paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0105260 | Choose an integer MATH so that MATH for the constant MATH given by REF . Fix MATH so that MATH. REF applied to MATH with MATH yields a constant MATH so that MATH for any NAME set MATH. In a same way, one can find constants MATH so that MATH for any MATH and any MATH. Take a NAME set MATH and MATH so that MATH. Write MATH with MATH. We have MATH which completes the proof of REF . |
math/0105260 | Denote by MATH the irreducible components of MATH that are not in MATH. For each MATH, pick a point MATH so that MATH, and fix MATH with MATH. One can find a constant MATH so that MATH for all MATH. Introduce the set MATH for a suitable MATH to be chosen later. Because of REF, it is a finite set. Let MATH and MATH. MATH . First assume MATH. Write MATH with MATH. We have MATH . By REF, we infer MATH . Set MATH. Note that MATH. We get MATH . Now for a fixed MATH we take MATH large enough to conclude MATH for some constant MATH. MATH . Now assume MATH. The set MATH is finite. By definition it does not intersect MATH, hence one can find constants MATH, MATH such that MATH for all MATH and MATH. Let MATH be the smallest integer such that MATH. Applying REF, we get MATH and the proof is complete. |
math/0105260 | The result is local. Pick MATH. We may assume that MATH and that MATH or MATH locally at MATH. Further, write MATH where MATH is the multiplicity of MATH as a critical point of MATH and where MATH are multivalued functions with MATH. Let MATH be the subset of the bidisk MATH where MATH. We will show that MATH which will prove REF. For the rest of the proof we will let MATH denote various positive constants. Let MATH. For MATH and fixed MATH we may estimate MATH . We can then use NAME 's Theorem to estimate the volume of MATH: MATH . If MATH, then the second term in REF can be estimated by MATH and so MATH in this case. If MATH then the second term in REF is instead bounded by MATH and so MATH. This proves REF. As for REF we notice that for MATH, all the critical points for MATH, except the ones at MATH, will have multiplicity MATH. Thus the above calculations imply REF. |
math/0105260 | The proof is similar to that of REF . First assume that MATH, with MATH from REF . We pick a ``stopping time" MATH defined by MATH with MATH from REF . Then we get MATH . It is now easy to iterate this estimate and arrive at REF. |
math/0105260 | Let MATH be an irreducible component of MATH, that is, a line. After replacing MATH by an iterate we may assume that MATH is fixed by MATH and hence MATH induces a rational map of MATH of degree MATH. Let MATH be a current as in the statement of the lemma. By REF we may define the probability measure MATH. We have MATH . By REF the measure MATH does not charge totally invariant orbits of MATH. On the other hand, one can find MATH such that for large MATH and any MATH. We conclude MATH . |
math/0105260 | We will use of REF on the behavior of NAME numbers when one weight tends to infinity. Let MATH and MATH be as above We cover MATH by a finite number of coordinates chart MATH such that MATH. In the open set MATH, the map MATH can be written in the form MATH . For a point MATH, we denote by MATH the NAME number of MATH at MATH with weights MATH associated to the coordinate systems MATH. Assume that we can prove the following result: For any point MATH and any MATH, we have: MATH . As before fix a constant MATH such that for large MATH and any point MATH, we have MATH. By REF does not charge MATH and MATH for any point MATH; hence for any MATH large enough we get MATH where the first inequality follows from REF (with MATH), the second from REF , and the last convergence from REF . This concludes the proof. |
math/0105260 | This is a local result so we may assume MATH and MATH with MATH. We easily check that there exist constants MATH such that for any MATH . We remark that this property is easy to verify but nevertheless central to the proof. Write MATH for some local psh potential MATH. We infer MATH which concludes the proof. |
math/0105260 | We argue by contradiction. Suppose that MATH is a positive closed current on MATH for which the assumptions, but not the conclusions, of Theorem A hold. As in REF we write MATH with MATH qpsh, and conclude that there exists a ball MATH, a positive number MATH and a sequence MATH such that MATH . We will get a contradiction from REF by estimating the volumes of the two sides. Fix small neighborhoods MATH, MATH of the exceptional sets MATH and MATH, respectively. By the superattracting nature of MATH and MATH we may assume that MATH for MATH. In order to reach a contradiction, it is sufficient to consider three different cases. MATH . Let us first assume that MATH avoids MATH for all MATH. Then REF applies and shows that MATH for some MATH. On the other hand, the NAME estimate REF shows that MATH for some MATH and for all MATH. This yields a contradiction. MATH . The second case is when MATH for all MATH. We then use the results from REF on the dynamics near the first exceptional set MATH. First, by REF there exists a constant MATH such that MATH for sufficiently large MATH. Second, by REF , for arbitrarily large MATH, one can find an integer MATH so that MATH. Hence by the NAME estimate REF one has MATH for large enough MATH. For MATH, REF then imply MATH . We get a contradiction by choosing MATH so that MATH and letting MATH. MATH . The third and last case is when MATH for all MATH. But by our assumption MATH is bounded at MATH and so REF clearly cannot hold. This completes the proof of Theorem A. |
math/0105260 | If MATH is the current of integration on a curve MATH of degree MATH, then MATH satisfies the assumptions of Theorem A unless CASE: MATH contains an irreducible component of MATH; or CASE: MATH. This concludes the proof as the set of curves MATH satisfying either of these conditions is a algebraic proper subset of MATH. |
math/0105260 | Let MATH be the set of holomorphic maps MATH of degree MATH for which MATH. By REF , MATH consists of at most three totally invariant lines and a totally invariant set whose cardinality is bounded by some integer MATH. It is to check from this that MATH defines an algebraic set in MATH. To conclude the proof we only have to exhibit one holomorphic map MATH with MATH. We follow a construction of NAME. Take a NAME map in MATH of degree MATH for instance MATH. Consider the holomorphic map MATH . It has topological degree MATH. The quotient MATH by the symmetry MATH is isomorphic to MATH and MATH induces a holomorphic map MATH on the quotient. The topological degree of MATH is MATH hence MATH. As MATH does not contain critical periodic points, the same is true for MATH and for MATH too. Hence MATH and we are done. |
math/0105260 | Assume that MATH. In homogeneous coordinates, MATH can be written MATH for homogeneous polynomials MATH, MATH, MATH of degree MATH with MATH. Hence, locally, MATH for some germ MATH with MATH. As MATH is contracting, one can define the map MATH and one checks the map MATH conjugates MATH to MATH. |
math/0105260 | Again assume MATH. We may assume that the set MATH is given by MATH or by MATH. In the first of these cases, MATH is a homogeneous point and MATH is locally conjugate to MATH by REF . In the second case, we have MATH in homogeneous coordinates, where MATH. Hence, locally, MATH for some germ MATH with MATH. As in the proof of REF we define MATH and conclude that the map MATH conjugates MATH to the desired form. |
math/0105260 | This follows immediately from the normal forms in REF . |
