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math/0004181 | MATH has the form MATH. Let MATH be the partial isometry defining MATH and note that MATH for all MATH because of REF . Thus MATH is a unitary in MATH which is MATH-invariant modulo MATH and satisfies that MATH where MATH. Thanks to REF we have that MATH for all MATH. Since MATH (with convergence in the strict topology... |
math/0004181 | We are going to use REF. Let MATH, where MATH is the element represented by the identity map of MATH and MATH is the stabilising MATH-homomorphism. By REF it suffices to identify the image of MATH under the NAME isomorphism MATH and show that the image of that element is not changed under the map we are trying to prove... |
math/0004181 | If MATH in MATH, there is a path MATH, of asymptotic homomorphisms MATH such that MATH and MATH and a unit sequence MATH in MATH such that MATH connects MATH to MATH. By REF we may assume that MATH is an equi-homotopy and it is then easy to see that REF is a strong homotopy. By REF we conclude from this that MATH in MA... |
math/0004181 | Consider an extension MATH and assume that MATH in MATH. With the notation from REF we find that MATH. But then REF implies that MATH in MATH. By REF this yields the conclusion that MATH in MATH. Thus MATH is injective. But MATH is weakly stable so the result follows. |
math/0004181 | Consider another discretization MATH of MATH and define MATH by MATH . There is then a MATH-extension MATH such that MATH. It suffices to show that MATH is unitarily equivalent to an asymptotically split MATH-extension. Define MATH such that MATH . There is then a MATH-extension MATH such that MATH. MATH is clearly uni... |
math/0004181 | CASE: Assuming REF there is an asymptotically split extension MATH such that MATH and MATH are unitarily equivalent. By REF this implies that MATH and MATH are strongly homotopic. Then MATH and MATH are also strongly homotopic, where MATH inverts the orientation of the suspension. REF follows by observing that MATH is ... |
math/0004181 | We may assume that MATH is not homotopic to MATH. Let MATH, be MATH-homomorphisms with orthogonal ranges, both homotopic to the identity map. Then MATH and MATH are homotopic to MATH, and in particular non-zero. Let MATH be a non-zero positive element in the range of MATH and let MATH be a positive lift of MATH. By spe... |
math/0004181 | It follows from REF that MATH is strongly homotopic to MATH for any MATH-homomorphism MATH. By using that the unitary group of MATH is norm-connected, it follows from this and REF that MATH is naturally isomorphic to MATH. Since MATH is naturally isomorphic to MATH by REF , this completes the proof. |
math/0004183 | Assume MATH is reducible. Let MATH be a sphere that does not bound a ball on either side. MATH cannot be disjoint from MATH or else it would bound a ball on the side that does not contain MATH. Assume MATH intersects MATH minimally and transversally. The intersection will consist of a union of circles. Let MATH be one ... |
math/0004183 | Because the MATH have linking number REF with MATH we can find a NAME surface for MATH disjoint from the MATH. Let MATH be a minimal genus NAME surface for MATH in the link complement. We supplement the notation introduced in REF. Recall MATH is the link exterior of MATH. Let MATH be the corresponding link of MATH comp... |
math/0004183 | Let MATH be strongly MATH-trivial with MATH-trivializers MATH. Let MATH be a minimal genus NAME surface for MATH disjoint from MATH as in REF . MATH has genus MATH. Each linking circle MATH bounds a disk MATH that intersects MATH in an arc running between the two points of MATH and perhaps some simple closed curves. Si... |
math/0004183 | The link consists of the arcs MATH, together with short segments from MATH connecting the end points of the segments (and disjoint from the end points of the other segments). The base case is trivial because, MATH contains a NAME link of MATH components: the NAME link. MATH is obtained from MATH by doubling one of the ... |
math/0004184 | We take the inner product of REF with MATH and integrate over MATH. By the divergence theorem and the incompressibility, we are left with MATH . The NAME and NAME inequalities give MATH so by cancellation of MATH, MATH . An integration over MATH, taking sup in MATH of MATH, gives MATH . Next we integrate the inequality... |
math/0004184 | The subscript MATH indicates that the solutions are only weakly continuos in MATH. We take the inner product of REF with MATH and integrate over MATH. Using NAME 's inequality we obtain MATH . Now MATH by the NAME inequalities, where MATH is a constant. Thus, MATH . An application of the inequality MATH, on the right h... |
