some edge MATH is bad in MATH. Since MATH is REF-connected, there is at most one MATH-segment in MATH. If there is no MATH-segment in MATH, we erase all but one MATH-points in MATH; otherwise, we erase all the MATH-points in MATH. In both cases, MATH remains an atlas of MATH and edge MATH becomes good in MATH while no good edge becomes bad in MATH. Consequently, we can make all bad edges good in MATH. |
cs/9910013 | To prove REF , suppose that MATH is not a clique. For each edge MATH, if nations MATH and MATH weakly touch in MATH, then we erase the MATH-point in MATH. Then, MATH is a layout of MATH and the holes in MATH are disjoint disc homeomorphs. So, MATH must be connected. To prove REF , suppose that MATH is a clique. The ``if" part is clear. To prove the ``only if" part, suppose that REF or REF does not hold. In REF is false, MATH, MATH and MATH meet at a point in MATH, and the well-formedness of MATH ensures that their union is a disc homeomorph, and so MATH is connected. Otherwise, suppose REF is true and REF is false. For each edge MATH, if nations MATH and MATH weakly touch in MATH, then we erase the MATH-point to get atlas MATH. Then MATH is a layout of MATH and the holes in MATH are disjoint disc homeomorphs. So, MATH is connected. Next, we prove REF . Since MATH is disconnected, REF hold. By this, there are exactly two holes MATH and MATH in MATH and they are disjoint. For MATH, let MATH be the set of nations that occupy MATH in atlas MATH. Each MATH is a connected component of MATH. Let MATH be the marked graph obtained from MATH by marking the edges in MATH. There is a unique hole in MATH and it is (strongly) touched only by the nations of MATH. So, modifying MATH by extending nation MATH to occupy its unique hole yields a well-formed atlas of MATH. Similarly, we can obtain a well-formed atlas of MATH. On the other hand, suppose that we are given a well-formed atlas MATH of MATH and one MATH of MATH. Let MATH. Since the edges in MATH are marked in MATH, each pair of nations of MATH strongly touch in MATH. Note that MATH is connected. Then by REF and the well-formedness of MATH, the nations in MATH meet at a REF-point MATH in MATH. Let MATH be a disc that is centered at MATH and touches no nation of MATH in atlas MATH. To obtain a well-formed atlas of MATH, we remove each MATH from MATH to obtain a connected region MATH, and then glue MATH and MATH together by identifying nations MATH, MATH, MATH in MATH with those in MATH, respectively. |
cs/9910013 | First observe that no hole-free REF-map graph has a REF-clique, by REF . Assume, on the contrary, that MATH has a MATH. Then it must be a hamantasch, and by REF displays MATH. Thus, MATH, contradicting REF . |
cs/9910013 | Let MATH be a well-formed atlas of MATH in which nations MATH, MATH, MATH, MATH meet at a point MATH in this order. After erasing the MATH-point MATH in MATH, we obtain a well-formed atlas of MATH in which nations MATH, MATH, and MATH meet at a REF-point and nations MATH, MATH, and MATH meet at a REF-point. Thus, by REF , both MATH and MATH are connected. Let MATH be a well-formed atlas of MATH. Since MATH is connected and the edges MATH, MATH, and MATH are marked in MATH, nations MATH, MATH, and MATH must meet at a REF-point MATH in MATH according to REF . Similarly, nations MATH, MATH, and MATH must meet at a REF-point MATH in MATH. Thus, the boundaries of MATH and MATH in MATH share a curve segment MATH with endpoints MATH and MATH. We modify MATH by contracting MATH to a point, obtaining a well-formed atlas of MATH. |
cs/9910013 | Let MATH be the atlas of MATH obtained from MATH by contracting those shrinkable segments whose ending nations are MATH and MATH. All edges except MATH are good in MATH. First, we claim that for every MATH such that MATH and MATH belong to different connected components of MATH, there is a point in MATH at which nations MATH, MATH, MATH, MATH meet at MATH cyclically in this order. Towards a contradiction, assume that such a point does not exist in MATH. By the definition of MATH, MATH is in MATH. There is no nation MATH adjacent to both MATH and MATH; otherwise, MATH would connect MATH and MATH in MATH by REF . By the fact that MATH has no hole, nations MATH and MATH only share a unique curve segment MATH in MATH and the endpoints of MATH can be touched only by MATH or MATH in MATH. Nation MATH cannot touch both endpoints of MATH; otherwise, since the edges MATH and MATH are still good in MATH, nations MATH, MATH, and MATH would have to occupy the whole sphere, a contradiction. Similarly, MATH cannot touch both endpoints of MATH. So, both endpoints of MATH are REF-points. In summary, one endpoints of MATH is touched by MATH and the other is touched by MATH. Then MATH would be a shrinkable segment whose ending nations are MATH and MATH, contradicting the choice of MATH. Second, we claim that there is no MATH-segment in MATH. Towards a contradiction, assume that a MATH-segment MATH exists in MATH. By the first claim, there is a MATH-point MATH in MATH. Note that MATH is not on MATH. Let MATH and MATH be the two nations of MATH that meet at MATH. Since MATH has no hole, MATH is REF-connected and MATH, there is a nation MATH that touches either MATH or MATH in MATH. If MATH touches MATH (respectively, MATH) in MATH, then MATH is not reachable from MATH (respectively, MATH) in MATH (respectively, MATH), contradicting REF . Third, we claim that there are at least two MATH-points in MATH. Otherwise, if there is only one MATH-point, then by the first claim, MATH would have at most one edge, and REF-connectivity of MATH would prevent the separation of MATH. Let MATH be the number of MATH-points. Since MATH and there is no MATH-segment, we see that the atlas MATH has a cyclic sequence of MATH-points MATH, , MATH. These points alternate with MATH REF-holes in MATH; REF displays MATH when MATH. For each MATH, let MATH and MATH be the nations meeting MATH and MATH. We claim that MATH is a MATH; otherwise to form a containing REF-clique would force MATH and MATH, contradicting the separation of MATH. So in fact each MATH corresponds to an edge MATH in MATH, and the components of MATH correspond to the set of nations in each hole. Now consider a particular edge MATH. To show that MATH is a correct REF-pizza, we must find a well-formed atlas of MATH where they meet as they do in MATH. This is easy to do: we simply erase all MATH-points in MATH except MATH, and the resulting atlas is well-formed. |
cs/9910013 | The ``if" part is obvious from the proof of REF . To prove the ``only if" part, suppose there is a shrinkable segment MATH in MATH with ending nations MATH and MATH. Let MATH be the atlas of MATH obtained from MATH by contracting MATH to a single point MATH. Similarly to the second claim in the proof of REF , we can claim that there is no MATH-segment in MATH. Thus, besides MATH, there is exactly one MATH-point MATH in MATH, inherited from MATH. Now, MATH has exactly two holes MATH and MATH. Let MATH (respectively, MATH) be the set of nations of MATH occupying MATH (respectively, MATH) in atlas MATH. Let MATH and MATH be the two nations that meet at MATH in MATH. Similarly, let MATH and MATH be the two nations that meet at MATH in MATH. By MATH, edges MATH and MATH are not marked in MATH and they are all the edges connecting nations of MATH to those of MATH. Moreover, since no REF-clique contains MATH, MATH and MATH are MATH's of MATH. Therefore, no connected component of MATH contains both the nations of MATH and those of MATH. In other words, MATH is a separating edge of MATH. |
cs/9910013 | Since MATH is a cycle and MATH is well-formed, there are exactly two holes MATH and MATH in MATH. For MATH, let MATH be the set of nations that occupy MATH in atlas MATH. Clearly, the nations in MATH are connected together in MATH. By this and the assumption that MATH is disconnected, both MATH and MATH are connected components of MATH and MATH has no other connected component. So, MATH and MATH must be disjoint. In turn, for each edge MATH in MATH, there is a MATH-segment in MATH. For each MATH, there is a unique hole in MATH and it is (strongly) touched only by the nations of MATH. So, modifying MATH by extending nation MATH to cover its unique hole yields a well-formed atlas of MATH in which nations MATH, MATH, and MATH meet at a REF-point and nations MATH, MATH, and MATH meet at a REF-point. So, by REF , both MATH and MATH are connected. Suppose that we are given an atlas MATH for each MATH. Since MATH is connected and the three edges MATH, MATH, and MATH are marked in MATH, nations MATH, MATH, and MATH meet at a REF-point in MATH, by REF . Similarly, nations MATH, MATH, and MATH must meet at a REF-point in MATH. Thus, by the well-formedness of MATH, REF displays MATH. By the figure, we can cut off a small area inside MATH around the MATH-segment; the nations in the other atlas can be embedded in the resulting open area. |
cs/9910013 | Let MATH and MATH. Since MATH is REF-connected, MATH is connected. So MATH is non-empty to disconnect MATH, and we may choose MATH such that MATH and MATH belong to different components of MATH. By definition of MATH, MATH is a MATH in MATH. We claim that nations MATH and MATH do not strongly touch in MATH. Assume, on the contrary, that a MATH-segment MATH exists in MATH. Since MATH has no hole, there are nations MATH, MATH in MATH such that MATH touches one endpoint of MATH and MATH touches the other. If MATH were not in MATH, then MATH would connect MATH and MATH in MATH in MATH by REF . Thus MATH, and similarly MATH. By the well-formedness of MATH and the fact that MATH, we can verify that there is no way for nation MATH or MATH to touch both endpoints of MATH. So, nation MATH touches one endpoint of MATH and MATH touches the other; REF displays MATH. By REF and the fact that edge MATH is not marked in MATH, MATH is a separating edge of MATH, a contradiction. Thus, the claim holds. By the claim, nations MATH and MATH weakly touch at a point MATH in MATH. Since MATH has no hole, there are two distinct nations MATH, MATH in MATH which meet at MATH in MATH. As before, we can show that MATH. Thus, the four nations a, MATH, b, MATH appear around MATH cyclically in this order in MATH. In turn, by the well-formedness of MATH, MATH is a correct REF-pizza of MATH. The discussions above actually prove that for every pair of adjacent nations MATH and MATH of MATH that belong to different connected components of MATH, nations MATH, MATH, MATH, MATH must meet at a REF-point cyclically in this order in MATH. Since MATH is the unique point in MATH at which MATH and MATH meet, MATH is the unique pair of adjacent nations of MATH that belong to different connected components of MATH. We now claim that the connected components of MATH are only MATH and MATH. Assume, on the contrary, that MATH has a connected component MATH other than MATH and MATH. Then, there exists a nation MATH which touches some nation MATH of MATH in MATH; otherwise, MATH would be a connected component of MATH, a contradiction. But now, MATH is a pair of adjacent nations of MATH that belong to different connected components of MATH, a contradiction. Thus, the connected components of MATH are only MATH and MATH. We may assume that MATH and MATH. MATH and MATH only touch at MATH; otherwise, nations MATH and MATH would have to meet at a point other than MATH in MATH, a contradiction against the well-formedness of MATH. Hence, MATH and MATH. |
