paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
cs/0106031 | 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 individual 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 CASE: adds fresh individuals to MATH and hence adds additional nodes to the tree MATH, or CASE: adds a constraint to a node MATH and deletes all subtrees rooted below MATH. Assume that algorithm does not terminate. Due to the mentioned facts this can only be because of an infinite number of deletions of subtrees. Each node can of course only be deleted once, but the successors of a single node may be deleted several times. The root of the constraint system cannot be deleted because it has no predecessor. Hence there are nodes that are never deleted. Choose one of these nodes MATH with maximum distance from the root, that is, which has a maximum number of ancestors in MATH. Suppose that MATH's successors are deleted only finitely many times. This can not be the case because, after the last deletion of MATH's successors, the ``new" successors were never deleted and thus MATH would not have maximum distance from the root. Hence MATH triggers the deletion of its successors infinitely many times. However, the MATH-rule is the only rule that leads to a deletion, and it simultaneously leads to an increase of MATH, namely by the missing concept which caused the deletion of MATH's successors. This implies the existence of an infinitely increasing chain of subsets of MATH, which is clearly impossible. |
cs/0106031 | Assume that MATH occur in MATH. Then they also occur in all intermediate NAME because, once an individual is deleted from the constraint system, it is never re-introduced. The proof is by induction on the number of rule applications necessary to derive MATH from MATH. If no rule must be applied, then MATH holds, and since application of the MATH-rule to MATH does not alter the concepts asserted for MATH, we are done. Now assume that the lemma holds for every ABox MATH derivable from MATH by MATH rule applications. Let MATH be derivable from MATH in MATH steps and let MATH be an ABox such that MATH. Since MATH holds by induction, also MATH holds as long as the rule application that derives MATH from MATH does not alter the concepts asserted for MATH. The MATH-rule does not alter the constraints for any individual that is already present in the ABox because it introduces a fresh individual. The MATH- or MATH-rule cannot be applicable to MATH because, if the rule is applicable in MATH, then, since MATH, it is also applicable in MATH and the MATH-rule that creates MATH is not applicable. Assume that an application of the MATH-rule asserts an additional concept for MATH. Any application of the MATH-rule that adds a constraint for MATH removes the individuals MATH from MATH. This includes MATH and hence MATH would not occur in MATH, in contradiction to the assumption that MATH occur in MATH. Since the concept assertions for MATH have not changed since the generation of MATH, it holds that MATH iff MATH and so MATH is ensured by the MATH-rule that creates MATH. The individual MATH still occurs in MATH and hence MATH holds, which implies that the MATH-rule cannot be applied for the constraint MATH in a way that adds MATH or MATH to MATH. |
cs/0106031 | Let MATH be a complete and clash-free ABox obtained by a sequence of rule applications starting from MATH. We show that the canonical interpretation MATH (as defined in REF ) is indeed a model of MATH that satisfies MATH. Please note that we need the condition ``MATH iff MATH", which is maintained by the algorithm, to make sure that all information from the ABox is reflected in the canonical interpretation. Every canonical interpretation trivially satisfies MATH and also every two different individuals are interpreted differently, which takes care of the additional assertions in MATH. So, it remains to show that MATH implies MATH for all individuals MATH in MATH and all concepts MATH. This is done by induction over the norm of concepts MATH. The only interesting cases that are different from the MATH-case are the qualifying number restrictions. CASE: MATH implies MATH because MATH is complete. Hence, there are MATH distinct individuals MATH with MATH and MATH for each MATH. By induction and REF , we have MATH and MATH and hence MATH. CASE: MATH implies, for any MATH that occurs in MATH and any MATH with MATH, MATH or MATH. For any predecessor of MATH, this is guaranteed by the MATH-rule. For any successor, this follows from REF . Hence, MATH is present in MATH by the time MATH is generated and the MATH-rule ensures MATH or MATH. We show that MATH: assume MATH. This implies the existence of some MATH with MATH and MATH but MATH. Due to the syntactic restriction on role expressions, MATH implies MATH for some MATH that occurs in MATH and and hence MATH must hold by construction of MATH. The MATH-rule and the MATH-rule then guarantee that MATH implies MATH. By induction this yields MATH in contradiction to MATH. |