math/0004184 | We consider MATH . The second equation in REF gives MATH . We multiply this equation by MATH, integrate over MATH and use the divergence theorem, to get MATH . Thus, by NAME 's inequality MATH where again MATH is the first eigenvalue of the negative Laplacian with vanishing boundary conditions on MATH. Consequently MAT... |
math/0004184 | We recall the relationship between MATH and MATH . The maximum principle in REF , for T, MATH, implies that MATH . Thus MATH . Now consider the system REF . We multiply the first equation by MATH and integrate over MATH. Integration by parts and NAME 's inequality give MATH . Thus by the use of the bound for MATH above... |
math/0004184 | The maximum principle in REF , for T, MATH, implies a maximum principle for MATH, MATH by the relationship between MATH and MATH . This implies that MATH . The rest of the proof, using this bound on MATH, is similar to the proof of the local existence of the NAME equation. CITE can be consulted for details. Then given ... |
math/0004184 | A proof of REF can be found in CITE. |
math/0004184 | The subscript MATH indicates that the weak solutions are only weakly continuos in MATH. We multiply the second equation in REF above by MATH and treat it in the same way as MATH in the proof of REF to get MATH and by adding and subtracting MATH and applying NAME 's inequality, we get MATH . Then integrating with respec... |
math/0004184 | REF follows from REF if we set the viscosity and heat conductivity in REF equal to MATH and MATH, respectively. |
math/0004184 | In the three-dimensional case the proof is similar to the proof of REF but simpler. First we multiply the first equation in REF by MATH and integrate by parts to get MATH after integration in MATH, where MATH. In CITE we show that MATH where MATH denotes the projection onto the divergence free part of MATH. Hence, MATH... |
math/0004184 | By REF we have MATH and MATH . All terms involving spatial derivatives will vanish by the MATH-periodicity on the unit torus and by the incompressibility of MATH. Moreover, MATH, for MATH, so the system reduces to MATH and MATH . We solve this system of ODE's with the initial condition MATH to get the trivial solution,... |
math/0004184 | We apply the NAME inequality on MATH which, by the result of REF , gives MATH . A change of variables MATH yields MATH . We note that the constant MATH will be the same for all MATH-cubes in the interior of MATH. For MATH-cubes intersecting MATH, MATH will vanish at, at least, one point and the usual NAME inequality ap... |
math/0004184 | Let us consider the first equation in REF . We multiply by MATH and integrate over MATH. By the incompressibility we get, by using the NAME 's inequality MATH . By REF , see also the proof of REF , MATH is bounded, and assuming that MATH is a time where MATH is finite, see REF , we can absorb the term MATH into the tim... |
math/0004184 | By the NAME 's inequality, REF it immediately follows that MATH . |
math/0004184 | REF follows from duality by REF follows by REF . |
math/0004184 | By assumption the initial data admit unique two-scale limits. Consider now the first equation of REF , MATH . Let MATH and choose test functions MATH. By the results of REF, all the sequences MATH, MATH, MATH and MATH are uniformly bounded in MATH. Therefore, according to REF , they admit two-scales limits. In order to... |
math/0004185 | For convenience we omit the index MATH and will write simply MATH and MATH instead of MATH and MATH. We belief that no confusion with the system REF can arise. Let MATH and MATH. The system MATH can be rewritten in the form MATH . Consider the system ``orthogonal" to REF MATH . It is evident that the right-hand vector ... |
math/0004185 | Obviously, MATH for MATH. It remains to prove that MATH. Let us use the notations from the proof of REF . Changing in REF coordinates MATH and MATH into the standard polar coordinates MATH, MATH, we get for MATH . Obviously, MATH and MATH. Then NAME 's lemma implies MATH where MATH. Hence, MATH. Let MATH denote the sol... |
math/0004185 | We shall find a comparison equation for MATH. We have MATH here MATH denotes inner product. This gives for MATH (we can assume that MATH for sufficiently large MATH, because MATH). Let us consider the comparison equation MATH . It is well - known (see CITE) that every function MATH satisfying REF does not exceed the so... |
math/0004185 | As it follows from REF the system REF written in ``quasi-polar" coordinates satisfies all conditions of REF . |
math/0004185 | We have MATH . Assumption MATH implies that MATH converges. Hence the limit MATH exists. The theorem is proved. |
math/0004185 | It follows immediately from REF. |