cs/9910013 | The ``if" part is obvious. For the ``only if" part, suppose MATH has a separating quadruple MATH. If MATH has a separating REF-cycle MATH, then as observed in the proof of REF , each pair of adjacent nations of MATH strongly touch in MATH. Otherwise, as observed in the modified proof of REF for REF , exactly one pair of adjacent nations of MATH weakly touch in MATH. |
cs/9910013 | Since MATH, either MATH or MATH is a MATH of MATH. In both cases, MATH. For the last part, such a REF-point would imply a REF-clique containing the MATH, contradicting its maximality. |
cs/9910013 | For REF , let MATH. Since MATH is connected, some edge in MATH connects MATH to an outside vertex, so to support the corresponding MATH, MATH must contain either MATH or MATH. If MATH, then MATH would be a separating edge of MATH, separating MATH from the rest. So, MATH. For REF , if on the contrary MATH for every MATH, then MATH would be a component of MATH, contradicting REF . |
cs/9910013 | Since MATH and MATH are disconnected in MATH, at least two of the triangle edges are removed. REF applied to these edges implies MATH. On the other hand, by REF at most one triangle edge is in each of MATH and MATH, so in fact exactly two edges are removed, and either edge MATH or MATH remains in MATH. We suppose MATH remains, the other case is similar (swap MATH and MATH). We also suppose MATH and MATH, the other case is similar (swap MATH and MATH). Then MATH, MATH, and MATH. On the other hand, MATH cannot have the edges MATH or MATH, since either would imply a REF-clique containing a MATH. So, the first assertion holds. For the second assertion, suppose on the contrary there is a MATH with MATH. As above, we suppose that both MATH and MATH are MATH's of MATH. Then neither MATH nor MATH is a subset of MATH, since otherwise MATH would extend one of these MATH's to a REF-clique. But then the edges from MATH would all survive in MATH, contradicting the disconnection of MATH and MATH. |
cs/9910013 | Assume, on the contrary, that REF displays MATH. Let MATH be the point in MATH at which nations MATH, MATH, and MATH meet. Let MATH (respectively, MATH) be the endpoint of the MATH-segment (respectively, MATH-segment) other than MATH in MATH. There must exist a nation MATH which touches MATH in MATH. By the well-formedness of MATH, nation MATH touches MATH only at MATH and MATH is not a marked edge in MATH. Let MATH be the connected component of MATH containing MATH. Let MATH be a connected component of MATH other than MATH such that some nation MATH of MATH touches some nation MATH of MATH in MATH; MATH exists by REF . We claim that nation MATH touches some nation of MATH in MATH. Assume, on the contrary, that the claim is false. Clearly, MATH or MATH is a MATH of MATH. Since no nation of MATH touches MATH in MATH, MATH. That is, MATH or MATH is a MATH of MATH. Since MATH, we have MATH; otherwise, MATH would be a REF-clique of MATH. We assume that MATH; the other case is similar. Then, since nation MATH cannot touch nation MATH in MATH and MATH has no hole, there is a point in MATH at which nations MATH, MATH and some MATH meet. By REF , MATH and there is no MATH such that MATH. Now, we see that REF displays MATH. There is no MATH with MATH; otherwise, MATH by REF , which is impossible by REF . Similarly, there is no MATH with MATH. Thus, REF is transformable to REF . By REF and the fact that MATH is not a marked edge in MATH, MATH is a separating quadruple of MATH, a contradiction. So, the claim holds. Next, we claim that for every connected component MATH of MATH, there is no point MATH in MATH at which two nations MATH and MATH of MATH together with two nations MATH and MATH of MATH meet cyclically in the order MATH, MATH, MATH, MATH. Assume, on the contrary, that such MATH exists in MATH. Then, by REF , MATH. By REF , MATH and hence MATH. So, MATH. In turn, MATH; otherwise, by REF , MATH would be a REF-clique of MATH, a contradiction. We assume that MATH; the other case is similar. Now, by REF , MATH and MATH. We assume that MATH and MATH; the other case is similar. In summary, REF displays MATH. There is no MATH with MATH; otherwise, by REF , MATH which is impossible by REF . Similarly, there is no MATH with MATH. So, REF is transformable to REF . By the latter figure, MATH would be a separating triple of MATH, a contradiction. So, the claim holds. By the above two claims and the fact REF that MATH, MATH or MATH is touched by both MATH and MATH in MATH. Suppose that MATH is touched by both MATH and MATH; the other case is similar. Let MATH (respectively, MATH) be the nation of MATH (respectively, MATH) touching MATH. Then, the boundaries of nations MATH and MATH in MATH share a curve segment MATH. One endpoint of MATH is MATH. Let MATH be the other endpoint of MATH. Neither nation MATH nor MATH touches MATH in MATH; otherwise, MATH or MATH would be a REF-clique of MATH. By the well-formedness of MATH, it is impossible that nation MATH or MATH touches MATH. In turn, since MATH has no hole, there is a nation MATH that touches MATH in MATH. Now, by REF , MATH and either REF MATH and MATH or REF MATH and MATH. In REF holds, REF displays MATH and MATH would be a separating quadruple, a contradiction. So, REF holds and only REF can possibly display MATH. There is no MATH with MATH; otherwise, by REF , MATH which is impossible by REF . By this, REF is transformable to REF , by which MATH would be a separating triple of MATH, a contradiction. This completes the proof. |
cs/9910013 | Assume, on the contrary, that REF displays MATH. We assume that MATH in the figure; the other case is similar. Define points MATH and MATH, nation MATH and MATH as in the proof of REF . By the well-formedness of MATH, nation MATH meets MATH only at MATH and MATH is not a marked edge in MATH. Let MATH be the endpoint of the MATH-segment other than MATH in MATH. We claim that MATH touches nation MATH in MATH. Assume, on the contrary, that MATH does not touch nation MATH. Let MATH be a connected component of MATH other than MATH such that some nation MATH of MATH touches some nation MATH of MATH in MATH. By REF , such MATH exists. Clearly, MATH or MATH is a MATH of MATH. CASE: MATH. Then MATH and MATH is a MATH of MATH. Moreover, there is no MATH with MATH; otherwise, since MATH, we would have MATH and MATH by REF , and in turn MATH would be a vertex of MATH that touches nation MATH in MATH, a contradiction. So, by the fact that MATH has no hole, the boundaries of nations MATH and MATH share a curve segment MATH in MATH, and both endpoints of MATH are REF-points one of which is touched by MATH and the other is touched by MATH in MATH. By this, MATH is a shrinkable segment in MATH, MATH and MATH fall into different connected components of MATH, and MATH would be a separating edge of MATH, a contradiction. CASE: MATH. Then MATH or MATH is a MATH of MATH. Since MATH, we have MATH; otherwise, MATH would be a REF-clique of MATH. So we have two sub-cases. CASE: MATH. Then MATH and MATH is a MATH of MATH. Moreover, since nation MATH cannot touch nation MATH in MATH and MATH has no hole, there is a point in MATH at which nations MATH, MATH and some MATH meet. By REF , MATH and there is no MATH such that MATH. Now, we see that REF displays MATH. There is no MATH with MATH; otherwise, MATH by REF , which is impossible by REF . Thus, REF is transformable to REF . By REF , MATH is a separating quadruple of MATH, a contradiction. CASE: MATH. If there is a MATH with MATH, then similarly to REF, we can prove that MATH would be a separating quadruple of MATH, a contradiction. Otherwise, the boundaries of nations MATH and MATH share a curve segment MATH in MATH, and both endpoints of MATH are REF-points one of which is touched by MATH and the other is touched by MATH in MATH; by this, MATH is a shrinkable segment in MATH, MATH and MATH fall into different connected components of MATH, and MATH would be a separating edge of MATH, a contradiction. Therefore, the claim holds: MATH touches MATH. Next, we claim that for every connected component MATH of MATH, there is no point MATH in MATH at which two nations MATH and MATH of MATH together with two nations MATH and MATH of MATH meet cyclically in the order MATH, MATH, MATH, MATH. Assume, on the contrary, that such MATH exists in MATH. Then, by REF , MATH. By REF , MATH and hence MATH. So, MATH. In turn, MATH; otherwise, by REF , MATH would be a REF-clique of MATH, a contradiction. We assume that MATH; the other case is similar. Now, by REF , MATH and MATH. We assume that MATH and MATH; the other case is similar. In summary, REF displays MATH. There is no MATH with MATH; otherwise, by REF , MATH which is impossible by REF . Similarly, there is no MATH with MATH. So, REF is transformable to REF . By the latter figure, MATH would be a separating triple of MATH, a contradiction. So, the claim holds. By the above two claims and the fact that MATH, MATH or MATH is touched by both MATH and MATH in MATH. By REF , MATH cannot be touched by both MATH and MATH. So, MATH is touched by both MATH and MATH. Let MATH (respectively, MATH) be the nation of MATH (respectively, MATH) touching MATH. Then, the boundaries of nations MATH and MATH in MATH share a curve segment MATH. One endpoint of MATH is MATH. Let MATH be the other endpoint of MATH. Neither nation MATH nor MATH touches MATH in MATH; otherwise, MATH or MATH would be a REF-clique of MATH. By the well-formedness of MATH, it is impossible that nation MATH or MATH touches MATH. So, there is a nation MATH that touches MATH in MATH. Now, by REF , MATH and either REF MATH and MATH or REF MATH and MATH. In REF holds, REF displays MATH; by the figure, it is impossible for nation MATH to touch all of nations MATH, MATH, and MATH in MATH, a contradiction. So, only REF can possibly display MATH. There is no MATH with MATH; otherwise, by REF , MATH which is impossible by REF . By this, REF is transformable to REF , by which MATH would be a separating quadruple of MATH, a contradiction. This completes the proof. |