cs/0106031 | Let MATH be a model of MATH that satisfies MATH, as required by our notion of satisfiability. We distinguish the different rules. For most rules MATH can remain unchanged, in all other cases we explicitly state how MATH must be modified in order to witness the satisfiability of the modified ABox. CASE: The MATH-rule: if MATH, then MATH. This implies MATH for MATH, and hence satisfiability is preserved. CASE: The MATH-rule: if MATH, then MATH. This implies MATH or MATH. Hence the MATH-rule can add a constraint MATH with MATH and maintains satisfiability. CASE: The MATH-rule: obviously, either MATH or MATH for any individual MATH in MATH. Hence, the rule can always be applied in a way that maintains satisfiability. Deletion of constraints as performed by the MATH-rule cannot cause unsatisfiability. CASE: The MATH-rule: if MATH, then MATH. This implies MATH. We claim that there is an element MATH such that MATH . We will prove this claim later. Let MATH be an enumeration of the set MATH. The MATH-rule can add the constraints MATH as well as MATH to MATH. If we set MATH, then MATH is a model of the differentiation of the ABox obtained this way that satisfies MATH. Why does there exists an element MATH that satisfies MATH? Let MATH be an individual with MATH and MATH that appears as an image of an arbitrary element MATH with MATH for some MATH. The requirement MATH implies that MATH and also MATH must hold. This can be shown as follows: Assume MATH. This implies MATH: either MATH, then in order for the MATH-rule to be applicable, no non-generating rules and especially the MATH-rule is not applicable to MATH and its ancestor, which implies MATH. If not MATH, then MATH must have been generated by an application of the MATH-rule to MATH. REF implies that at the time of the generation of MATH already MATH held and hence the MATH-rule ensures MATH. In any case MATH holds, which implies MATH, in contradiction to MATH. Together this implies that, whenever an element MATH with MATH and MATH is assigned to an individual MATH with MATH, then it must be assigned to an individual that contributes to MATH. Since the MATH-rule is applicable, there are less than MATH such individuals and hence there must be an unassigned element MATH as required by MATH. |
cs/0106031 | From REF , it follows that MATH is satisfiable and every model of MATH is also a model of MATH because MATH and MATH contains no role assertions, which implies MATH and every interpretation trivially satisfies MATH for MATH. |
cs/0106031 | Termination has been shown in REF . As mentioned before, REF are trivially satisfied due to the chosen notion of ABox satisfiability. REF has been shown in REF ' in REF . |
cs/0106031 | Consider the algorithm in REF , where MATH denotes all role expressions that occur in the input concept MATH. Like the algorithm for MATH, the MATH-algorithm re-uses the space used to check for the existence of a complete and clash-free ``subtree" for each successor MATH of an individual MATH and keeps only a single path in memory at one time. Counter variables are used to keep track of the values MATH for all MATH and MATH. Resetting a node and restarting the generation of its successors is achieved by jumping to the label MATH in the algorithm, which re-initializes all successor counters for a node MATH. Note, how the predecessor of a node is taken into account when initializing the counter individuals. Since MATH is a tree, every newly generated node has a uniquely determined predecessor and since only safe role expressions occur in MATH, it is sufficient to take only this predecessor node into account when initializing the counter. Let MATH. For every node MATH of a path in MATH, MATH bits suffice to store the constraints of the form MATH and MATH suffice to store the counters (in binary representation) because MATH, MATH, and the out-degree of MATH is bounded by MATH (by REF , which also holds for MATH). Also by REF , the length of a path in MATH is bounded by MATH, which yields an overall memory requirement of MATH for a path. |
cs/0106031 | The DL MATH is a syntactic restriction of the DL MATH, where we do not allow for inverse roles and in number restrictions MATH, MATH must be a conjunction of positive roles and MATH the tautological concept MATH. Hence, the MATH-algorithm can immediately be applied to MATH-concepts, which yields decidability in NAME. |
cs/0106031 | The cardinality of MATH is bounded by MATH. The cardinality of MATH is bounded by MATH and hence MATH. Finally, the cardinality of MATH is bounded by MATH, where MATH is an upper bound for MATH if numbers are coded binarily in the input. Summing up, we get MATH as a bound for MATH and MATH as a bound for MATH, which dominates MATH. |