math/0004185 | Let MATH denote any bounded solution of REF . REF implies the convergence MATH as MATH. Every solution of REF lying on the invariant torus MATH has the form MATH, where MATH is the initial value. It is necessary and sufficient to prove the existence of a MATH such that MATH . Thus it is sufficient to prove that the lim... |
math/0004188 | Multiplying both sides of REF by MATH, we reduce REF to the identity MATH . By NAME 's formula MATH the Right-hand side of REF can be transformed into MATH . Thus, we need to verify that MATH which is equivalent to MATH which is true because [REF ] MATH . |
math/0004188 | Multiplying both sides of REF by MATH and summing on MATH, we find: MATH where we used the two handy formulae: MATH . Formulae REF convert the identity REF into MATH where MATH. Each side of the equality REF vanishes for MATH, so it's equivalent to the MATH-derivative MATH of it: MATH where we used REF . Since the Righ... |
math/0004188 | Each side of the identity REF vanishes for MATH, so we apply MATH to it to make it simpler-looking. We get: MATH because, as is easily verified, MATH . We thus arrive at the identify MATH . Denote MATH . We shall prove that MATH REF then results from REF when MATH. To prove REF we use induction on MATH. For MATH, REF i... |
math/0004188 | By REF , the LHS of REF is MATH . |
math/0004188 | Since MATH we have MATH . Now, MATH . Since, MATH, the exponents MATH are, modulo MATH, just a permutation of the exponents MATH. Therefore, the expressions MATH are, since MATH, just a permutation of the expressions MATH modulo MATH. This proves REF . Now, for any MATH, coprime to MATH or not, MATH so that, by REF , f... |
math/0004188 | First, since MATH is a prime, MATH . Next, it's easy to verify that MATH . Taking MATH and using REF , we get MATH . Therefore, MATH . |
math/0004188 | The LHS of the congruence REF is MATH . Since MATH is coprime to MATH, the set MATH is, modulo MATH, a permutation of the set MATH. |
math/0004188 | The congruence MATH has exactly two solutions: MATH so that MATH . Otherwise, MATH . For each such pair MATH we have: MATH . Therefore, for all such MATH pairs combined, MATH where MATH . Thus, MATH . In addition, MATH and MATH . Combining formulae REF - REF we obtain REF . Since REF follows from REF . |
nlin/0004008 | REF follows from REF - REF, and REF. Similarly, REF follows from REF - REF, and REF. Relations REF - REF are obvious from REF. |
nlin/0004008 | REF are verified using REF - REF, and REF follows by combining REF, and REF. Clearly MATH is meromorphic on MATH by REF. Since MATH one infers that MATH is meromorphic on MATH if the zeros of MATH are all simple. This follows from REF by restricting MATH to a sufficiently small neighborhood MATH of MATH such that MATH ... |
nlin/0004008 | The existence of these asymptotic expansions (in terms of local coordinates MATH near MATH and local coordinate MATH near MATH) is clear from the explicit form of MATH in REF. Insertion of the polynomials MATH, MATH, and MATH, then in principle, yields the explicit expansion coefficients in REF - REF. However, this is ... |
nlin/0004008 | REF follow from REF noting MATH . |
nlin/0004008 | REF , and REF imply MATH . Using MATH by REF, one concludes REF. Similarly, one derives from REF, and REF, MATH . Since MATH by REF, one arrives at REF are derived analogously. In order to conclude REF, one first needs to investigate the case where MATH hits one of the branch points MATH and hence the right-hand sides ... |
nlin/0004008 | REF follow from REF by comparing powers of MATH and MATH, using REF follow from taking MATH in REF, using again REF. Finally, REF follow from MATH, MATH and REF. |
nlin/0004008 | It suffices to insert REF into the system REF - REF. |
nlin/0004008 | Define the polynomial MATH . Using REF and MATH (by differentiating REF with respect to MATH) one then computes MATH . In order to investigate the leading-order term with respect to MATH of MATH we first study the leading-order MATH-behavior of MATH, and MATH. Writing (compare REF - REF) MATH a comparison of leading po... |
nlin/0004008 | Abbreviating MATH one infers from MATH (compare REF), and REF, that MATH . Next, observing the fact that MATH it becomes a straightforward matter deriving REF - REF. For simplicity we just focus on the expansion of MATH as MATH, the rest is completely analogous. Using MATH and REF - REF, one computes by comparison with... |
nlin/0004008 | Given the expansions REF of MATH near MATH and MATH, REF are standard facts following from NAME interpolation results of the type (see, for example, CITE) MATH . |