cs/9910013 | Assume, on the contrary, that REF displays MATH. Define points MATH, MATH and MATH as in the proof of REF . Let MATH be the endpoint of the MATH-segment other than MATH in MATH. Similarly to the proof of REF , we can prove that for every connected component MATH of MATH, there is no point MATH in MATH at which two nations MATH and MATH of MATH together with two nations MATH and MATH of MATH meet cyclically in the order MATH, MATH, MATH, MATH. Let MATH and MATH be two connected components of MATH such that MATH. By the above claim, REF , and REF , MATH or MATH is touched by both MATH and MATH in MATH. We assume that MATH is touched by both MATH and MATH in MATH; the other case is similar. Let MATH (respectively, MATH) be the nation of MATH (respectively, MATH) touching MATH. Similarly to the proof of REF , we can prove that there is a nation MATH such that only REF or REF can possibly displays MATH. If REF displays it, then MATH would be a separating quadruple (indeed, a separating REF-cycle) of MATH, a contradiction. So, suppose that REF displays it. Then, there is no MATH with MATH; otherwise, by REF , MATH which is impossible by REF . By this, REF is transformable to REF . By the latter figure, MATH would be a separating quadruple of MATH, a contradiction. This completes the proof. |
cs/9910013 | Let MATH be one hole of MATH, and MATH be the other. Let MATH (respectively, MATH) be the set of nations in MATH that occupy hole MATH (respectively, MATH) in atlas MATH. Let MATH be the point at which nations MATH and MATH together with some nation of MATH meet in MATH. Define points MATH and MATH similarly. First, we observe that MATH for every connected component MATH of MATH. If MATH, then this is clear from REF . Otherwise MATH contains some other component MATH adjacent to MATH in MATH, and now our argument resembles that for REF . That is, let MATH. Since an edge between MATH and MATH is absent in MATH, MATH contains either MATH or MATH. Assume MATH; the MATH case is similar. Then, in case MATH is also a connected component of MATH, it is clear that MATH would be a separating edge in MATH, separating MATH from MATH. In case MATH is not a connected component of MATH, there is exactly one edge MATH with MATH and MATH; moreover, the four nations MATH must meet at point MATH in atlas MATH cyclically in this order (so, the MATH-segment in the layout in REF should be contracted to a point). If MATH is a MATH of MATH, then MATH would be a connected component of MATH, a contradiction. Otherwise, there is a REF-clique MATH containing MATH. The nation MATH must belong to MATH and touch nation MATH, in order to touch MATH and MATH. By this, edge MATH remains in MATH, and MATH, contradicting the fact that MATH is a connected component of MATH. So, MATH. Similarly, we have MATH for every connected component MATH of MATH. We assume that MATH; the other case is similar. We want to prove that MATH. Towards a contradiction, assume that MATH. Then, since MATH has no hole, there is a connected component MATH of MATH with MATH. First, we claim that MATH and a nation of MATH must meet at MATH, MATH, or MATH. Assume, on the contrary, that the claim does not hold. Then, since MATH by the above observation, there must exist a point MATH in MATH at which two nations MATH and MATH of MATH together with MATH and some MATH meet cyclically in the order MATH, MATH, MATH, MATH. REF ensures that either REF MATH and MATH or REF MATH and MATH. In either case, we have MATH. We assume that REF holds; the other case is similar. Then, REF displays MATH. There is no MATH with MATH; otherwise, by REF , MATH which is impossible by REF . Similarly, there is no MATH with MATH. So, REF is transformable to REF . By the latter figure and REF , MATH and each of MATH and MATH strongly touch in MATH. In turn, by the fact that MATH is not a separating triple of MATH, nations MATH, MATH and MATH meet at a point MATH in MATH. By REF , MATH must be a REF-point; so, MATH and MATH strongly touch in MATH. In summary, REF is transformable to REF . By the latter, MATH would be a separating quadruple of MATH, a contradiction. So, the claim holds: MATH meets MATH at MATH, MATH, or MATH. Next, we use the above claim to get a contradiction. By the above claim, MATH and a nation MATH of MATH must meet at MATH, MATH, or MATH in MATH. By REF , MATH and MATH cannot meet at MATH. So, they meet at MATH or MATH. We assume that they meet at MATH; the other case is similar. Then, the boundaries of nations MATH and MATH in MATH share a curve segment MATH. One endpoint of MATH is MATH. Let MATH be the other endpoint of MATH. Nation MATH cannot touch MATH in MATH; otherwise, MATH would be a REF-clique of MATH. By the well-formedness of MATH, it is impossible that only nations MATH and MATH touch MATH. So, there is a nation MATH that touches MATH in MATH. Now, by REF , MATH. Thus, REF or REF displays MATH. Actually, the former does not display MATH or else MATH would be a separating triple of MATH, a contradiction. So, only REF can possibly display MATH. There is no MATH with MATH; otherwise, by REF , MATH which is impossible by REF . By this, REF is transformable to REF . The latter is further transformable to REF , because each pair of nations in MATH must strongly touch in MATH by REF and the fact that MATH is not a separating triple of MATH. By REF , MATH would be a separating quadruple of MATH, a contradiction. This completes the proof that MATH. Now, MATH. Thus, by REF (MATH has no separating triple), REF displays MATH. By the figure and the fact that MATH constitutes a connected component of MATH, there are exactly two distinct nations MATH and MATH of MATH such that MATH and MATH. By the figure, both MATH and MATH are correct REF-pizzas. Since MATH, finding MATH and MATH is easy. This completes the proof of REF . |
cs/9910013 | Define sets MATH and MATH and points MATH, MATH, and MATH as in REF . As in the proof of REF , we observe that MATH for every connected component MATH of MATH or MATH. We claim that for every connected component MATH of MATH, there is no point MATH in MATH at which two nations MATH and MATH of MATH together with two nations MATH and MATH of MATH meet cyclically in the order MATH, MATH, MATH, MATH. Assume, on the contrary, that such MATH exists in MATH. Then, by REF with MATH, MATH. By REF , MATH and hence MATH. So, MATH. In turn, MATH; otherwise, by REF , MATH would be a REF-clique of MATH, a contradiction. We assume that MATH; the other case is similar. Now, by REF , MATH and MATH. We assume that MATH and MATH; the other case is similar. In summary, REF displays MATH. There is no MATH with MATH; otherwise, by REF , MATH which is impossible by REF . Similarly, there is no MATH with MATH. So, REF is transformable to REF . The latter figure is further transformable to REF , because REF MATH and MATH are not separating triples of MATH and REF both MATH and MATH are MATH's of MATH. By REF , MATH would be a strongly separating triangle of MATH, a contradiction. So, the claim holds. Next, we claim that MATH is connected. Assume, on the contrary, that MATH is disconnected. Then, since MATH has no hole, there are two distinct connected components MATH and MATH of MATH such that MATH. Since MATH and MATH, some nation MATH of MATH and some nation MATH of MATH have to meet at MATH, MATH or MATH in MATH, by the claim of the previous paragraph and REF . By REF , MATH and MATH cannot meet at MATH in MATH. We assume that MATH and MATH meet at MATH in MATH; the other case is similar. Similarly to the proof of REF , we can prove that there is a nation MATH such that only REF or REF can possibly display MATH. Actually, REF does not display it or else MATH would be a separating triple of MATH. So, only REF can possibly display it. Since MATH is not a separating triple of MATH, REF is transformable to REF . By the latter figure, MATH would be a strongly separating triangle of MATH, a contradiction. So, the claim holds. Similarly, we can prove that MATH is connected. Since both MATH and MATH are connected, both are connected components of MATH and MATH has no other connected component. So, by REF , the figure obtained from REF by contracting the bold MATH-segment to a single point does not display MATH. In turn, the bold MATH-segment in REF should be contracted to a single point; otherwise, MATH would be a separating triple of MATH. Similarly,the bold MATH-segment in REF should be contracted to a single point. Thus, REF displays MATH. Let MATH (respectively, MATH) be the point where nations MATH and MATH (respectively, nations MATH and MATH) meet in MATH. By the figure, a unique nation MATH and a unique nation MATH meet at MATH, and MATH is a MATH of MATH. Similarly, a unique nation MATH and a unique nation MATH meet at MATH, and MATH is a MATH of MATH. Moreover, both MATH and MATH are correct REF-pizzas. By the figure, other than MATH and MATH, there is no MATH with MATH and MATH. |
cs/9910013 | The ``if" part is obvious. For the ``only if" part, suppose MATH has a separating triangle MATH. If MATH is not REF-connected, then by REF , there is a REF-clique MATH in MATH such that the nations of MATH do not meet at a point in MATH and every pair of nations of MATH strongly touch in MATH. So, we may assume that MATH is REF-connected. Then, by the proof of REF , in case MATH has a separating triple MATH, the nations of MATH do not meet at a point in MATH and at most one pair of nations of MATH weakly touch in MATH. Thus, we may further assume that MATH has no separating triple. Then, by the layouts found in REF through REF, the nations of MATH do not meet at a point in MATH and at least one pair of nations of MATH strongly touch in MATH. |