cs/0106031 | Assume MATH, MATH is a tree accepted by MATH, and MATH is an arbitrary run of MATH on MATH with MATH. From MATH, we will construct a model MATH for MATH and MATH, which proves satisfiability of MATH with respect to MATH. For every path MATH with MATH, we define MATH, and MATH. The domain MATH of MATH is defined by MATH. Hence, MATH contains only ``right successors" and the root. For concept names MATH, we define MATH . For the interpretation of roles, we define MATH . Before we prove that MATH is indeed a model for MATH and MATH, we state some general properties of the automaton and this construction. CASE: Due to the construction of MATH, for every MATH, MATH and hence MATH. CASE: ``Once MATH, always MATH." For a path MATH, if MATH, then, for all MATH with MATH, MATH and MATH. CASE: ``A left successor is either MATH or an auxiliary state, in which case it is labelled with the same set from MATH." For a path MATH, if MATH, then, for all MATH, if MATH then MATH is of the form MATH. CASE: ``MATH and MATH are a lower an upper bounds on the number of successors of a node." For a path MATH with MATH, MATH, and MATH, MATH . This property is less obvious than the others and we give a proof by induction on MATH . If MATH, then MATH for all MATH and MATH and hence, by REF , for all ancestors MATH of MATH, MATH and MATH. Thus MATH holds for all MATH and MATH. If MATH then there is a MATH and a MATH with MATH, MATH, and MATH. Hence, MATH and MATH by REF and we can use the induction hypothesis for MATH. For all MATH and MATH, MATH where the steps marked with MATH use the induction hypothesis. This is what we needed to show. CASE: For two paths MATH and a role expression MATH, if MATH then MATH or MATH. Because of the syntactic restriction to safe role expressions in MATH, for MATH to hold there must be a role MATH such that MATH. By construction of MATH, this can only be the case if MATH or MATH. CASE: For two paths MATH with MATH and a role MATH, MATH iff MATH and MATH iff MATH. For every MATH, MATH iff MATH holds as follows. For a (non-inverse) role MATH, immediately by the construction of MATH, MATH iff MATH. For an inverse role MATH with MATH, MATH iff MATH iff MATH. Hence, MATH iff MATH. Similarly, for every MATH, MATH iff MATH, and hence MATH iff MATH. Using these properties we can now show: For all MATH and MATH, MATH. The proof is by induction on the norm MATH of the concepts (as defined REF ). The base cases are MATH or MATH for a concept name MATH. For MATH this is immediate by the definition of MATH. For the case MATH, since MATH, MATH implies MATH and hence MATH. For the induction step, we distinguish the different concept operators of MATH. CASE: If MATH then, since MATH, also MATH. Hence, by induction, MATH, MATH and thus MATH. CASE: The case MATH is similar to the previous one. CASE: Now assume MATH. For every MATH, MATH and hence MATH iff MATH. Since MATH, by induction, MATH iff MATH holds for every MATH. If MATH is the root of MATH then, by REF , MATH and hence, by REF , MATH . If MATH, then MATH is a ``right successor". Let MATH be the unique path in MATH with MATH, that is, MATH's ``predecessor" in MATH. MATH . If MATH or MATH, then MATH and MATH holds because of induction, REF , and REF . If MATH and MATH, then MATH and MATH again holds by REF , and REF . If MATH then MATH and hence MATH. If MATH then MATH and hence MATH. This finishes the proof of the claim, which yields the only-if direction of the lemma: if MATH then there exists a tree MATH and a corresponding interpretation MATH that satisfies the claim. Since MATH, MATH and hence MATH. Also, for every MATH and every MATH, MATH. Hence MATH and MATH. For the if-direction, let MATH be satisfiable with respect to MATH and MATH a model of MATH with MATH. We construct a tree MATH from MATH that is accepted by MATH. To this purpose, we define a function MATH and maintain an agenda of paths MATH whose successors still need consideration. Let MATH be an arbitrary element such that MATH. Set MATH and MATH with MATH. Initialize the agenda with MATH. Pick the first element MATH off the agenda. For MATH, let MATH and let MATH be a set such that CASE: MATH . CASE: For every MATH there are MATH with MATH, MATH for MATH and MATH for MATH . CASE: MATH is minimal with respect to set cardinality with these properties. Such a set MATH exists, is finite, possibly empty, and not necessarily uniquely defined. Let MATH be an enumeration of MATH. CASE: For every MATH, we set MATH and MATH. CASE: For every MATH, we set MATH and MATH . Put MATH at the end of the agenda. CASE: Finally, for all MATH we define MATH and MATH. REF illustrates this construction. Continuing this process until the agenda runs empty (or indefinitely if it never does) eventually defines MATH for every MATH (since the agenda is organised as a queue, every element will eventually be taken off the agenda). The proof that MATH (and hence MATH) is relatively simple and omitted here. |