math/0105260 | We first consider the case MATH and write MATH in the skew product form REF, which we may rewrite as MATH where MATH are multi-valued with MATH and MATH. Fix MATH small. It follows from REF that there exists a constant MATH such that for any MATH and any NAME set MATH we have MATH. Further, for any NAME set MATH we have MATH. Iterating these estimates yields MATH and MATH, where MATH. Now pick a NAME set MATH with MATH. After iterating forward we may assume that MATH. For MATH we write MATH. There exists MATH and a set MATH with MATH such that MATH for MATH. But then the previous estimates imply that MATH and MATH for MATH, so by NAME 's Theorem we get MATH as desired. The remaining case, when MATH is a homogeneous point, is similar. We use the skew product structure REF. The only new observation that we need is that if MATH is a rational map of degree MATH, then there exists MATH such that MATH for any NAME set MATH. |
math/0105260 | The implication MATH is relatively easy. If MATH puts mass on a totally invariant curve MATH, say MATH, then for all MATH we have MATH. Since MATH has bounded potential we cannot have convergence towards MATH. Similarly, if MATH with MATH, then one immediately checks that MATH. Hence, by REF , MATH for some MATH, which also prevents the sequence to converge towards MATH. Conversely, suppose that the current MATH satisfies REF of Theorem A'. To prove that MATH we follow the proof of Theorem A up to the third case, that is, when MATH for all MATH. We pick a constant MATH small enough. As MATH for all MATH, by REF one can find a constant MATH such that MATH for all MATH. Combining REF and the hypothesis MATH with REF , we get MATH hence MATH by letting MATH. But MATH is fixed and MATH is arbitrarily small, so this yields a contradiction. |
math/0105260 | Since MATH is totally invariant, the current MATH is in MATH. This current need not be extremal, but can be decomposed into currents in MATH supported on MATH. |
math/0105260 | Assume MATH preserves the pencil of lines through MATH. Then MATH induces a rational map MATH of MATH (the set of lines) of degree MATH. Let MATH be the measure of maximal entropy for MATH. This satisfies MATH and if we define MATH where MATH denotes the current of integration on the line through MATH corresponding to MATH, then MATH, so MATH. Assume MATH with MATH. From the local structure of MATH, we infer the existence of a positive measure MATH such that MATH. The equation MATH is equivalent to MATH. As MATH has no atoms, this forces MATH for some constant MATH. Hence MATH, and MATH. Finally, a direct computation yields MATH. |
math/0105260 | Pick MATH. Introduce the homogeneous real analytic function on MATH . It vanishes exactly on the ray MATH and by homogeneity the positive closed current MATH can be pushed down to MATH as a current MATH, smooth outside MATH, with a pole at MATH whose NAME number is MATH. Define MATH where MATH is a lift a MATH to MATH. Then MATH induces by projection a function on MATH which is real analytic outside MATH. We claim there exists constants MATH such that MATH for any point in MATH. Indeed, as MATH is totally invariant, MATH and MATH both vanishes exactly along MATH, hence the inequality has to be checked only in a neighborhood MATH of MATH in the chart MATH. To do so, you may decompose MATH in two sets MATH and MATH for a well chosen MATH. In each of these sets, the estimates follow from a direct computation we leave to the reader. By normalizing MATH, we can assume MATH. We now follow the standard construction of the NAME current. We have MATH, hence MATH for all MATH. The sequence of function MATH is decreasing converging uniformly on compact sets. Hence the limit MATH is a MATH function, continuous outside MATH, and bounded everywhere. The positive closed current MATH belongs to MATH, has a continuous potential outside MATH, and a singularity at MATH with NAME number MATH. More precisely, in the coordinates MATH, the NAME number of MATH with weight MATH is given by MATH . To conclude we show MATH is extremal in the cone MATH. Assume MATH with MATH for MATH. NAME numbers behave additively hence MATH . On the other hand, the following inequalities are standard (see CITE,CITE) MATH . Whence MATH and we infer MATH for MATH. Pick a global potential MATH defined in MATH. By definition of NAME numbers we have in a neighborhood of the origin MATH for some constant MATH. Together with REF , we deduce that MATH is globally bounded from above. As MATH, and MATH is bounded, we conclude that MATH are also bounded everywhere. Hence MATH . This shows that MATH. |
math/0105261 | By the separability assumption, the result follows if one knows that there are no non-trivial embeddings of MATH over MATH. But if MATH is any such injection then one has MATH as MATH. Thus MATH. |
math/0105261 | Suppose that MATH has characteristic MATH. Now, for some MATH one knows that MATH is separable over MATH. As MATH the result follows from the proposition. |
math/0105261 | Let MATH be an injection of MATH into MATH over MATH. Then MATH . The first and third terms on the right have the same absolute value. Moreover, by assumption, if the second term is non-zero then its absolute value is the greatest of the three; thus it is also the absolute value of MATH. The result now follows. |
math/0105261 | The functions MATH are continuous. Moreover, let MATH be the additive valuation associated to MATH. Then, from CITE, one also has exponential lower bounds on MATH which are independent of MATH. The result follows directly. |
math/0105261 | Let MATH be a non-negative integer and (in the notation of CITE) MATH . The main result in CITE is to establish a formula originated by NAME for MATH (this formula is then used to compute MATH and the NAME polygon of MATH). The formula expresses MATH in terms of a certain MATH-tuple, called the ``greedy element," MATH of non-negative integers such that MATH in such a way that there is no carry-over of MATH-adic digits and such that the first MATH elements are both positive and divisible by MATH. From this formula one obtains a formula for MATH by choosing MATH sufficiently close to MATH (see REF). The result follows simply by noting that this formula is linear and involves only the first MATH-terms of the given MATH-tuple. |
math/0105261 | The fact that MATH is positive follows easily from general theory. To see that it is divisible by MATH, note that NAME 's work shows that the NAME polygon of MATH only has segments of vertical length MATH (that is, their projection to the MATH-axis has unit length). Thus the divisibility follows immediately from the proposition. |
math/0105261 | The first part follows immediately from NAME 's construction of the greedy element (see for example, the proof of REF). The second part follows from the first part and the fact that all segments of the NAME polygon of MATH are known to have projections to the MATH-axis of unit length. |
math/0105261 | This again follows from NAME 's construction of the greedy element. |
math/0105261 | Clearly the trivial zero of MATH has valuation MATH and it is easy to see that this is the unique zero of MATH of highest valuation. The result now follows as before. |
math/0105261 | Suppose that MATH is almost real, where MATH is non-zero and MATH. Put MATH; it is simple to check that with this MATH the NAME functional equation holds. Conversely, assume the NAME functional equation and let MATH and MATH be two non-negative integers such that MATH. Then MATH thus MATH is real. Now let MATH be the smallest non-negative integer with MATH. Then MATH with MATH real, and the result is established. |