cs/9910013 | For the ``only if" direction, suppose that REF displays MATH. Then, MATH has four holes, and each hole is touched by exactly three nations of MATH. For each REF-subset MATH of MATH, let MATH be the hole touched by the nations of MATH, and let MATH be the nations of MATH that occupy MATH in atlas MATH. We want to prove that for each REF-subset MATH of MATH, MATH. To this end, first observe that for each connected component MATH of MATH, there is a REF-subset MATH of MATH with MATH. This observation follows from REF immediately. Consequently, MATH; we claim they are equal. Towards a contradiction, assume that MATH has a connected component MATH with MATH. Let MATH. If MATH, then MATH would be a connected component of MATH, a contradiction. If MATH, then MATH is a connected component of MATH, and the vertices of MATH define a separating edge, a contradiction. So, the claim holds. By this claim, the above observation and REF , we have MATH for each REF-subset MATH of MATH. In turn, by REF through REF hold. For the ``if" direction, suppose that REF through REF hold. We first prove that REF does not display MATH. Towards a contradiction, assume that REF displays MATH. We may assume that MATH in REF ; the other cases are similar. Let MATH be a REF-subset of MATH. We claim that there is no point in MATH at which two nations MATH and MATH of MATH together with two nations MATH and MATH of MATH meet cyclically in the order MATH, MATH, MATH, MATH. This claim holds; otherwise, MATH by REF , so MATH or MATH belongs to MATH for some REF-subset MATH of MATH other than MATH, and in turn MATH would be a subset of MATH, a contradiction against REF in the lemma. By this claim and REF , the nations of MATH form a disc homeomorph in MATH. Thus, by REF displays MATH. By this figure, there is a REF-point MATH in MATH such that for each REF-subset MATH of MATH, exactly one nation MATH touches MATH. Since the nations MATH meet at MATH but no two of them belong to the same connected component of MATH, MATH. In turn, by REF-connectedness of MATH, each MATH would equal MATH, contradicting REF . Therefore, REF does not display MATH. We next prove that REF does not display MATH. Towards a contradiction, assume that REF displays MATH. As in the last paragraph, we may assume MATH, and we claim that the nations of each MATH form a disc homeomorph in MATH. Thus, by REF displays MATH. By this figure, there is a REF-point MATH in MATH at which nation MATH, some MATH, some MATH, and some MATH meet. Since MATH, MATH and MATH meet at MATH but no two of them belong to the same connected component of MATH, MATH, MATH, and MATH. In turn, by REF , MATH. Moreover, nations MATH, MATH, MATH, MATH meet at a point in MATH, and nations MATH, MATH, MATH, MATH meet at a point in MATH. Thus, MATH or else MATH would be a separating triple of MATH, a contradiction. Similarly, MATH. In a similar way, we can also prove that MATH. In summary, we have MATH, a contradiction. Therefore, REF does not display MATH. Since both REF do not display MATH, only REF can display MATH. This completes the proof. |
cs/9910014 | For the remainder of this proof, we consider a fixed EUF formula MATH. We will only consider a function to be a parsing if it is a parsing when the set of g-function symbols is MATH and the set of p-function symbols is MATH. We prove this theorem by induction on the syntactic structure of MATH. Our induction hypothesis consists of four assertions, two for terms and two for formulas: CASE: For MATH such that MATH or MATH is a function application with a function symbol in MATH, there is a parsing of MATH as a g-term. CASE: For MATH, there is a parsing of MATH as a p-term. CASE: For MATH satisfying one of the following conditions: CASE: MATH is MATH or MATH, CASE: MATH is a formula of the form MATH, CASE: MATH is a predicate application, CASE: MATH is in MATH, there is a parsing of MATH as a g-formula. CASE: For MATH, there is a parsing of MATH as a p-formula. Recall that the syntax of PEUF allows any g-formula to be promoted to a p-formula, and any g-term to be promoted to a p-term. These promotion rules will be used several times in the proof. For the base cases, we consider expressions having no proper subexpressions: CASE: For a domain variable MATH, if MATH, then MATH, so there is a parsing of MATH as a g-term and a parsing as a p-term. CASE: For a domain variable MATH, MATH is in MATH, so there is a parsing of MATH as a p-term. CASE: EUF formulas MATH and MATH can be parsed as either g-formulas or p-formulas. CASE: For a propositional variable MATH, there is a parsing of MATH as a g-formula or as a p-formula. For the inductive argument, we prove the following cases for EUF expressions, assuming that all proper subexpressions obey the induction hypothesis. CASE: Terms in MATH: CASE: Consider MATH. If MATH, then by definition, MATH and MATH. Thus, by the inductive hypothesis, there are parsings of MATH as a g-formula and of MATH and MATH as g-terms. This means there is a parsing of MATH as a g-term. If MATH, then by the inductive hypothesis, there are parsings of MATH as a g-formula and of MATH and MATH as p-terms. Thus there is a parsing of MATH as a p-term. CASE: Consider MATH. By the inductive hypothesis, there are parsings of MATH as p-terms. When MATH, there are parsings of MATH as a g-term and, by promotion, as a p-term. When MATH, there is a parsing of MATH as a p-term. Thus, there is a parsing of MATH as a p-term in either case. In addition, when MATH, we must have MATH, and hence there is also a parsing of MATH as a g-term. CASE: Formulas in MATH: CASE: Consider MATH. We have MATH, so there is a parsing of MATH as a g-formula. Hence MATH can be parsed as a g-formula or a p-formula. CASE: Consider MATH. If MATH is in MATH, then MATH are in MATH, so MATH can be parsed as g-formulas and MATH can be parsed as a g-formula or as a p-formula. If MATH is in MATH, then MATH can be parsed as p-formulas, so MATH can be parsed as a p-formula. CASE: Consider MATH. Similar to previous case. CASE: Consider MATH. If MATH, then MATH and hence MATH and MATH can be parsed as g-terms, so MATH can be parsed as a g-formula or as a p-formula. If MATH, then MATH and MATH can be parsed as p-terms, so MATH can be parsed as a p-formula. CASE: Consider MATH. By the inductive hypothesis, there are parsings of MATH as p-terms. Thus there is a parsing of MATH as a g-formula, and by promotion, as a p-formula. The theorem follows directly from the induction hypothesis. |
cs/9910014 | Let MATH be a term occurring in MATH of the form MATH, where MATH is a p-function symbol. Let MATH be a term occurring in MATH of the form MATH, where MATH may be either a p-function or a g-function symbol. Assume furthermore that MATH and MATH both equal MATH, but that either symbols MATH and MATH differ, or MATH for some value of MATH. Let MATH be a value not in MATH, and define a new domain MATH. Our strategy is to construct an interpretation MATH over MATH that partitions the terms in MATH in the same way as MATH, except that it splits the class containing terms MATH and MATH into two parts - one containing MATH and evaluating to MATH, and the other containing MATH and evaluating to MATH. Define function MATH to map elements of MATH back to their counterparts in MATH, that is, MATH, while all other values of MATH give MATH equal to MATH. For p-function symbol MATH, define MATH as: MATH . For other function and predicate symbols, MATH is defined to preserve the functionality of interpretation MATH, while also treating argument values of MATH the same as MATH. That is, MATH for function symbol MATH having MATH equal to MATH is defined such that MATH. Similarly, MATH for predicate symbol MATH having MATH equal to MATH is defined such that MATH. We claim the following properties for the different forms of subexpressions occurring in MATH: CASE: For every g-formula MATH: MATH CASE: For every g-term MATH: MATH CASE: For every p-term MATH: MATH CASE: For every p-formula MATH: MATH CASE: MATH and MATH. Informally, interpretation MATH maintains the values of all g-terms and g-formulas as occur under interpretation MATH. It also maintains the values of all p-terms, except those in the class containing terms MATH and MATH. These p-terms are split into some having valuation MATH and others having valuation MATH. With respect to p-formulas, consider first an equation of the form MATH where MATH and MATH are p-terms. The equation will yield the same value under both interpretations except under the condition that MATH and MATH are split into different parts of the class that originally evaluated to MATH, in which case the equation will yield MATH under MATH, but MATH under MATH. Thus, although this equation can yield different values under the two interpretations, we always have that MATH. This implication relation is preserved by conjunctions and disjunctions of p-formulas, due to the monotonicity of these operations. We will now present this argument formally. Most of the cases are straightforward; we indicate those that are ``interesting." We prove REF to REF above by simultaneous induction on the expression structures. For the base cases, we have: CASE: NAME: MATH, MATH, and MATH for any propositional variable MATH. CASE: NAME: If MATH is a g-function symbol of zero order, then MATH. CASE: NAME: If MATH is a p-function symbol of zero order, then by the definition of MATH, MATH. CASE: NAME: same as g-formula. For the inductive step, we prove that REF through REF hold for an expression given that they hold for all of its subexpressions. CASE: NAME: There are several cases, depending on the form of MATH. CASE: Suppose MATH has one of the forms MATH, MATH, MATH, where MATH and MATH are g-formulas. By the inductive hypothesis, MATH, and MATH. It follows that MATH, MATH, and MATH. CASE: Suppose MATH has the form MATH, where MATH are g-terms. By the inductive hypothesis on g-terms, MATH, and MATH. It follows that MATH. CASE: The remaining case is that MATH is a predicate application of the form MATH, where MATH is a predicate symbol of order MATH, and MATH, are p-terms. By the inductive hypothesis for p-terms, we have MATH, for MATH. By the definition of MATH, MATH . CASE: NAME: There are two cases. CASE: Suppose MATH has the form MATH, where MATH is a g-formula, and MATH and MATH are g-terms. By the inductive hypothesis, we have MATH, MATH, and MATH. Then MATH. CASE: Suppose MATH has the form MATH, where MATH is a g-function symbol of