cs/0106031 | NAME was established in REF . By REF , generating MATH and testing MATH for emptiness decides satisfiability of MATH with respect to MATH. Due to REF this can be done in time exponential in MATH. |
cs/0106031 | NAME is immediate from REF . It remains to show that these problems can be decided in exponential time. Let MATH be a MATH-knowledge base, MATH the set of roles that occur in MATH together with their inverse, and MATH defined as above. Let MATH. Only NAME MATH with no more individuals than MATH are candidates for pre-completions because the mapping MATH must be surjective. The number of individuals in MATH is bounded by MATH. For an ABox MATH with MATH individuals, concept assertions ranging over MATH, and role assertions ranging over MATH, there are at most MATH different possibilities, and each such ABox contains at most MATH assertions. For an ABox MATH with MATH individuals there are at most MATH different possibilities of mapping the individuals from MATH (of which there are at most MATH many) into the MATH individuals of MATH. Given a fixed MATH and a fixed mapping MATH, testing whether the requirement of REF are satisfied can be done in polynomial time in MATH and hence certainly in time MATH. Summing up, it is possible to enumerate all potential pre-completions of MATH, generate all possible mappings MATH, and test whether all requirements from REF are satisfied in time bounded in MATH . Due to REF , MATH is satisfiable iff this enumeration yields a pre-completion MATH such that MATH is satisfiable with respect to MATH for every MATH that occurs in MATH. Since all candidate pre-completions MATH from the enumeration contain at most MATH assertions, this can be checked for in time exponential in MATH for every candidate pre-completion MATH. This yields an overall decision procedure that runs in time exponentially bounded in MATH. Instance checking is at least as hard as concept satisfiability with respect to general NAME and not harder than knowledge base satisfiability, hence NAME of instance checking for MATH is immediate from what we have just proved. |
cs/0106031 | Let MATH be a MATH-knowledge base, where the individuals in the ABox coincide with the individuals of MATH. The ABox MATH is transformed into a NAME as follows. We define MATH . MATH is satisfiable iff MATH is satisfiable. If MATH is satisfiable with MATH, it is easy to verify that MATH, which is obtained from MATH by setting MATH and preserving the interpretation of the concept and role names, is a model for MATH. Conversely, any model MATH of MATH can be turned into a model MATH of MATH by setting, for every individual MATH, MATH for the unique MATH and preserving the interpretation of concept and role names. |
cs/0106031 | It is obvious that the cardinality restrictions MATH and MATH enforce the interpretation of a concept name MATH to be a singleton, which can now serve as a substitute for a nominal. Also, an interpretation satisfies a general axiom MATH iff it satisfies MATH. In this manner, every nominal can be replaced by a concept and every general axiom by a cardinality restriction, which yields the reduction from reasoning with nominals and NAME to reasoning with cardinality restrictions. For the converse direction, the reduction works as follows. Let MATH be a MATH-CBox. Without loss of generality, we assume that MATH contains no cardinality restriction of the form MATH because these are trivially satisfied by any interpretation. The translation of MATH, denoted by MATH, is the MATH-TBox defined by: MATH where MATH is defined depending on whether MATH or MATH. MATH where MATH are fresh and distinct nominals and we use the convention that the empty disjunction is interpreted as MATH to deal with the case MATH. Assuming unary coding of numbers, the translation of a NAME MATH is obviously computable in polynomial time. MATH is satisfiable iff MATH is satisfiable. If MATH is satisfiable then there is a model MATH of MATH and MATH for each MATH. We show how to construct a model MATH of MATH from MATH. MATH will be identical to MATH in every respect except for the interpretation of the nominals MATH (which do not appear in MATH). If MATH, then MATH implies MATH. If MATH, then we have not introduced new nominals, and MATH contains MATH. Otherwise, we define MATH such that MATH. This implies MATH and hence, in either case, MATH. If MATH, then MATH must hold, and MATH implies MATH. Let MATH be MATH distinct elements from MATH with MATH. We set MATH. Since we have chosen distinct individuals to interpret different nominals, we have MATH for every MATH. Moreover, MATH implies MATH and hence MATH. We have chosen distinct nominals for every cardinality restrictions, hence the previous construction is well-defined and, since MATH satisfies MATH for every MATH, MATH. For the converse direction, let MATH be a model of MATH. The fact that MATH (and hence the satisfiability of MATH) can be shown as follows: let MATH be an arbitrary cardinality restriction in MATH. If MATH and MATH, then we have MATH and, since MATH, we have MATH and MATH. If MATH and MATH, we have MATH. From MATH follows MATH. If MATH, then we have MATH. From the first set of axioms we get MATH. From the second set of axioms we get that, for every MATH, MATH. This implies that MATH. |