math/0105261 | We know that MATH. Thus we see MATH consequently, MATH. On the other hand, the functional equation immediately gives us MATH . Consequently we deduce that MATH . The only if part now follows upon substituting in the power series for MATH. The if part follows since these calculations are reversible. |
math/0105261 | This follows directly from REF . |
math/0105262 | By NAME 's theorem, a line through any REF singularities MATH and MATH implies MATH. Thus at most one singularity can have multiplicity MATH, and the result follows. |
math/0105262 | Since MATH is irreducible over MATH, its absolutely irreducible factors are conjugate over MATH. If MATH is a root of one of the factors, it must be a root of the others. But MATH only vanishes to order REF, thus MATH has one factor, and is absolutely irreducible. |
math/0105263 | CASE: Since MATH is Killing and MATH is selfdual NAME, we have MATH . The antiselfdual part of this is what we want. CASE: Differentiating the twistor equation again and skew symmetrizing gives MATH . Contracting with another vector field MATH and taking the trace over MATH and MATH, we obtain MATH . The right hand side is skew, so MATH is skew, MATH is a Killing field and MATH. |
math/0105263 | CASE: Since MATH, we have MATH. The results follow by contracting with MATH and MATH respectively. CASE: Since MATH and MATH, we have MATH and hence MATH. We deduce that MATH, and, in the same way, MATH. |
math/0105263 | Contracting the formula of REF with MATH, we obtain MATH. Now MATH and MATH is nonzero, so we deduce that MATH. This implies that MATH which vanishes by the above lemma. A similar argument shows that the other twist scalar also vanishes. |
math/0105263 | Contracting the equation with MATH, we deduce that MATH. Hence MATH, which may be rewritten MATH. This means that MATH is parallel with respect to the NAME connection of MATH. |
math/0105263 | Given a twistor MATH, we can use the freedom in the choice of half-space model to set MATH. In these half-space coordinates MATH satisfies the equation MATH . Since MATH direct calculation yields MATH, so MATH satisfies the NAME equation. |
math/0105263 | Suppose that MATH is a selfdual NAME metric of nonzero scalar curvature on MATH with two commuting Killing fields. Let us review what we have proven so far about MATH. Firstly, by REF , the Killing fields are surface orthogonal, and therefore, by CITE, the conformal class of MATH is a NAME space. (If MATH is conformally hyperkähler, it must be conformally flat, that is, locally isometric to MATH or MATH, but then the Killing fields of MATH cannot all be triholomorphic with respect to the flat hyperkähler metric.) The quotient of MATH by one of its Killing fields is an NAME - NAME space MATH with an abelian monopole MATH CITE. The work of Tod CITE shows that the choice of Killing field determines a compatible scalar-flat kähler metric on MATH, a NAME structure on MATH, and a coordinate MATH on MATH such that the monopole MATH is MATH. On the other hand, we showed in REF that MATH is an NAME - NAME space with an axial symmetry. Since the NAME structure is invariant under this symmetry, it is one of the NAME structures determined by a point at infinity on MATH. Introducing compatible half-space coordinates we may write the solution of the NAME equation corresponding to this NAME - NAME space as MATH and then MATH for some function MATH on MATH with MATH and MATH. Now set MATH. Then MATH and MATH is obtained by applying MATH to the twistor MATH as in REF . On the other hand, applying MATH to MATH yields the solution MATH, which is precisely the solution needed to construct the MATH monopole on MATH. Hence MATH generates the pencil of solutions of the NAME equation yielding the underlying conformal structure of the selfdual NAME metric MATH. The distinguished scalar-flat kähler metric is MATH and rescaling this by MATH, according to REF , we recover the selfdual NAME metric MATH. The explicit REF is obtained from REF by direct substitution. As we remarked in the introduction, the reader who is not convinced that we really have encoded the entire selfdual NAME condition in the construction can easily verify this directly. Such calculations amount to reproving Tod's result CITE in this special case. |
math/0105263 | For the first part we show that MATH is the metric on MATH induced by MATH. Let MATH be the matrix MATH, so that the selfdual NAME metric is MATH . Then the metric on the torus induced by MATH is MATH and MATH by the NAME - NAME theorem. This is what we want, because MATH. For the momentum maps, we must compute the contraction of the MATH-valued symplectic form MATH with MATH and MATH. The contraction with any vector field MATH in the torus is MATH. It is straightforward to compare this with MATH or MATH when MATH or MATH: in particular note that MATH and MATH. |
nlin/0105012 | A direct calculation based on REF . |
nlin/0105012 | Let MATH. Then from MATH we get: MATH . Hence MATH which yields REF . Similarly, we derive from REF : MATH which implies REF . |
nlin/0105012 | The right reduced NAME - NAME REF have in the present setup the following form: MATH where MATH . To calculate the derivatives of MATH, we use the following formulas: MATH . Indeed, the first one of these expressions follows from: MATH . To prove the second one, proceed similarly: MATH . Finally, for the third expression we have: MATH . With the help of REF we find: MATH and MATH . Comparing the latter formula with the first equation of motion in REF , we find that it can be rewritten as MATH which is the first equation in REF . To obtain the second one, derive from REF : MATH . Adding these two equations, we find: MATH . (On the last step we took into account that MATH.) This is nothing but the second equation of motion in REF . The NAME properties of the map REF are assured by the version of REF for right-invariant NAME. It remains to demonstrate that the function REF is indeed an integral of motion. This is done by the following derivation: MATH . (In this calculation we used the identity MATH for MATH, MATH; notice that the scalar products on the both sides of this identity are defined on different spaces!) The theorem is proved. |
nlin/0105012 | Direct verification. Notice that it is most convenient to check REF in the form MATH when it becomes polynomial in MATH. |
nlin/0105012 | The derivation of REF is straightforward, like in REF . Obviously, due to MATH, the equations of motion REF approximate the continuous time ones REF , while the ``NAME transformation" REF approximates REF . We discuss now the inversion of the ``NAME transformation" REF . Obviously, it is trivially invertible if MATH: then MATH . For a general MATH we derive from REF : MATH . Indeed, we have: MATH . Adding these two equations, we find: MATH which yields REF . Now plugging the expression for MATH through MATH, MATH into the second term on the right-hand side of REF , we see that this equation is uniquely solvable for MATH. Turning to the last statement of REF , we have: MATH . (On the last but one step we used REF .) |