order MATH and MATH are p-terms. By the inductive hypothesis, MATH, for MATH. Then we have, MATH . CASE: NAME: There are three cases. CASE: Suppose MATH is a g-term. By the inductive hypothesis, MATH. Since MATH cannot be equal to MATH, it must be the case that MATH. CASE: Suppose MATH has the form MATH, where MATH is a g-formula, and MATH and MATH are p-terms. By the inductive hypothesis, MATH, MATH, and MATH. It follows that MATH . CASE: [Important case:] Suppose that MATH has the form MATH, where MATH is a p-function symbol of order MATH and MATH are p-terms. Here, we have to consider two cases. The first case is that the following two conditions hold: CASE: MATH is the function symbol MATH, that is, the function symbol of the term MATH mentioned at the beginning of the proof of this lemma, and REF MATH, for MATH. If these two conditions hold, then by the definition of MATH, MATH, while MATH. Since MATH, we have MATH. The second case is when one of the two conditions mentioned above does not hold. The proof of this case is identical to the proof of REF above. CASE: NAME: There are three cases. CASE: If the p-formula MATH is a g-formula, then by the inductive hypothesis, MATH, so MATH. CASE: Suppose MATH has one of the forms MATH, or MATH, where MATH are p-formulas. By the inductive hypothesis, MATH, and MATH. Thus we have MATH so MATH. The proof for MATH is the same. CASE: [Important case:] Finally, we consider the case that MATH is a p-formula of the form MATH, where MATH and MATH, are p-terms. By the inductive hypothesis, we have that if MATH, then MATH, for MATH. Also, by the definition of MATH, we have that if MATH does not equal MATH, then MATH. Now, we consider cases depending on whether MATH or MATH are equal to MATH. If both terms are equal to MATH in MATH, then both MATH and MATH must be equal to MATH, so the equation is true in both MATH and MATH. If neither MATH nor MATH is equal to MATH, then MATH and MATH, so the equation has the same truth value in MATH and MATH. The last case is that exactly one of the p-terms is equal to MATH in MATH. In this case, the equation is false in MATH, so we have MATH. This completes the inductive proof. REF above, which implies that MATH is a proper refinement, is a consequence of the definition of MATH and the inductive REF . First, we show that MATH. By definition, MATH. By REF on p-terms, we can assume MATH, for all MATH in the range MATH. By the definition of MATH, we have MATH. The proof that MATH is in two cases, depending on whether MATH and MATH are applications of the same function symbol. CASE: First, consider the case that MATH and MATH, where MATH and MATH are different function symbols. In this case, MATH . CASE: Finally, we have the case that MATH and MATH are the same function symbol, and there is some value of MATH with MATH, such that MATH does not equal MATH. Here, we have: MATH . By REF , MATH, for all MATH such that MATH. Since MATH does not equal MATH, the value of the above application of MATH is: MATH . |
cs/9910014 | Starting with interpretation MATH equal to MATH, we define a sequence of interpretations MATH by repeatedly applying the construction of REF . That is, we derive each interpretation MATH from its predecessor MATH by letting MATH and letting MATH. NAME MATH is a proper refinement of its predecessor MATH such that MATH. At some step MATH, we must reach a maximally diverse interpretation MATH, because our set MATH is finite and therefore can be properly refined only a finite number of times. We then let MATH be MATH. We can see that MATH, and hence MATH. |
cs/9910014 | For value MATH in the range MATH define MATH as the minimum value of MATH in the range MATH such that MATH for all MATH in the range MATH. By this definition MATH. Observe also that if MATH then MATH. In addition, for any value MATH in the range MATH, if MATH, then MATH. We prove by induction on MATH that MATH, where MATH. The base case of MATH is trivial, since MATH, and MATH. Assuming the property holds for MATH, we consider two possibilities. First, if MATH, we have MATH, and hence the top-level MATH operation in MATH REF will select its first term argument MATH, giving MATH. On the other hand, if MATH, we must have MATH, and hence the top-level MATH operation in MATH will select its second term argument MATH, giving MATH, which by the inductive hypothesis equals MATH for MATH. Since MATH, we must also have MATH, and hence MATH, where MATH. Since MATH is defined as MATH, our induction argument proves that MATH for MATH. |
cs/9910014 | We provide a somewhat more general construction of MATH than is required for the proof of this lemma in anticipation of using this construction in the proof of REF . Given MATH defined over domain MATH, we define MATH over a domain MATH such that MATH. We define MATH for the function and predicate symbols occurring in MATH based on their definitions in MATH. For any function symbol MATH in MATH having MATH, and any argument values MATH, we define MATH. For argument values MATH such that for some MATH, MATH, we let MATH be an arbitrary domain value. Similarly, for predicate symbol MATH, we define MATH to yield the same value as MATH for arguments in MATH and to yield an arbitrary truth value when at least one argument is not in MATH. One can readily see that MATH for every subexpression MATH of MATH. This takes care of the second equality in the statement of the lemma, and hence we can concentrate on the relation between MATH and MATH for the remainder of the proof. Recall that MATH are the domain variables introduced when generating the nested MATH terms MATH. Our strategy is to define interpretations of these variables such that each MATH mimics the behavior of the original MATH-application term MATH in MATH. We consider two cases. For the case where MATH, we define MATH, that is, the value of the MATH-application term in MATH under MATH. Otherwise, we let MATH be an arbitrary domain value - we will show that its value does not affect the valuation of any expression MATH in MATH having a counterpart MATH in MATH. We argue by induction on MATH that MATH for any subexpression MATH of MATH. For the case where MATH, this hypothesis implies that MATH. The base case of MATH is trivial, since MATH is defined to be MATH. Suppose that for every MATH in the range MATH and every subexpression MATH of MATH, we have MATH, and consequently that MATH for the case where MATH. We must show that for every subexpression MATH of MATH, we have MATH. We first focus our attention on term MATH in MATH and its counterpart MATH in MATH, showing that MATH. The MATH-application terms for all MATH such that MATH have MATH, and hence we can assume that MATH for these values of MATH. Furthermore, any argument MATH to a MATH-application term for MATH and MATH has MATH, and hence we can assume MATH. We consider two cases: MATH, and MATH. In the former case, we have by REF that MATH. Our definition of MATH gives MATH. Otherwise, suppose that MATH. REF shows that MATH. We can see that MATH, and hence MATH is defined to be MATH. By the definition of MATH we have MATH for MATH. By the induction hypothesis we have MATH, since MATH, and similarly that MATH. By transitivity we have MATH for all MATH such that MATH, that is, the arguments to MATH-application terms MATH and MATH have equal valuations under MATH. Function consistency requires that MATH. From this we can conclude that MATH. Combining these cases gives MATH. For any subexpression MATH its form MATH differs from MATH only in that all instances of term MATH have been replaced by MATH. We have just argued that MATH, and by the induction hypothesis we have that MATH, giving by transitivity that MATH. REF implies that MATH, and our induction hypothesis gives MATH. By transitivity we have MATH. To complete the proof, we observe that our induction argument implies that for any subexpression MATH of MATH, MATH, including for the case where MATH, giving MATH. |
cs/9910014 | We define MATH to be identical to MATH for any symbol occurring in MATH. This implies that MATH for every subexpression MATH of MATH. This takes care of the second equality in the statement of the lemma, and hence we can concentrate on the relation between MATH and MATH for the remainder of the proof. For function symbol MATH, we define MATH for domain elements MATH as follows. Suppose there is some value MATH such that MATH for all MATH such that MATH, and such that MATH. Then we define MATH to be MATH. If no such value of MATH exists, we let MATH be some arbitrary domain value. We argue by induction on MATH that MATH for any subexpression MATH of MATH. For the case where MATH, this hypothesis implies that MATH. The base case of MATH is trivial, since MATH is defined to be MATH. Suppose that for every MATH in the range MATH and every subexpression MATH of MATH, we have MATH, and consequently that MATH for the case where MATH. We must show that for every subexpression MATH of MATH, we have MATH. We focus initially on term MATH in MATH and its counterpart MATH in MATH, showing that MATH. Any MATH-application term MATH for MATH has MATH, and hence we can assume that MATH. Furthermore, any argument MATH to a MATH-application term for MATH and MATH has MATH, and hence we can assume that MATH. We consider two cases: MATH, and MATH. In the former case, we have by REF that MATH. In addition, MATH is defined such that MATH, giving MATH. Otherwise, suppose that MATH. REF shows that MATH. We can see that MATH, and hence MATH is defined such that MATH. For any MATH such that MATH, we also have by the definition of MATH that MATH. By the induction hypothesis we have MATH, since MATH, and similarly that MATH. By transitivity we have MATH, that is, the arguments to MATH-application terms MATH and MATH have equal valuations under MATH. Functional consistency requires that MATH. Putting this together gives MATH. For any subexpression MATH its form MATH differs from MATH only in that all instances of term MATH have been replaced by MATH. We have just argued that MATH, and by the induction hypothesis we have that MATH, giving by transitivity that MATH. REF implies that MATH, and our induction hypothesis gives MATH. By transitivity we have MATH. To complete the proof, we observe that our induction argument implies that for any subexpression MATH of MATH, MATH, including for the case where MATH, giving MATH. |