nlin/0004008 | Using REF (compare REF) one obtains MATH and hence MATH . Here we used REF to convert the directional derivatives MATH and MATH, MATH into MATH and MATH derivatives. Since by REF exactly the same REF apply to MATH, insertion of REF, and REF (and their MATH analogs) into REF proves REF - REF. |
nlin/0004008 | A comparison of REF - REF - REF yields MATH and hence MATH proves REF. Insertion of REF into the leading asymptotic term of REF - REF then yields REF - REF. |
nlin/0004008 | Introducing MATH with an appropriate normalization MATH (which is MATH-independent) to be determined later, we next intend to prove that MATH . A comparison of REF, and REF shows that MATH and MATH share the identical essential singularity near MATH. Next we turn to the local bahavior of MATH with respect to its zeros ... |
nlin/0004008 | Since MATH for all MATH, there is nothing to prove in the special case MATH. Hence we assume MATH. Let MATH be a fixed branch point of MATH and suppose that MATH is special. Then by REF there is a pair MATH such that MATH where MATH. Let MATH so that MATH. Then MATH and hence by REF , MATH . Since by REF is nonspecial ... |
nlin/0004041 | Let us take in previous theorem MATH. Since MATH it follows that all conditions of REF are satisfied. |
physics/0004057 | Taking the derivative with respect to MATH, for given MATH and MATH, one obtains MATH since the marginal distribution satisfies MATH. MATH are the normalization NAME multipliers for each MATH. Setting the derivatives to zero and writing MATH, we obtain REF . When varying the normalized MATH, the variations MATH and MAT... |
physics/0004057 | First we note that the conditional distribution of MATH on MATH follows from the NAME chain condition MATH. The only variational variables in this scheme are the conditional distributions, MATH, since other unknown distributions are determined from it through NAME 's rule and consistency. Thus we have MATH and MATH . T... |
physics/0004057 | For lack of space we can only outline the proof. First we show that the equations indeed are satisfied at the minima of the functional MATH (known for physicists as the ``free energy"). This follows from REF when applied to MATH with the convex sets of MATH and MATH, as for the BA algorithm. Then the second part of the... |
quant-ph/0004017 | . At deposit time NAME sends NAME one qubit MATH, which might be entangled with the qubits MATH that NAME holds. Let us denote the reduced density matrix of MATH by MATH. At revealing time, NAME may choose whether she wants to bias the result towards MATH, in which case she applies the generalized measurement MATH, or ... |
quant-ph/0004017 | We first represent the strategy of a honest NAME in quantum language. Consider two maximally parallel purifications MATH and MATH of MATH and MATH, where MATH and MATH are density matrices of the register MATH, and the purifications are states on a larger NAME space MATH. By CITE, MATH. At preparation time, NAME prepar... |
quant-ph/0004017 | We first describe a general scenario. NAME is honest and sends MATH to NAME. NAME has an ancilla MATH. NAME applies some unitary transformation MATH acting on the registers MATH and MATH. Let us denote MATH NAME then sends register MATH to NAME, and keeps register MATH to himself. We want to show that if MATH contains ... |
quant-ph/0004017 | . We will prove that all the unprimed MATH vectors lie in one bunch of small width, using the unitarity of MATH. The unitarity of MATH implies that MATH. We can express MATH as in REF . We get: MATH . Substituting the values MATH for actual values, and noticing that MATH, we in particular get the following equations: M... |
quant-ph/0004017 | We will show MATH . Thus, MATH, where the last equality is due to REF . Since, MATH we get MATH as desired. |
quant-ph/0004017 | . We express MATH, where MATH is a pure state. We further express each MATH in the eigenbasis MATH: MATH . Applying MATH, this state is taken to: MATH . The reduced density matrix to the register MATH, in case of event MATH is: MATH and altogether, MATH. To complete the proof we just notice that MATH. The proof for MAT... |
quant-ph/0004017 | Say NAME sent NAME the state MATH. We can express it as MATH where MATH and MATH. NAME applies MATH on MATH and gets MATH . Therefore, if we measure the last qubit, then with probability MATH we end up in MATH and with probability MATH we end up in MATH normalized. Thus the density matrix of MATH after tracing out the ... |
quant-ph/0004032 | Let MATH and denote by MATH the orthogonal projection onto MATH. For all MATH and MATH we have MATH . Since MATH and MATH are in MATH and MATH is square integrable modulo MATH, we have MATH . Explicitly, MATH . For any MATH the function MATH defined in REF is in MATH and we get MATH . We claim that MATH . Indeed MATH a... |