nlin/0105012 | Derivation of equations of motion is based on REF . We have: MATH which proves the second equation of motion in REF , and MATH which proves the first one. As for the NAME representation, it is verified by a direct calculation. It is most convenient to check it in the form MATH where MATH . Let us indicate the main steps of this calculation. Expanding REF in powers of MATH, we come to the following equations: MATH . Here REF follows from the first equation in REF . REF yields REF , which follows from the skew-symmetry of MATH, MATH. To prove REF we notice that the first equation of motion in REF easily yields the following expressions: MATH while the second equation of motion in REF shows that MATH . This is a companion formula to REF . Now REF follows directly from REF . Another consequence of these three formulas is: MATH and this proves REF , since the right-hand side of this formula vanishes due to the skew-symmetry of MATH, MATH. |
nlin/0105021 | REF follows using REF of MATH as well as relations REF - REF. The other relations, REF - REF, are easy consequences of MATH in addition to REF - REF. By REF - REF, MATH is meromorphic on MATH away from the poles MATH of MATH. By REF, and REF, MATH and hence MATH is meromorphic on MATH by REF as long as 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 for all MATH and all MATH. Since MATH is meromorphic on MATH by REF, MATH is meromorphic on MATH by REF. The remaining properties of MATH can be verified by using REF - REF as well as relations REF - REF. In particular, REF follows by inserting the definition of MATH, REF, into REF, using REF. |
nlin/0105021 | We only prove REF since the proof of REF follows in an identical manner. Inserting MATH into REF , one concludes from REF, MATH proving REF. The smoothness REF is clear as long as MATH stays away from the branch points MATH. In case MATH hits such a branch point, one can use the local chart around MATH (with local coordinate MATH, MATH) to verify REF. |
nlin/0105021 | REF follows by considering the coefficient of MATH in MATH in REF which yields MATH . The constant MATH can be determined by considering the coefficient of the term MATH in REF, which results in MATH . |
nlin/0105021 | The existence of the asymptotic expansions of MATH in terms of the appropriate local coordinates MATH near MATH and MATH near MATH is clear from the explicit form of MATH in REF. Insertion of the polynomials MATH, MATH, and MATH into REF then, in principle, yields the explicit expansion coefficients in REF. However, a more efficient way to compute these coefficients consists in utilizing the NAME REF . Indeed, inserting the ansatz MATH into REF and comparing the leading powers of MATH immediately yields the first line in REF. Similarly, the ansatz MATH inserted into REF then yields the second line in REF. Finally, the ansatz MATH inserted into REF yields REF - REF then follow from REF, and REF. |
nlin/0105021 | Let MATH. Then, using MATH (compare REF) one obtains MATH using REF is just a special case of REF follows as in REF using REF. |
nlin/0105021 | First we temporarily assume that MATH where MATH is open. Since by REF, MATH, and MATH by hypothesis, one can apply REF to conclude that MATH is nonspecial. This argument is of course symmetric with respect to MATH and MATH. Thus, MATH is nonspecial if and only if MATH is. The representation REF for MATH, subject to REF, then follows by combining REF, and REF since MATH and MATH are nonspecial. The representation REF for MATH on MATH follows from the trace REF (taking MATH). By continuity, REF extend from MATH to MATH. Assuming MATH, MATH, in addition to REF, the constraint REF follows by combining REF, and REF is clear from REF. Again the extra REF can be removed by continuity and hence REF extend to MATH. |
nlin/0105021 | Given the solutions MATH, MATH of REF we introduce MATH on MATH. The NAME equations imply MATH . Thus MATH . Furthermore MATH, and hence there exists a polynomial MATH such that MATH . Computing the coefficient of the term MATH in REF one finds MATH . Next, one defines a polynomial MATH by MATH . Using REF, and REF one infers that indeed MATH has degree at most MATH. Multiplying REF by MATH, and replacing the term MATH with the result obtained upon differentiating REF with respect to MATH, yields MATH and hence MATH on MATH. Restricting MATH temporarily to MATH, where MATH one infers that MATH on MATH. Since MATH has degree at most MATH, one concludes MATH and hence REF, that is, MATH on MATH. Differentiating REF with respect to MATH and using REF one finds MATH on MATH. In order to extend these results to MATH one must prove that MATH does not pause once it hits MATH. Hence we suppose MATH for some MATH, MATH. Introducing MATH for some MATH in an open interval centered around MATH, the NAME REF for MATH becomes MATH . Thus, MATH does not pause at MATH for MATH in a small interval centered around MATH, and hence relations REF - REF extend to MATH. We have now established relations REF - REF on MATH, and one can now proceed as in REF to obtain REF. |
nlin/0105021 | REF - REF are proved as in REF . To prove REF one first observes that MATH using REF and relations REF - REF repeatedly. Thus, MATH where the left-hand side is meromorphic in a neighborhood of MATH, while the right-hand side is meromorphic near MATH only if MATH. This proves REF. Using REF one obtains MATH . Combining this result with REF one concludes that REF holds. |
nlin/0105021 | We prove REF by using REF which shows that MATH . However, the left-hand side of REF also equals MATH using REF, and REF. Combining REF proves REF. Similarly, to prove REF, we use REF to write MATH . Here the left-hand side can be expressed as MATH using REF, and REF. Combining REF, using REF, proves REF. Finally, REF follows by differentiating REF, that is, MATH, with respect to MATH, and using REF. |
nlin/0105021 | By REF, MATH is meromorphic on MATH away from the poles MATH of MATH and MATH of MATH. That MATH is meromorphic on MATH if MATH has only simple zeros is a consequence of (compare REF) MATH and MATH using REF, and REF. This follows from REF by restricting MATH to a sufficiently small neighborhood MATH of MATH such that MATH for all MATH and all MATH and by simultaneously restricting MATH to a sufficiently small neighborhood MATH of MATH such that MATH for all MATH and all MATH. By REF and the fact that MATH is meromorphic on MATH one concludes that MATH is meromorphic on MATH as well. Relations REF follow as in REF , while the time evolution REF is a consequence of the definition of MATH in REF as well as REF, rewriting MATH using REF. To prove REF we recall REF , that is, MATH using the calculation leading to REF show that MATH which inserted into REF yields REF. Evaluating REF at the points MATH and MATH and multiplying the resulting expressions yields REF. The remaining statements are direct consequences of REF - REF. |
nlin/0105021 | It suffices to prove REF since the argument for REF is analogous and that for REF has been given in the proof of REF . Inserting MATH into REF, observing REF, yields MATH . The rest is analogous to the proof of REF . |
nlin/0105021 | Let MATH. Since REF, and REF are proved as in in the stationary context of REF , we focus on the proofs of REF, and REF. Then, using REF, and REF, and REF one obtains MATH . REF is just a special case of REF follows as in REF using again REF. |
nlin/0105021 | First, let MATH be defined by requiring that MATH, MATH, are distinct and nonvanishing on MATH and MATH on MATH, MATH. The representation REF for MATH on MATH then follows by combining REF, and REF since MATH and MATH are simultaneously nonspecial as discussed in the proof of REF . The representation REF for MATH on MATH follows from the trace REF (taking MATH). By continuity, REF extend from MATH to MATH. The constraint REF then holds on MATH by combining REF - REF, and REF are clear from REF. Again by continuity, REF - REF extend from MATH to MATH. |