cs/9910014 | Assume MATH is universally valid, and consider any interpretation MATH of the symbols in MATH. We construct a sequence of interpretations MATH, where each interpretation MATH is generated by extending its predecessor MATH by letting MATH and MATH in REF or a similar one for predicate applications. The effect is to include in MATH interpretations of the domain or propositional variables introduced when eliminating the MATH function or predicate symbol. We then define interpretation MATH to be identical to MATH for every variable appearing in MATH. By induction, we have MATH. Since MATH is universally valid, we have MATH. Since this construction can be performed for any interpretation MATH, MATH must also be universally valid. Only if: Assume MATH is universally valid. Starting with an interpretation MATH of the domain and propositional variables of MATH, we can define a sequence of interpretations MATH, using the construction in the proof of REF (or a similar one for predicate applications) to generate an interpretation of each function or predicate symbol in MATH. We then define interpretation MATH to be identical to MATH for every function or predicate symbol appearing in MATH. By induction, we have MATH. Since MATH is universally valid, we have MATH. Since this construction can be performed for any interpretation MATH, MATH must also be universally valid. |
cs/9910014 | Given interpretation MATH defined over domain MATH, we define interpretation MATH over a domain MATH. Each MATH is a unique value, that is, MATH for any MATH, and MATH. The proof of this lemma is based on a refinement of the proof of REF . Whereas the construction in the earlier proof assigned arbitrary values to the new domain variables in some cases, we select an assignment that is diverse in these variables. As in the construction in the proof of REF , we define MATH for any function or predicate symbol in MATH to be identical to that of MATH when the arguments are all elements of MATH. When some argument is not in MATH, we let the function (respectively, predicate) application yield an arbitrary domain (respectively, truth) value. For domain variable MATH introduced when generating term MATH, we consider two cases. For the case where MATH, we define MATH, that is, the value of the MATH-application term in MATH under MATH. For the case where MATH, we define MATH. We saw in the proof of REF that we could assign arbitrary values in this latter case and still have MATH. In fact, for every subexpression MATH of MATH, we have that its counterpart MATH in MATH satisfies MATH. We must show that MATH is diverse for MATH with respect to MATH. We first observe that MATH is identical to MATH for all function application terms in MATH, and hence MATH must be diverse with respect to MATH for MATH. We also observe that MATH assigns to each variable MATH either a unique value MATH or the value yielded by MATH-application term MATH in MATH under MATH. Suppose there were distinct variables MATH and MATH such that MATH. This could occur only for the case that MATH. Since MATH is diverse, we can have MATH only if MATH. We cannot have both MATH and MATH, and hence either MATH or MATH would have been assigned unique value MATH or MATH, respectively. Thus, we can conclude that MATH for distinct variables MATH and MATH. In addition, we must show that interpretation MATH does not create any matches between a new variable MATH and a function application term MATH in MATH that does not have MATH as the topmost function symbol. Since MATH is diverse with respect to MATH for MATH and MATH, any function application term MATH in MATH that does not have function symbol MATH as its topmost symbol must have MATH for all MATH. In addition, we have MATH for all MATH. Hence, we must have MATH. |
cs/9910014 | The proof of this lemma is based on a refinement of the proof of REF . Whereas the construction in the earlier proof assigned arbitrary values to some of the new domain variables, we select an assignment such that we do not inadvertently violate the diversity of the other function symbols. We define MATH to be identical to MATH for any symbol occurring in MATH. For each domain variable MATH introduced when generating term MATH, we define MATH. This differs from the interpretation defined in the proof of REF only in giving fixed interpretations of domain variables that could otherwise be arbitrary, and hence we have have MATH. In fact, for every subexpression MATH of MATH, we have that its counterpart MATH in MATH satisfies MATH. We must show that MATH is diverse for MATH with respect to MATH. We first observe that MATH is identical to MATH for all function application terms in MATH, and hence MATH must be diverse for MATH with respect to MATH. We also observe that MATH assigns to each variable MATH the value of MATH-application term MATH. For term MATH having the application of function symbol MATH as the topmost operation, we must have MATH. Hence, we are assured that the values assigned to the new variables under MATH do not violate the diversity of the interpretations of the symbols in MATH. |
cs/9910014 | By REF , the universal validity of MATH implies that of MATH, and hence it must be true for every interpretation. If: The proof in the other direction follows by inducting on the number of function and predicate symbols in MATH having nonzero order. For the induction step we use REF when eliminating all applications of a p-function symbol, and REF when eliminating all applications of a g-function symbol. When eliminating a predicate symbol, we do not introduce any new domain variables. |
cs/9910014 | Consider any interpretation MATH of the variables in MATH that is diverse over MATH. We show that we can construct an isomorphic interpretation MATH that satisfies the restrictions of the corollary. Let MATH (respectively, MATH) be the range of MATH considering only variables in MATH (respectively, MATH). The function MATH must be a bijection and hence have an inverse MATH. Furthermore, we must have MATH. Let MATH be REF mapping MATH defined for any MATH in MATH, as MATH. Let MATH be an arbitrary REF mapping MATH. We now define MATH such that for any variable MATH in MATH (respectively, MATH) we have MATH equal to MATH (respectively, MATH). Finally, for any propositional variable MATH, we let MATH equal MATH. For any EUF formula, isomorphic interpretations will always yield identical valuations, giving MATH. Hence the set of interpretations satisfying the restrictions of the corollary form a sufficient set to prove the universal validity of MATH. |
cs/9910014 | For each propositional variable MATH occurring in MATH, we define MATH. For each pair of variables MATH and MATH such that MATH, we define MATH to be MATH iff MATH. We can see that MATH must obey transitivity, because it is defined in terms of a transitive relation in MATH. We prove the following hypothesis by induction on the expression depths: CASE: For every formula MATH in MATH: MATH. CASE: For every term MATH in MATH and all MATH such that MATH: MATH iff MATH. The base cases hold as follows: CASE: Formulas of the form MATH, MATH, and MATH have MATH and MATH. CASE: Term MATH has MATH iff MATH, and MATH iff MATH. Assuming the induction hypothesis holds for formulas MATH and MATH, one can readily see that it will hold for formulas MATH, MATH, and MATH, by the definition of MATH . Assuming the induction hypothesis holds for formula MATH and for terms MATH and MATH, consider term MATH of the form MATH. For the case where MATH, we have MATH, and also MATH. The induction hypotheses for MATH gives MATH iff MATH. The induction hypothesis for MATH gives MATH, and hence MATH. From all this, we can conclude that MATH iff MATH. A similar argument holds when MATH, but based on the induction hypothesis for MATH. Finally, assuming the induction hypothesis holds for terms MATH and MATH, consider the equation MATH. Suppose that MATH and MATH. Our induction hypothesis for MATH and MATH give MATH. Suppose either MATH or MATH. Then we will have MATH iff MATH. In addition, the right hand part of REF will hold under MATH iff MATH. Otherwise, suppose that MATH. We will have MATH iff MATH. In addition, the left hand part of REF will hold under MATH iff MATH . |
cs/9910014 | We define interpretation MATH over the domain of integers MATH. For propositional variable MATH, we define MATH. For MATH we let MATH be the minimum value of MATH such that MATH. For MATH we let MATH. Observe that this interpretation gives MATH for all MATH, since MATH, and MATH for MATH. We claim that for MATH, if MATH, then we must have MATH as well. If instead we had MATH, then we must have MATH. Combining this with MATH, the transitivity requirement would give MATH, but this would imply that MATH. We prove the following hypothesis by induction on the expression depths: CASE: For every formula MATH in MATH: MATH. CASE: For every term MATH in MATH and all MATH such that MATH: MATH iff MATH. The base cases hold as follows: CASE: Formulas of the form MATH, MATH, and MATH have MATH and MATH. CASE: Term MATH has MATH iff MATH and MATH iff MATH. Assuming the induction hypothesis holds for formula MATH and for terms MATH and MATH, consider term MATH of the form MATH. For the case where MATH, we have MATH. The induction hypothesis for MATH gives MATH iff MATH. The induction hypothesis for MATH gives MATH, giving MATH, and also MATH. Combining all his gives MATH iff MATH. A similar argument can be made when MATH, but based on the induction hypothesis for MATH. Finally, assuming the induction hypothesis holds for terms MATH and MATH, consider the equation MATH. Let MATH and MATH. In addition, let MATH and MATH. Our induction hypothesis gives MATH, and MATH. REF gives MATH and MATH. By our earlier argument, we must also have MATH and MATH. We consider different cases for the values of MATH, MATH, MATH, and MATH. CASE: Suppose MATH. Then we must have MATH. REF will hold under MATH iff MATH, and this will hold iff MATH. In addition, the right hand part of REF will hold under MATH iff MATH. CASE: Suppose MATH. By an argument similar to the previous one, we will have equation MATH holding under interpretation MATH and REF holding under interpretation MATH iff MATH. CASE: Suppose MATH. Since MATH we must have MATH. Similarly, since MATH we must have MATH. CASE: Suppose MATH, and hence MATH holds under MATH. Then we have MATH. Our transitivity requirement then gives MATH, and hence the left hand part of REF will hold under MATH. CASE: Suppose MATH, and hence MATH does not hold under MATH. We must have MATH. This condition is illustrated in the left hand diagram of REF . In this figure we use solid lines to denote equalities and dashed lines to denote inequalities. We argue that we must also have MATH by the following case analysis for MATH: CASE: For MATH, we get the case diagrammed in the middle of REF where the diagonal line creates a triangle with just one dashed line (inequality). This represents a violation of our transitivity requirement, since it indicates MATH, but MATH. CASE: For MATH and MATH, we have the case diagrammed on the right side of REF . Again we have a triangle with just one dashed line indicating a violation of our transitivity requirement, with MATH, but MATH. With MATH, REF will not hold under MATH. From this case analysis we see that MATH holds under MATH iff REF holds under MATH. |