quant-ph/0004032 | From REF above it follows that the set MATH is a resolution of the identity of MATH and MATH is an isotypic representation. Hence, MATH is a resolution of the identity in MATH if and only if MATH and, in this case, MATH is isotypic. Conversely, assume that MATH is an isotypic representation. Let MATH be an irreducible ... |
quant-ph/0004032 | It is known, see for example CITE, that MATH is informationally complete if and only if it separates the set of trace class operators. Let MATH be a trace class operator, then MATH for any MATH if and only if MATH for MATH-almost all MATH. Observing that MATH for all MATH such that MATH, this last condition is equivale... |
quant-ph/0004032 | Let MATH be a trace class operator, and consider the decompositions of MATH and MATH as given in REF, that is, MATH and MATH . Since MATH, it follows that MATH for all MATH. Given MATH, using the orthogonality relations REF , one has MATH since MATH converges in MATH to MATH, as shown in REF, and MATH is bounded. Hence... |
quant-ph/0004072 | This fact is a simple consequence of the linearity of quantum mechanics. Suppose we had such an operation and MATH and MATH are distinct. Then, by the definition of the operation, MATH . (Here, and frequently below, I omit normalization, which is generally unimportant.) But by linearity, MATH . This differs from REF by... |
quant-ph/0004072 | Suppose MATH is a basis for MATH and MATH is a basis for MATH. By setting MATH and MATH equal to the basis elements and to the sum and difference of two basis elements (with or without a phase factor MATH), we can see that REF is equivalent to MATH where MATH is a Hermitian matrix independent of MATH and MATH. Suppose ... |
cond-mat/0005026 | MATH is a continuous and positive function that satisfies the variational equation MATH with MATH. Let MATH. Since MATH we see that MATH on MATH, that is, MATH is subharmonic on MATH. Hence MATH achieves its maximum on the boundary of MATH, where MATH, so MATH is empty. |
cond-mat/0005026 | Obviously, the infimum is achieved for a multiple of a characteristic function for some measurable set MATH. If MATH denotes the NAME measure of MATH, then MATH . Now MATH, with equality for MATH . |
cond-mat/0005026 | Immediate consequences of REF , using that MATH and MATH. |
cond-mat/0005026 | It is clear that MATH. For the other direction, we use MATH as a test function for MATH, where MATH . Note that MATH and MATH. Therefore MATH where we have used convexity for the last term. Moreover, MATH as long as MATH. So we have MATH . Optimizing over MATH gives as a final result MATH . |
cond-mat/0005026 | With the demanded properties of MATH, REF holds. Using this and REF one easily verifies REF . Moreover, MATH is a minimizing sequence for the functional in question, so we can conclude as in REF that it converges to MATH strongly in MATH, proving REF . (Remark: In REF there is a misprint, instead of MATH one should hav... |
cond-mat/0005026 | The upper bound is trivial. Because MATH, defined in REF , converges uniformly to MATH and MATH as MATH, we have the lower bound MATH for some MATH. |
cond-mat/0005026 | Since MATH for all MATH we have MATH with MATH . Choosing MATH gives the desired result. |
cs/0005001 | The lemma can be proved directly by considering the worst case: namely we can always divide the rectangle at worst case into MATH regions of unit size, which means the above difference always vanishes. |
cs/0005001 | REF follows immediately for MATH because for any combination of partitioning, each of the anti-A-noise-concentrated block of size MATH can at best ``contaminate" MATH equal size regions of MATH. It is a simple matter to confirm that the conclusion is valid for MATH as well. REF comes from the fact that at most MATH of ... |
cs/0005001 | To prove REF , we must show that among all the possible MATH different partitions that the Shifting Strategy can possibly generate, each of MATH size anti-A-noise-concentrated block is capable of contaminating the total of MATH different regions. Once this is done, the NAME Hole Principle CITE ensures that there is at ... |
cs/0005009 | For a variable MATH, we define MATH as the maximum depth of nested modal operators in MATH. Obviously, MATH holds. Also, if MATH then MATH. Hence each path MATH in MATH induces a sequence MATH of natural numbers. MATH is a tree with root MATH, hence the longest path in MATH starts with MATH and its length is bounded by... |
cs/0005009 | The sequence of rules induces a sequence of trees. The depth and the out-degree of these trees is bounded in MATH by REF . For each variable MATH the label MATH is a subset of the finite set MATH. Each application of a rule either CASE: adds a constraint of the form MATH and hence adds an element to MATH, or CASE: adds... |