nlin/0105021 | Given solutions MATH, MATH of REF, we define polynomials MATH, MATH, and MATH on MATH as in the stationary case, compare REF , with properties MATH treating MATH as a parameter. Define polynomials MATH and MATH by MATH respectively. We claim that MATH . To prove REF we compute from REF that MATH . Using REF we see that REF is equivalent to MATH . REF is proved in REF . This in turn proves REF. Next, taking the derivative of REF with respect to MATH and inserting REF, yields MATH . On the other hand, by differentiating REF with respect to MATH, using REF we obtain MATH . Combining REF we conclude MATH . Next, we take the derivative of REF with respect to MATH and use REF for MATH and MATH, respectively, to obtain MATH . Finally, we compute MATH in two different ways. Differentiating REF with respect to MATH, using REF, and REF, one finds MATH . Differentiating REF with respect to MATH, using REF, results in MATH . Combining REF one concludes MATH which is equivalent to REF. |
nlin/0105021 | 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 and MATH, REF contradicts REF . Thus, MATH is nonspecial. |
nlin/0105021 | Dividing MATH by MATH (temporarily fixing the branch of MATH as MATH near infinity), one obtains MATH for some coefficients MATH to be determined next. Dividing REF by MATH, and inserting the expansion REF into the resulting equation then yields the recursion relation REF (with MATH replaced by MATH). More precisely, for MATH one originally obtains the relation MATH . Thus, MATH for some MATH, and hence the requirement MATH then yields MATH. The open sign of MATH has been chosen such that MATH. For MATH one obtains similarly MATH and hence, MATH for some MATH. Again the requirement MATH then yields MATH, MATH. Introducing MATH by REF with MATH, MATH, and MATH by REF, a straightforward computation shows that MATH . Hence, MATH for some constants MATH, MATH. Since MATH by inspection, we next proceed by induction on MATH and suppose that MATH . Thus, REF imply MATH where MATH denotes the expression on the right-hand side of REF in terms of MATH, MATH. Hence, MATH for some constants MATH. Since MATH, one concludes once more that MATH. Moreover, since MATH contains no constants by construction, one concludes MATH and hence MATH . Thus, we proved MATH and hence REF. A comparison of coefficients in REF then proves REF. Next, multiplying REF, a comparison of coefficients of MATH yields MATH . Thus, one computes MATH applying REF. Hence one obtains REF and thus REF. |
nlin/0105021 | By REF . Consider first the case MATH. Then MATH and hence MATH using REF. In the case where MATH we find (applying REF) MATH . Hence MATH using REF again. |
nlin/0105021 | It suffices to prove REF for the homogeneous case where MATH is replaced by MATH. Using MATH with the convention MATH one computes for MATH, MATH applying REF, and REF. For MATH one obtains from REF, and REF, MATH . |
nlin/0105021 | Let MATH be a nonspecial divisor on MATH, MATH. Introducing MATH we can render MATH single-valued on MATH where MATH denotes the union of cuts MATH with MATH (respectively, MATH) a cut connecting MATH and MATH (respectively, MATH) through the open interior MATH (that is, avoiding all curves MATH, MATH, with the exception of the point MATH), avoiding the points MATH, MATH. The left and right side of the cut MATH is denoted by MATH and MATH. The oriented boundary MATH of MATH, in obvious notation, is then given by MATH that is, it consists of MATH together with the piece from MATH to MATH along the left side of the cut MATH and then back to MATH along the right side of MATH, plus the corresponding pieces from MATH to MATH and back to MATH along the cut MATH, preserving orientation. Introducing the meromorphic differential, MATH the residue theorem applied to MATH yields MATH . Investigating separately the items occurring in REF then yields the following facts: MATH applying REF. Here MATH denotes the end point of MATH, MATH. In addition, the cut MATH produces the contribution MATH since (by an application of the residue theorem) MATH where MATH and MATH are on opposite sides of the cut MATH. Recalling the well-known results, MATH REF - REF imply MATH . This proves REF. In the following we will apply REF to MATH, MATH satisfying the first-order system REF on some open connected set MATH such that MATH, MATH, remain distinct on MATH and MATH on MATH, MATH. Using REF, and REF one computes MATH . Thus, REF imply MATH . We replaced MATH by MATH to arrive at REF using REF of MATH. If MATH, MATH, are distinct and MATH, MATH, we can choose MATH, MATH, and obtain REF. The general case where MATH is nonspecial, then follows from REF by continuity, choosing MATH such that there exists a sequence MATH with MATH as MATH. Finally, invoking the normal differential of the third kind in REF, MATH, corresponding to MATH, a simple computation, combining REF, and the normalization MATH, MATH, yields MATH . REF complete the proof of REF. |
quant-ph/0105036 | The trivial phase MATH is commutative. Let MATH be a phase observable with the matrix MATH. For any MATH and MATH, the map MATH is a complex measure. For any MATH, MATH . If MATH is commutative, then MATH for all MATH, so that MATH, for MATH, and MATH, for MATH. Let MATH, and let MATH be the smallest integer for which MATH. Then MATH. But this means that MATH for all MATH, that is, MATH. |
quant-ph/0105036 | For the phase observable MATH with the matrix MATH, let MATH and define MATH . Then MATH and therefore MATH for all MATH, so that MATH . |
quant-ph/0105036 | Consider a vector MATH. In view of REF , it remains to be shown that MATH implies MATH. For any MATH and MATH we define MATH . For a fixed MATH, the partial map MATH is a complex measure. For any MATH, MATH . Again, the map MATH is a complex measure, and we may carry out the integration MATH for all MATH. If MATH, then the value of the above integral is zero for all MATH. This implies MATH . On writing MATH these conditions are equivalent to: MATH . It remains to be shown that if MATH, then MATH for all MATH. Assume, therefore, that MATH for some MATH. Then MATH . Putting MATH, we get MATH for all MATH. If MATH, then MATH for all MATH. Assume next that MATH. Since MATH is strong, we get MATH for all MATH and MATH for all MATH, MATH. Further, for MATH and MATH, we get from the first of the above equalities that MATH. Thus it suffices to show that MATH. For that, assume the contrary: MATH. Fix MATH, and take MATH. Then MATH . Since MATH is strong, this equation is equivalent to the following: MATH . Therefore, if MATH, then also MATH, and MATH . We have thus shown that MATH for all MATH. But this is impossible since, by choosing MATH and MATH in REF , we have MATH. Therefore, MATH. |
quant-ph/0105036 | The minimal NAME dilation MATH of MATH in MATH is the canonical spectral measure MATH, with MATH acting as multiplication by the characteristic function MATH. The NAME space MATH is identified with the NAME space MATH in MATH. If MATH for some unit vector MATH and for some MATH for which MATH, then MATH vanishes on the complement set MATH which has positive measure. It follows from CITE that MATH is zero, which is a contradiction. |
quant-ph/0105036 | For the phase observable MATH with the phase matrix MATH, put MATH, for any unit vector MATH. Let MATH be a sequence of unit vectors such that MATH where MATH is the NAME measure concentrated on the point MATH. This implies that MATH for all MATH (see for example, CITE). Then MATH . The left hand side converges to REF for all MATH, and so MATH for all MATH if and only if MATH for all MATH. |