cs/0106031 | REF states that satisfiability of MATH-TBoxes and satisfiability of MATH-CBoxes are mutually polynomially reducible problems. Hence, both the lower and the upper complexity bound follow from REF . |
cs/0106031 | A simple inspection of the reduction used to prove REF , and especially of the proof of REF shows that only a single nominal, which marks the upper right corner of the torus, is sufficient to perform the reduction. If MATH is an individual name and MATH is a role name, then the following NAME defines a torus of exponential size: MATH . Since this reduction uses only a single individual name, the unique name assumption is irrelevant in this case. |
cs/0106031 | For the if-direction let MATH be a model of MATH with MATH. This implies MATH. Let MATH be defined by MATH and MATH. For every MATH and every MATH-concept MATH, we have MATH iff MATH. We proof this claim by induction on the structure of MATH. The only interesting case is MATH. In this case MATH. We have MATH where the equivalence MATH holds because, if MATH then MATH and MATH by induction. Also, if MATH, then MATH iff MATH and hence the sets MATH and MATH are equal. By construction, for every MATH, MATH. Due to REF , this implies MATH and hence MATH. For the only-if-direction, let MATH be an interpretation with MATH. We pick an arbitrary element MATH and define an extension MATH of MATH by setting MATH and MATH. Since MATH and MATH do not occur in MATH, we still have that MATH. For every MATH and every MATH-concept MATH that does not contain MATH or MATH, MATH iff MATH. Again, this claim is proved by induction on the structure of concepts and the only interesting case is MATH. MATH . The equivalence MATH holds because, by construction of MATH, MATH holds for every element MATH of the domain and MATH iff MATH holds by induction. Since, MATH, REF yields that MATH and hence MATH . |
cs/0106031 | From REF , we get that the function mapping a MATH-TBox MATH to MATH is a reduction from satisfiability of MATH-TBoxes to satisfiability of MATH concepts. From REF we know that the former problem is NAME. Obviously, MATH can be computed from MATH in polynomial time. Hence, the lower complexity bound transfers. The NAME upper bound is a consequence of REF and the fact that a MATH concept MATH is satisfiable iff, for an individual MATH that does not occur in MATH, the NAME MATH is satisfiable. |
cs/0106031 | For a role expression MATH, we define MATH inductively by MATH and set MATH. This translation is obviously polynomial and satisfies, for every interpretation MATH and concept MATH, MATH . Hence, a concept MATH is satisfiable iff MATH is satisfiable. NAME can be reduced to MATH as shown in REF . This yields the desired reductions. |
cs/0106031 | The lower bound is immediate from REF because the set of MATH-concepts is strictly included in the set of MATH- and MATH-concepts. In the case of unary coding of numbers in the input, the upper bound follows from REF . |
cs/0106031 | Let MATH be a MATH-CBox and MATH a role that does not occur in MATH. We transform MATH into a MATH concept MATH by setting MATH . MATH is satisfiable iff MATH is satisfiable. Let MATH be a model for MATH . We define a model MATH of MATH by setting MATH and preserving the interpretation of all other names. Since MATH does not occur in MATH, MATH holds for every MATH. Since MATH is interpreted by the universal relation, MATH holds. Also, again since MATH is the universal relation, for every MATH, MATH. Thus, if MATH, then MATH. Hence, from MATH is follows that MATH, which proves its satisfiability. For the converse direction, if MATH is satisfiable with MATH for an interpretation MATH, then, since MATH, MATH must hold and hence MATH. It immediately follows that MATH. Obviously, the size of MATH is linear in the size of MATH, which proves this lemma. |
cs/0106031 | Concept satisfiability for MATH is NAME by REF , it can be decided in NAME by REF . For MATH the situation is slightly more complicated because REF yields NAME only for binary coding of numbers. Yet, CITE show that concept satisfiability even for MATH, that is, MATH extended with Boolean role expressions, is NAME, which yields the lower bound also for the case of unary coding of numbers. The matching upper bound (in the case of unary coding) again follows from REF . |