cs/0005009 | MATH for any rule MATH implies MATH, hence each model of MATH is also a model of MATH. Consequently, we must show only the other direction. CASE: Let MATH be a model of MATH and let MATH be the constraint that triggers the application of the MATH-rule. The constraint MATH implies MATH. This implies MATH for MATH. Hence... |
cs/0005009 | The satisfiability of MATH implies that also MATH is satisfiable. By REF there is a sequence of applications of the optimised rules which preserves the satisfiability of the c.s. By REF any sequence of applications must be finite. No generated c.s. (including the last one) may contain a clash because this would make it... |
cs/0005009 | Let MATH be a complete and clash-free c.s. generated by applications of the optimised rules. We will show that the canonical model MATH together with the identity function is a model for MATH. Since MATH was generated from MATH and the rules do not remove constraints from the c.s., MATH. Thus MATH is also a model for M... |
cs/0005009 | Let MATH be the MATH-formula to be tested for satisfiability. We can assume MATH to be in NNF because the transformation of a formula to NNF can be performed in linear time and space. The key idea for the NAME implementation is the trace technique CITE, that is, it is sufficient to keep only a single path (a trace) of ... |
cs/0005009 | The sequence of rule applications induces a sequence of trees. As before, the depth and out-degree of this tree is bounded in MATH by REF . For each variable MATH, MATH is a subset of the finite set MATH. Each application of a rule either CASE: adds a constraint of the form MATH and hence adds an element to MATH, or CA... |
cs/0005009 | Let MATH be a complete and clash-free c.s. obtained by a sequence of rule applications from MATH. We show that the canonical structure MATH is indeed a model of MATH, where the canonical structure for MATH is defined as in REF . Please note, that we need the condition ``MATH iff MATH" to make sure that all information ... |
cs/0005009 | Let MATH be a model for MATH and MATH the set of relations that occur in MATH together with their inverse. We use MATH to guide the application of the non-deterministic completion rules by incremently defining a function MATH mapping variables from the c.s. to elements of MATH. The function MATH will always satisfy the... |
cs/0005009 | Consider the algorithm in REF , where MATH denotes all intersections of relations that occur in MATH. As the algorithm for MATH, it re-uses the space used to check for the existence of a complete and clash-free ``subtree" for each successor MATH of a variable MATH. Counter variables are used to keep track of the values... |
cs/0005009 | The DL MATH is a syntactic restriction of the DL MATH, which, in turn, is a syntactical variant of MATH. Hence, the MATH-algorithm can immediately be applied to MATH-concepts. |
cs/0005010 | Note that the deductive closure of the reduct MATH is a closure, and note that for every MATH that is a closure, the deductive closure of MATH is a subset of MATH. |
cs/0005010 | Note that the operator MATH is monotonic and that it has a fixed point MATH for any closure MATH. Hence, MATH for all MATH and therefore MATH. Since MATH, taken as a function of MATH, defines a closure, MATH. Thus, MATH follows. |
cs/0005010 | Let MATH be a logic program that does not contain any choice rules and let MATH be two stable models of MATH. As MATH since MATH is anti-monotonic in its first argument, MATH. |
cs/0005010 | Let MATH be the atoms not covered by MATH. We prove the claim by induction on the size of MATH. Assume that the set MATH. Then, MATH covers MATH by REF and MATH returns true if and only if MATH returns false. By REF, CREF, and CREF, this happens precisely when there is a stable model of MATH agreeing with MATH. Assume ... |
cs/0005010 | Recall from REF that MATH is monotonic and that its least fixed point is equal to MATH. Hence, if MATH is a stable model of MATH, then for any MATH, MATH. Let the stable model MATH agree with the set MATH. As MATH, the first claim follows. Notice that MATH . If for all MATH, MATH, then for all MATH, MATH and consequent... |
cs/0005010 | Observe that the function MATH is monotonic and that the function MATH is anti-monotonic. Hence, MATH are monotonic with respect to MATH. Assume that there exists MATH such that MATH for only one MATH and MATH. If MATH and MATH, then MATH . Consequently, both MATH and therefore MATH . It follows that MATH. Thus, MATH i... |
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