quant-ph/0105036 | We prove the first equality; the proof of the second is similar. Let MATH. There is a MATH such that MATH, whenever MATH. Since MATH, it follows that MATH if MATH. |
quant-ph/0105036 | By changing the values of MATH on a countable set if necessary, we may assume that MATH is left continuous. Denote MATH. If MATH, for MATH write MATH so that MATH. If MATH, then MATH and it is easily seen that MATH . Since the difference of the left and right extremes is MATH, we see that for any MATH, MATH can be chosen such that MATH and thus MATH . Let MATH be the two-dimensional NAME measure and denote MATH for MATH, and MATH. Then by the assumption we get MATH . Letting MATH we get the claim. |
quant-ph/0105036 | CASE: By REF we may choose MATH such that MATH and MATH whenever MATH. By REF we then have MATH, and by an analogous argument we also get MATH. Since MATH, there is MATH such that MATH whenever MATH. Thus MATH for all MATH. CASE: The NAME points of MATH form a set whose complement has measure zero (see CITE). Since MATH, there is such a point MATH. Using a translation, we may reduce the proof of this part to REF . |
quant-ph/0105036 | Suppose that MATH for all MATH, where MATH. The NAME coefficients of the probability measure MATH are of the form MATH and MATH for all MATH and MATH. Next we show that MATH for all MATH. Fix MATH and MATH. Since REF holds, one may choose such a MATH that MATH for all MATH. Since MATH, MATH, one gets MATH . Choose MATH such that MATH when MATH to get MATH for all MATH. Thus, MATH and MATH for all MATH when MATH. The condition MATH, MATH, implies that MATH for all MATH. |
quant-ph/0105079 | Define MATH for all MATH and MATH. Let MATH, and calculate MATH . If MATH for all MATH then MATH and, thus, MATH for all MATH. Conversely, if MATH, MATH, holds then MATH for all MATH and MATH. Therefore, MATH for all MATH. |
quant-ph/0105079 | Consider a sequence MATH of vectors of MATH and put MATH. If MATH is a sequence for which MATH for only finitely many MATH, then MATH the sums being finite. Suppose then that MATH is positive semidefinite. It follows that MATH, MATH, and MATH for all MATH. Especially, if MATH, then MATH. Then the doubles series MATH converges in the weak operator topology to a bounded and positive operator MATH. Let MATH be its square root and, for all MATH, MATH . Then, taking into account that MATH, one gets, for all MATH such that MATH, MATH . If MATH, for example MATH, then MATH and, for all MATH, MATH. |
quant-ph/0105079 | Fix MATH, and let MATH. Now MATH . Suppose that MATH and MATH are such that MATH. Then MATH, and we can define MATH so that MATH . On the other hand, for all MATH, MATH, and MATH . This shows that MATH, MATH. Since MATH, one has MATH. Thus, for all MATH, for which MATH, the number MATH is the same, and we may define MATH for all MATH and MATH. If MATH, MATH, MATH, REF gives MATH . Thus, if we define MATH we get MATH for all MATH and MATH. Let MATH. Now one gets MATH which implies that MATH. Thus the measure MATH is non-atomic, that is, MATH which implies that its distribution function MATH is continuous. From REF it follows that for all MATH, MATH, MATH . Since MATH is continuous, and the set MATH, MATH is dense in MATH, it follows that for all MATH . By the NAME extension theorem MATH for all MATH and MATH. |
quant-ph/0105079 | Denoting, in REF , MATH and MATH . REF follows. If MATH is self-adjoint, then from REF , one gets MATH for all MATH. Suppose that MATH is positive and, thus, self-adjoint. Hence, if REF does not hold, one may choose a MATH and a MATH such that MATH, and define a function MATH . Due to the continuity of MATH one can choose a MATH such that MATH. Thus, denoting MATH, MATH which contradicts the positivity of MATH. |
quant-ph/0105079 | Let MATH and MATH. Using the equations MATH and MATH, MATH, one gets MATH where the equality sign holds only when MATH for all MATH. On the other hand, if MATH, then MATH, MATH, for all MATH, since MATH is the structure matrix of MATH CITE. Define the following unitary transformations: MATH and MATH, where MATH, MATH. Now MATH is unitarily equivalent to the canonical spectral measure MATH, MATH, MATH, MATH, that is, MATH, and, thus, MATH is projection valued. |
cs/0106001 | From our preliminary work we have: MATH . But MATH and MATH, as MATH we get: MATH . The second equality is justified by the following fact: MATH . |
cs/0106001 | Let us consider MATH, MATH is the sum of indicator variables: MATH . By symmetry, MATH where MATH is any vector of weight MATH. For each MATH, consider the subset MATH of MATH formed by the indexes of the unit coordinates of MATH. Then MATH is in the kernel of MATH if and only if each row of the submatrix of MATH formed by the MATH columns whose indexes are in MATH has an even number MATH of unit coefficients. Thus, we have MATH possible rows for a matrix MATH in MATH such that MATH, and we deduce: MATH therefore, MATH . As MATH, the conclusion follows. |
cs/0106001 | Let us show how we get the leading term. Observe that: MATH where MATH is a polynomial of degree MATH. But, MATH . The study of the second term is left to the reader. |
cs/0106001 | From REF and from the mean value theorem we have when MATH: MATH with MATH lying between MATH and MATH. As MATH and MATH we get MATH. Thus, REF gives MATH which shows that MATH. Since we also have MATH, we can conclude that MATH are MATH. This leads to MATH and the rest of the proof is now pure routine. |
cs/0106001 | From REF and from the mean value theorem we get: MATH with MATH lying between MATH and MATH. Under the lemma's assumptions we get MATH. Thus, REF gives: MATH which shows that there exits MATH such that for sufficiently large MATH . As MATH we get MATH and with MATH, the first inequality leads to: MATH . |
cs/0106001 | The matrices we consider, in MATH, have three units by row. Let MATH be such a matrix. For each row MATH, MATH, of MATH there are three distinct indices MATH such that MATH if MATH or MATH, and MATH otherwise; let MATH denote the set MATH. So, to each row we can associate its MATH-weight, MATH defined as the following multiset: MATH . Now, let us introduce three random variables MATH, MATH and MATH. For any random matrix MATH, let MATH count the number of all-zero columns in MATHve that these columns do not contribute to the rank of MATH. Let MATH be defined by MATH . Observe that each row contributing to MATH underlines three columns such that only one of them contributes to the rank of MATH. Finally, let MATH be defined by MATH . Observe that each row contributing to MATH underlines three columns such that two of them contribute to the rank of MATH. Thus, we have MATH . Hence, MATH . Therefore, MATH . Observe that MATH . Thus, for any integer MATH, MATH . Hence, MATH . Thus REF gives for MATH . Let us estimate MATH. The random variables MATH and MATH are the sum of indicator variables: MATH where MATH iff the MATH-th column of MATH is all-zero; MATH where MATH iff there exists MATH such that MATH, and MATH. Now, MATH hence MATH and MATH therefore, MATH . Finally, let us estimate MATH. In the same way MATH where MATH iff there exists MATH such that MATH, MATH, MATH, and MATH. Let us introduce the random variables MATH such that MATH iff there exists MATH such that MATH, MATH and MATH. MATH . But, MATH . Therefore, MATH . And finally, MATH . Finally, REF follows from REF . |