cs/9910014 | This theorem follows directly from REF . |
cs/9910016 | Suppose MATH. Then we have either MATH or MATH. By condition REF defining a feasible probabilistic status set, we know that MATH. By condition (MATH REF), MATH and hence, MATH. Therefore, MATH. |
cs/9910016 | By the well known NAME/NAME theorem, REF. follows from REF. To show REF., let MATH. But then MATH because of the monotonicity of MATH (see CITE). This implies MATH. REF. follows similarly from the continuity of MATH and the fact that MATH . |
cs/9910016 | It is easy to see that the while loop of the algorithm can be executed in polynomial time (data-complexity). Checking if MATH satisfies the three conditions at the end of the algorithm are each polynomial time checks (assuming the existence of a polynomial oracle to compute code call conditions). |
cs/9910016 | The first two statements are immediate. Feasibility requires checking conditions (MATH) - (MATH), and therefore reduces to the first two statements. Rational and reasonable status sets are handled in a completely analogous manner. That our algorithms reduce to the non-probabilistic case under our general assumption is trivial: the difference is only the satisfaction relation MATH which, by the first statement, coincides with MATH. |
cs/9910016 | Let MATH be the original NAME structure, and MATH the result of executing the action. We just need to show that MATH. Using REF : MATH . |
cs/9910016 | All definitions for weak p-feasibility trivially coincide with those for feasibility if p=REF. The distinction between weak and strong p-feasibility is just in the definition of p-consistency, and we can easily see that for p=REF they coincide, since a probability greater or equal to REF cannot be but equal to REF. |
cs/9910016 | Identical to the proof of REF . |
cs/9910016 | CASE: Suppose MATH. Then, as MATH is feasible, we know that MATH, and hence MATH. As MATH is feasible, and hence deontically consistent, the third condition of deontic consistency specifies that MATH's precondition is true in state MATH. CASE: This follows immediately because as MATH is feasible, we have MATH. The second condition defining MATH, when written in contrapositive form, states that MATH implies that MATH. CASE: As MATH is feasible, MATH. The first condition specifying MATH allows us to infer that MATH implies that MATH. The result follows immediately from REF of this proposition. CASE: From the above argument, as MATH, we can conclude that MATH implies that MATH. By the deontic consistency requirement, MATH. |
cs/9910016 | In order to show that a reasonable probabilistic status set MATH of MATH is a rational status of MATH, we have to verify REF that MATH is a feasible probabilistic status set and REF that MATH is grounded. Since MATH is a reasonable probabilistic status set of MATH, it is a rational probabilistic status set of MATH, that is, a feasible and grounded probabilistic status set of MATH. Since the conditions MATH - MATH of the definition of feasible probabilistic status set depend only on MATH and MATH but not on the program, this means that for showing REF it remains to check that MATH (closure under the program rules) is satisfied. Let thus MATH be a ground instance of a rule from MATH. Suppose the body MATH of MATH satisfies REF . - REF. of MATH. Then, by the definition of MATH, we have that the reduct of the rule MATH, obtained by removing all literals of MATH from the body, is in MATH. Since MATH is closed under the rules of MATH, we have MATH. Thus, MATH is closed under the rules of MATH, and hence MATH is satisfied. As a consequence, REF holds. For REF , we suppose MATH is not grounded, that is, that some smaller MATH satisfies MATH - MATH for MATH, and derive a contradiction. If MATH satisfies MATH for MATH, then MATH satisfies MATH for MATH. For, if MATH is a rule from MATH such that REF. - REF. of MATH hold for MATH, then there is a ground rule MATH of MATH such that MATH is obtained from MATH in the construction of MATH and, as easily seen, REF. - REF. of MATH hold for MATH. Since MATH satisfies MATH for MATH, we have MATH. It follows that MATH satisfies MATH for MATH. Furthermore, since MATH and MATH do no depend on the program, also MATH and MATH are satisfied for MATH with respect to MATH. This means that MATH is not a rational probabilistic status set of MATH, which is the desired contradiction. Thus, REF hold, which proves the result. |
cs/9910016 | By definition of rationality, we know that if MATH is a rational status set of MATH then it must be a feasible probabilistic status set as well. Suppose MATH has a feasible probabilistic status set. Then the set of all feasible probabilistic status sets of MATH on MATH has a non-empty set of inclusion-minimal elements. Indeed, from the grounding of the probabilistic agent program, we can remove all rules which violate REF of the operator MATH, and can remove literals involving code calls from the remaining rules. Moreover, the deontic and action closure conditions can be incorporated into the program via rules. Thus, we end up with a set MATH of propositional clauses, whose models are feasible probabilistic status sets of MATH. Since MATH has a feasible probabilistic status set, MATH has a model, that is, an assignment to the propositional atoms which satisfies all clauses in MATH. Now, each satisfiable set of clauses in a countable language posseses at least one minimal model (with respect to inclusion, i.w., a MATH-minimal set of atoms is assigned the value MATH); this can be shown applying the same technique which proves that every such set of clauses can be extended to a maximal satisfiable set of clauses. Thus, MATH has at least one minimal model. As easily seen, any such model is a minimal feasible probabilistic status set of MATH. Suppose now MATH is one of the minimal feasible probabilistic status sets of MATH on MATH. Then (as we show below) MATH is grounded, and hence a rational probabilistic status set. To show that MATH is grounded, we need to show that MATH satisfies conditions (MATH REF) - REF of feasible probabilistic status set - this is true because MATH is feasible. In addition, we need to show that no strict subset MATH of MATH satisfies conditions (MATH REF) - (MATH REF). Suppose there is a strict subset MATH of MATH satisfying conditions (MATH REF) - (MATH REF). Then, as MATH, MATH also satisfies condition REF of feasibility, and hence MATH is a feasible probabilistic status set. But this contradicts the inclusion minimality of MATH, and hence, we may infer that MATH has no strict subset MATH of MATH satisfying conditions (MATH REF) - (MATH REF). Thus, MATH is grounded, and we are done. |
cs/9910016 | For each random variable MATH returned by some ground code call condition in the probabilistic state MATH, let us define its normalized version MATH where: MATH that is, we delete the zero-probability elements and add the extra one MATH (which stands for ``none of the above") whenever the distribution MATH is incomplete. Now we can see that each tuple MATH in the Cartesian product MATH corresponds to a distinct compatible state MATH with respect to MATH. In MATH, a ground code call returns an object MATH iff in the probabilistic state MATH it returns a variable MATH such that MATH. Let associate to each state MATH of this kind the value MATH and set MATH for all the other states. We can easily verify that MATH is a compatible probabilistic NAME structure for MATH: for each random variable MATH returned in state MATH and each object MATH if MATH then MATH, so it could appear only in the zero-probability states; otherwise: MATH . Both cases satisfy the condition for compatibility. Finally, it is easy to verify that MATH is really a probabilistic NAME structure: MATH . |
cs/9910016 | Let us consider the compatible NAME structure described in the proof of REF , and let us assume that MATH and MATH are the variables required in the thesis. The corresponding completed versions MATH and MATH will then contain at least two non zero-probability objects (one of them could be the extra object MATH), respectively MATH and MATH. Now let choose an arbitrary real number MATH such that: MATH . We can build a NAME structure MATH, where MATH is defined in the same way as MATH but replacing MATH by MATH, which in turn is defined in the following way: MATH . It is easy to verify that it is a compatible NAME structure. Since MATH can be arbitrarily chosen within a non-point interval, we can obtain an infinite number of distinct compatible NAME structures. |
cs/9910016 | MATH . Suppose MATH a rational probabilistic status set of MATH on MATH. Then, MATH is feasible by definition of rational probabilistic status set. By REF , MATH is a pre-fixpoint of MATH. Since MATH is monotone, it has by the NAME Theorem a least pre-fixpoint, which coincides with MATH (see CITE). Thus, MATH. Clearly, MATH satisfies MATH and MATH; moreover, MATH satisfies MATH, as MATH satisfies MATH and this property is hereditary. By the definition of rational probabilistic status set, it follows MATH. MATH . Suppose MATH is a feasible probabilistic status set. Since every probabilistic status set MATH which satisfies MATH - MATH is a pre-fixpoint of MATH and MATH is the least prefix point, MATH implies MATH. It follows that MATH is rational. Notice that in case of positive programs, MATH always satisfies the conditions MATH and MATH of a feasible probabilistic status set (that is, all closure conditions), and thus is a rational probabilistic status set if it satisfies MATH and MATH, that is, the consistency criteria. The uniqueness of the rational probabilistic status set is immediate from the previous theorem. |