cs/0106026 | The proof is straightforward by principle REF and the laws for equality and quantifiers having in mind the biconditional: MATH . |
cs/0106031 | CASE: For every node MATH, MATH. Hence, MATH. CASE: For every node MATH we define MATH as the maximum nesting of existential or universal restrictions in a concept in MATH. Obviously, MATH. Also, if MATH, then MATH. Hence, any path MATH in MATH induces a sequence MATH of non-negative integers. Since MATH is a tree rooted at MATH, the longest path starts with MATH and is bounded by MATH. CASE: Successors of a node MATH are only generated by an application of the MATH-rule, which generates at most one successor for each concept of the form MATH in MATH. Together with REF , this implies that the out-degree is bounded by MATH. |
cs/0106031 | A sequence of rule-applications induces a sequence of trees whose depth and out-degree is bounded by the size of the input concept by REF . Moreover, every rule application adds a concept to the label of a node or adds a node to the tree. No nodes are ever deleted from the tree and no concepts are ever deleted from the label of a node. Hence, an unbounded sequence of rule-applications would either lead to an unbounded number of nodes or to an unbounded label of one of the nodes. Both cases contradict REF . |
cs/0106031 | Termination is required as a precondition of the theorem. The tableau algorithm is sound because a complete and clash-free constraint system is satisfiable REF which implies satisfiability of the initial constraint system (either by REF and induction over the number of rule applications of directly by REF ') and hence (by REF ) the satisfiability of the input concept (or knowledge base). It is complete because, given a satisfiable input concept (or knowledge base), the initial constraint system is satisfiable REF . Each rule can be applied in a way that maintains the satisfiability of the constraint system REF and, since the algorithm terminates, any sequence of rule-applications is finite. Hence, after finitely many REF satisfiable and complete constraint system can be derived from the initial one. This constraint system must be clash-free because (by REF ) a clash would imply unsatisfiability. |
cs/0106031 | It is obvious that, for an arbitrary ABox MATH, the canonical interpretation satisfies all assertion of the form MATH. MATH does not contain any assertions of the form MATH. By induction on the structure of concepts occurring in MATH, we show that the canonical interpretation MATH satisfies any assertion of the form MATH and hence is a model of MATH. CASE: For the base case MATH with MATH, this holds by definition of MATH. CASE: For the case MATH, since MATH is clash free, MATH and hence MATH. CASE: If MATH, then, since MATH is complete, also MATH. By induction this implies MATH and MATH and hence MATH. CASE: If MATH, then, again due the completeness of MATH, either MATH or MATH. By induction this yields MATH or MATH and hence MATH. CASE: If MATH, then completeness yields MATH for some MATH. By construction of MATH, MATH holds and by induction we have MATH. Together this implies MATH. CASE: If MATH, then, for any MATH with MATH, MATH must hold due to the construction of MATH. Then, due to completeness, MATH must hold and induction yields MATH. Since this holds for any such MATH, MATH. |
cs/0106031 | CASE: Since MATH is a subset of MATH, satisfiability of MATH immediately implies satisfiability of MATH. CASE: Let MATH be a model of MATH. We distinguish the different rules: CASE: The application of the MATH-rule is triggered by an assertion MATH. Since MATH, also MATH. Hence, MATH is also a model for MATH. CASE: The MATH-rule is applied due to an assertion MATH. Since MATH is a model of MATH, there exists a MATH with MATH and MATH. Hence, the interpretation MATH, which maps MATH to MATH and behaves like MATH on all other names, is a model of MATH. Note, that this requires MATH to be fresh. CASE: The MATH-rule is applied due to an assertions MATH. Since MATH, MATH must hold. Hence, MATH is also a model of MATH. CASE: Again, let MATH be a model of MATH. If an assertion MATH triggers the application of the MATH-rule, then MATH must hold. Hence, at least for one of the possible choices for MATH, MATH holds. For this choice, adding MATH to MATH leads to an ABox that is satisfied by MATH. |
cs/0106031 | Termination was shown in REF . In REF , we have established the conditions required to apply REF , which yields correctness of the MATH-algorithm. |
cs/0106031 | Let MATH be the MATH-concept to be tested for satisfiability. We can assume MATH to be in NNF because transformation into NNF can be performed in linear time. REF sketches an implementation of the MATH-algorithm that uses the trace-technique to preserve memory and runs in polynomial space. The algorithm generates the constraint system in a depth-first manner: before generating any successors for an individual MATH, the MATH- and MATH-rule are applied exhaustively. Then successors are considered for every existential restriction in MATH one after another re-using space. This has the consequence that a clash involving an individual MATH must be present in MATH by the time generation of successors for MATH is initiated or will never occur. This also implies that it is safe to delete parts of the constraint system for a successor MATH as soon as the existence of a complete and clash-free ``sub" constraint system has been determined. Of course, it then has to be ensured that we do not consider the same existential restriction MATH more than once because this might lead to non-termination. Here, we do this using the set MATH that records which constraints still have to be considered. Hence, the algorithm is indeed an implementation of the MATH-algorithm. Space analysis of the algorithm is simple: since MATH is reset for every successor that is generated, this algorithm stores only a single path at any given time, which, by REF , can be done using polynomial space only. |
cs/0106031 | Satisfiability of MATH-concepts is known to be CITE, which is shown by reduction from the well-known NAME problem QBF CITE. REF together with the fact that MATH (NAME 's theorem CITE) yields the corresponding upper complexity bound. |
cs/0106031 | As mentioned before, MATH is a syntactic variant of the propositional modal logic MATH CITE. As a simple consequence of the proof of NAME of MATH with a universal modality CITE] (that is, in DL terms, a role linking every two individuals), one obtains that the global satisfaction problem for MATH is an NAME problem. The global satisfaction problem is defined as follows: Given a MATH-formula MATH, is there a NAME model MATH such that MATH holds at every world in MATH? Using the correspondence between MATH and MATH, this can be re-stated as an NAME problem for MATH: Given a MATH-concept MATH, is there an interpretation MATH such that MATH? Obviously, this holds iff the tautological concept MATH is satisfiable with respect to the (non-simple) NAME MATH, which implies that satisfiability of MATH-concepts (and hence of NAME) with respect to general NAME is NAME. |
cs/0106031 | This is an immediate consequence of the fact that MATH which can be shown as follows. Obviously, the set MATH contains MATH and is closed under the application of MATH (Note that, for a sub-concept MATH of a concept in NNF, MATH). Closure under sub-concepts for the concepts in MATH is also immediate, and can be established for MATH by considering the various possibilities for MATH-concepts. |
cs/0106031 | The linear bound on the length of a path in MATH is established as for the MATH-algorithm using the fact that the nesting of qualifying number restrictions strictly decreases along a path in MATH. Successors in MATH are only generated by the MATH-rule. For an individual MATH this rule will generate at most MATH successors for each MATH. There are at most MATH such concepts in MATH. Hence the out-degree of MATH is bounded by MATH, where MATH is a limit for the biggest number that may appear in MATH if binary coding is used. |
cs/0106031 | The sequence of rules induces a sequence of trees. The depth and the out-degree of these trees is bounded by some function in MATH by REF . For each individual MATH the label MATH is a subset of the finite set MATH. Each application of a rule either CASE: adds a new constraint of the form MATH and hence adds an element to MATH, or CASE: adds fresh individuals to MATH and hence adds additional nodes to the tree MATH. Since constraints are never deleted and individuals are never deleted or identified, an infinite sequence of rule application must either lead to an infinite number of nodes in the trees which contradicts their boundedness, or it leads to an infinite label of one of the nodes MATH which contradicts MATH. |
cs/0106031 | Let MATH be a complete and clash-free ABox generated by applications of the optimal rules and MATH its differentiation. We show that the canonical interpretation MATH, as defined in REF , is a model of MATH. By definition of MATH, all constraints of the form MATH are trivially satisfied. Also, MATH implies MATH by construction of MATH. Thus, all remaining assertions in MATH are of the form MATH and are also present in MATH. Thus, it is sufficient to show that MATH implies MATH, which we will do by induction on the norm MATH of a concept MATH. Note that, by the definition of MATH, MATH for every individual MATH that occurs in MATH. CASE: The first base case is MATH for MATH. MATH immediately implies MATH by the definition of MATH. The second base case is MATH. Since MATH is clash-free, this implies MATH and hence MATH. This implies MATH CASE: For the conjunction and disjunction of concepts this follows exactly as in the proof of REF . CASE: MATH implies MATH because otherwise the MATH-rule would be applicable and MATH would not be complete. By induction, we have MATH for each MATH with MATH. Hence MATH and thus MATH. CASE: MATH implies MATH because MATH is clash-free. Hence it is sufficient to show that MATH holds. On the contrary, assume MATH holds. Then there is an individual MATH such that MATH and MATH but MATH. The application of the MATH-rule is suspended until the propositional rules are no longer applicable to MATH and hence, by the time MATH is generated by an application of the MATH-rule, MATH contains the assertion MATH. Hence, the MATH-rule ensures MATH or MATH. Since we have assumed that MATH, this implies MATH and, by the induction hypothesis, MATH holds, which is a contradiction. |
cs/0106031 | MATH for any rule MATH implies MATH and, by the definition of MATH, MATH, hence, if MATH is satisfiable then so is MATH. For the other direction, the MATH- and MATH-rule can be handled as in the proof for MATH in REF . It remains to consider the MATH-rule. Let MATH be a model of MATH and let MATH be the constraint that triggers the application of the MATH-rule. Since the MATH-rule is applicable, we have MATH. We claim that there is a MATH with MATH . Before we prove this claim, we show how it can be used to finish the proof. The element MATH is used to ``select" a choice of the MATH-rule that preserves satisfiability: let MATH be an enumeration of the set MATH. We set MATH . Obviously, MATH, the interpretation that maps MATH to MATH and agrees with MATH on all other names, is a model for MATH, since MATH is a fresh individual and MATH satisfies MATH. The ABox MATH is a possible result of the application of the MATH-rule to MATH, which proves that the MATH-rule can indeed be applied in a way that maintains satisfiability of the ABox. We will now come back to the claim. It is obvious that there is a MATH with MATH and MATH that is not contained in MATH, because MATH. Yet MATH might appear as the image of an individual MATH such that MATH but MATH. Now, MATH and MATH implies MATH. This is due to the fact that the constraint MATH must have been generated by an application of the MATH-rule because it has not been an element of the initial ABox. The application of this rule was suspended until neither the MATH- nor the MATH-rule were applicable to MATH. Hence, if MATH is an element of MATH now, then it has already been in MATH when the MATH-rule that generated MATH was applied. The MATH-rule guarantees that either MATH or MATH is added to MATH, hence MATH. This is a contradiction to MATH because under the assumption that MATH is a model of MATH this would imply MATH while we initially assumed MATH. |
cs/0106031 | Let MATH be a MATH-concept to be tested for satisfiability. We can assume MATH to be in NNF because the transformation of a concept to NNF can be performed in linear time. The key idea for the NAME implementation is the trace technique CITE we have already used for the MATH-algorithm in REF, and which is based on the fact that it is sufficient to keep only a single path (a trace) of MATH in memory at a given MATH is generated in a depth-first manner. This idea has been the key to a NAME upper bound for MATH and MATH in CITE. To do this we need to store the values for MATH for each individual MATH in the path, each MATH that appears in MATH, and each MATH. By storing these values in binary form, we are able to keep information about exponentially many successors in memory while storing only a single path at a given stage. Consider the algorithm in REF , where MATH denotes the set of role names that appear in MATH. It re-uses the space needed to check the satisfiability of a successor MATH of MATH once the existence of a complete and clash-free ``subtree" for the constraints on MATH has been established. This is admissible since, as was the case for MATH, the optimal rules will never modify this subtree once it is completed. Constraints in this subtree also have no influence on the completeness or the existence of a clash in the rest of the tree, with the exception that constraints of the form MATH for MATH-successors MATH of MATH contribute to the value of MATH. These numbers play a role both in the definition of a clash and for the applicability of the MATH-rule. Hence, in order to re-use the space occupied by the subtree for MATH, it is necessary and sufficient to store these numbers. Let us examine the space usage of this algorithm. Let MATH. The algorithm is designed to keep only a single path of MATH in memory at a given stage. For each individual MATH on a path, constraints of the form MATH have to be stored for concepts MATH. The size of MATH is bounded by MATH and hence the constraints for a single individual can be stored in MATH bits. For each individual, there are at most MATH counters to be stored. The numbers to be stored in these counters do not exceed the out-degree of MATH, which, by REF , is bounded by MATH. Hence each counter can be stored using MATH bits when binary coding is used to represent the counters, and all counters for a single individual require MATH bits. Due to REF , the length of a path is limited by MATH, which yields an overall memory consumption of MATH. |
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