cs/9910019 | Sufficiency. We prove that if MATH is satisfied then MATH can be included in a consistent global checkpoint. Let us consider the global checkpoint defined as follows: CASE: if MATH, we take MATH; CASE: if MATH, for each MATH we consider the integer MATH (with MATH if MATH or if this set is empty). Then we take MATH with MATH. Let us note that, from that definition, it is possible that MATH (in that case, MATH is an initial data checkpoint). By construction, this global checkpoint satisfies the two following properties : MATH . We show that MATH is consistent. Assume the contrary. So, there exists MATH and MATH and a dependence edge MATH that starts after MATH and arrives before MATH. So, it follows that: MATH . Four cases have to be considered: CASE: MATH, MATH. REF is contradicted by REF. CASE: MATH, MATH. Since MATH, from REF we have: MATH. As, at data MATH both the dependence edge ending the path MATH, and the dependence edge starting the path MATH belong to the same interval, we conclude from REF that MATH which contradicts the assumption MATH. CASE: MATH, MATH. REF contradicts REF . CASE: MATH, MATH. Since MATH, from REF we have: MATH. As in REF , we can conclude that MATH which contradicts REF . Necessity. We prove that, if there is a consistent global checkpoint MATH including MATH, then property MATH holds for any MATH. Assume the contrary. So, there exist MATH and MATH such that MATH. From the definition of MATH, there exists a sequence of dependence edges MATH such that: We show by induction on MATH that, MATH, MATH and MATH cannot belong to the same consistent global checkpoint. Base step. MATH. In this case, MATH starts after MATH and arrives before MATH, and consequently the pair REF cannot belong to a consistent global checkpoint. Induction step. We suppose the result true for some MATH and show that it holds for MATH. We have: From the assumption induction applied to the path of dependence edges MATH, we have : for any MATH, MATH and MATH cannot belong to the same consistent global checkpoint. Moreover, MATH starts in MATH and arrives in MATH imply that, for any MATH and for any MATH, MATH and MATH cannot belong to the same consistent checkpoint. Since MATH, it follows that no checkpoint of MATH can be included with MATH and MATH to form a consistent global checkpoint. |
cs/9910024 | For a point MATH, let MATH denote the distance from MATH to the line MATH through MATH and MATH. We have that MATH. Because MATH is on the small arc MATH, MATH. Hence, MATH . By similar reasoning, MATH . Note that MATH, by REF . For MATH, MATH, so MATH. Because MATH is an increasing function, MATH, and MATH, by REF . Both MATH and MATH are MATH. Hence, by REF , both MATH and MATH are bounded from above by MATH, as MATH by assumption. But this is strictly less than MATH by REF . |
cs/9910024 | Suppose, for contradiction, the petal is in a restricted configuration with MATH. Consider the two triangles MATH and MATH. These triangles share the common side MATH, and MATH. Two sides of MATH are equal length with two sides of MATH. Moreover the included angles satisfy MATH, since by definition of restricted configuration REF , joint MATH lies on the small arc MATH of MATH. Applying the cosine law to the remaining side in each triangle (or using NAME 's REF , NAME I) we see MATH implies MATH. By definition REF MATH, which is less than MATH since MATH is an increasing function. A direct computation shows that MATH, so we have MATH. By REF , we have also MATH and MATH. All four points MATH, MATH, MATH and MATH are strictly inside the circle of radius MATH centred at MATH. Joint MATH is of course on this circle, so it cannot be inside MATH. This contradicts REF of a restricted configuration. |
cs/9910024 | Note that the points MATH, MATH, MATH, MATH, and the circle MATH, are defined by the positions of the joints MATH and MATH; as the joints move, so do the points MATH, MATH, etc. For simplicity, we omit displaying this dependence on time. Consider, in turn, the three conditions required for a restricted configuration. REF holds throughout the motion by assumption. In any configuration, MATH must be on MATH. Since MATH starts the motion on the small arc MATH, for REF to be violated, MATH must pass through point MATH or through point MATH. MATH and MATH are on the interior of links MATH and MATH, respectively, and since MATH, MATH always properly intersects these links. Thus MATH may not move through MATH or MATH, and hence REF holds throughout the motion. Given that MATH remains on the small arc MATH, points MATH and MATH are well defined. As MATH starts the motion inside MATH, REF is violated only if MATH passes through one of the sides of this quadrilateral. Sides MATH and MATH are portions of links, so MATH may not pass through them. By REF , MATH and MATH are both strictly less than MATH, so MATH cannot pass through side MATH or side MATH. Thus, REF holds throughout the motion. |
cs/9910024 | Suppose, to the contrary, a motion exists that takes some petal angle out of the range MATH. Let MATH be the first instant that some petal angle, say for petal MATH, reaches MATH. Let MATH be the first instant that some petal angle reaches MATH. If MATH, then at time MATH all angles are strictly greater than MATH and at least one is equal to MATH. This means MATH a contradiction. Hence MATH. During the supposed motion, the joint angles change continuously in time. Since MATH by REF , and MATH approaches MATH from above as MATH approaches MATH from below, we may choose MATH such that MATH at time MATH. Note that during the motion up to time MATH, all petal angles are strictly less than MATH, as MATH. By REF all petals remain in a restricted configuration before time MATH, so REF applies to petal MATH. This means MATH, contradicting the choice of MATH. |
cs/9910024 | Consider the linkage in REF , in which there are MATH copies of an eight-petal lockable tree connected by long links joining the MATH joints of the subtrees. The connecting links are long enough that when they are stretched out to form a straight chain, each subtree can be in either an open or a closed configuration without crossing links. Consider simple configurations in which the long links form a straight chain, and each subtree is in either an open or a closed configuration. Label such configurations by a MATH-bit vector, specifying for each subtree, whether its configuration is open or closed. Configurations with different labels are clearly not equivalent, as a motion of the entire linkage that would take some subtree in a closed configuration to an open configuration would imply, by removing links outside the subtree, the existence of a motion that would make petal angle of the subtree inferior to MATH. Hence the number of inequivalent configurations is at least MATH, as MATH. |
cs/9910024 | We show that MATH and MATH are feasible link lengths, where MATH is feasible for MATH and any MATH is feasible for MATH. The proof consists of checking, in turn, each of the constraints mentioned above. Constraint REF: MATH . Given our choice of MATH, this is satisfied as long as MATH. Recall that MATH and MATH REF , so MATH. Since MATH, the constraint is satisfied. Constraint REF: MATH . By the definition of MATH REF , and our choice of MATH, we have MATH. Since MATH, the constraint is satisfied. Constraint REF: MATH . Since MATH, MATH. Using the definition MATH, we see that MATH. Since MATH, MATH. Now, MATH is an increasing function on MATH, and MATH, so the constraint is satisfied. We summarize, for future reference, some inequalities derived so far, MATH . Constraint REF: MATH . This constraint is the only one for which we distinguish cases based on MATH. For MATH and MATH, a direct calculation shows that the given link lengths (with MATH) satisfy the constraint. For the case MATH, we show that any MATH yields MATH, whence the constraint follows since MATH (the equality is an identity, the inequality uses MATH). Plugging our choice of MATH into the definition of REF , MATH . Cosine is a decreasing function on MATH, so MATH given MATH (REF and MATH). Under the radical of REF, substituting MATH gives the expression MATH. Using the identity MATH, this expression becomes MATH. The last two inequalities follow by noting that MATH, and sine is increasing on the interval MATH. Using these bounds for the two terms in REF , MATH . Constraint REF: MATH . Define function MATH on MATH. We know MATH is a strictly increasing function on MATH, hence so is MATH. Furthermore, MATH, so MATH for MATH. From REF, we see MATH, so MATH. By definition of MATH, this becomes MATH. Since MATH, we obtain that MATH as desired. Constraint REF: MATH . By definition, MATH REF . This shows MATH. Rewriting MATH as MATH, and MATH as REF , the equation for MATH becomes MATH. Combining all this with REF , we find MATH . From REF we see that MATH. By definition REF , MATH is also in MATH. Since cosine is decreasing on this interval, the constraint MATH holds if, and only if, MATH. Using the definition of MATH REF , this latter inequality becomes MATH. We collect the terms in MATH to one side, MATH, and complete the square to get MATH . Since MATH (by definition of MATH, REF ), the right hand side can be written as MATH, which is positive since MATH by REF . Since both sides of the inequality are positive, we can take square roots which leads to MATH so MATH and we deduce that MATH (which equals MATH) must satisfy MATH . Comparing the left inequality of this with the definition of REF , we see that the former can be written MATH. From REF we note that MATH is less than MATH and both quantities are in the range MATH. The increasing property of MATH ensures that the left inequality of REF is satisfied. For the upper bound, note that by REF , MATH and both quantities are in the range MATH. The increasing property of MATH ensures that MATH. By the definition of REF , MATH. Given REF , the decreasing nature of cosine on this interval ensures MATH. Putting this together, we find MATH, so the upper bound of REF is satisfied. |
gr-qc/9910103 | The derivative MATH can be written as MATH . |
gr-qc/9910104 | This can directly be proved by using MATH. |
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