cs/0106031 | For the if-direction, if MATH is a tableau for MATH with MATH, a model MATH of MATH can be defined as follows: MATH where MATH denotes the transitive closure of MATH. Transitive roles are interpreted by transitive relations by definition. By induction on the structure of concepts, we show that, if MATH, then MATH. This implies MATH because MATH. Let MATH: CASE: If MATH is a concept name, then MATH by definition. CASE: If MATH for MATH then MATH (due to REF ), so MATH. CASE: If MATH, then, due to REF , MATH and MATH, and hence, by induction, MATH and MATH. Thus, MATH. CASE: The case MATH is analogous to the previous one. CASE: If MATH, then, due to REF , there is some MATH such that MATH and MATH. By definition of MATH, MATH holds as follows. It is immediate, if MATH. If MATH for MATH, then MATH implies MATH by REF . Hence, MATH and MATH holds. By induction, MATH and hence MATH. CASE: If MATH and MATH, then either CASE: MATH and MATH (due to REF ), or CASE: MATH. Due to MATH, this can only be the case if MATH is transitive and there exists a path of length MATH such that MATH. Due to REF , MATH for all MATH, and we have MATH, again due to REF . In both cases, by induction MATH holds, and hence MATH. For the converse direction, if MATH is a model of MATH, then a tableau MATH for MATH can be defined by: MATH . It remains to demonstrate that MATH is a tableau for MATH: CASE: MATH satisfies REF - REF as a direct consequence of the semantics of MATH concepts and of inverse roles. CASE: If MATH, MATH and MATH, then MATH unless there is some MATH such that MATH and MATH. However, if MATH, MATH and MATH, then MATH, which would imply MATH. MATH therefore satisfies REF . |
cs/0106031 | Obviously, MATH holds for the only node MATH of the initial tree. Subsequently, whenever a concept MATH is added to MATH by an application of one of the rules, then it is always also added to MATH. |
cs/0106031 | Let MATH. Obviously, MATH is linear in the length of MATH. Termination is a consequence of the following properties of the expansion rules: CASE: The expansion rules never remove concepts from node labels. CASE: Successors are only generated for concepts of the form MATH and, for any node, each of these concepts triggers the generation of at most one successor. Since MATH contains at most MATH concepts of the form MATH, the out-degree of the tree is bounded by MATH. CASE: Nodes are labelled with nonempty subsets of MATH. If a path MATH is of length MATH, then there are REF nodes MATH on MATH, with MATH and MATH, and blocking occurs. Since a path on which nodes are blocked cannot become longer, paths are of length at most MATH. An infinite run of the completion algorithm can thus only occur due to an infinite number of deletions of nodes of the tree. That this can never happen can be shown in exactly the same way this has been done in the proof of REF . |
cs/0106031 | Let MATH be the complete and clash-free completion tree constructed by the tableau algorithm for MATH. A tableau MATH can be defined by MATH . We now show that MATH is a tableau for MATH. By definition of MATH, we have MATH for all MATH, so it is sufficient to establish the required properties for the function MATH. CASE: MATH for the root MATH of MATH and, since MATH has no predecessors, it cannot be blocked. Hence MATH for some MATH and REF holds. CASE: REF is satisfied because MATH is clash-free. CASE: REF are satisfied because neither the MATH-rule nor the MATH-rule apply to any MATH. Hence MATH satisfies the required properties. CASE: REF is satisfied because, for all MATH, if MATH, then the MATH-rule ensures that there is either: CASE: a MATH-predecessor MATH with MATH (see REF ). Because MATH is a predecessor of MATH (which is an unblocked node), it cannot be blocked, so MATH and MATH. CASE: a MATH-successor MATH with MATH (again, see REF ). If MATH is not blocked, then MATH and MATH. Otherwise, MATH is blocked by some MATH with MATH. Hence MATH, MATH and MATH. CASE: To show that REF is satisfied for all MATH, if MATH and MATH, we have to consider three possible cases: CASE: MATH is a MATH-neighbour of MATH. The MATH-rule guarantees MATH. CASE: MATH, MATH blocks MATH. Then by the MATH-rule we have MATH and, by the definition of blocking, MATH. Hence MATH. CASE: MATH, MATH blocks MATH. From the definition of blocking we have that MATH. Hence MATH and the MATH-rule guarantees MATH. In all three cases, MATH holds. CASE: For REF , let MATH with MATH, MATH, and MATH. There are three possible cases: CASE: MATH is a MATH-neighbour of MATH. The MATH-rule guarantees MATH. CASE: MATH, MATH blocks MATH. Then, by the MATH-rule, we have MATH and, by the definition of blocking, MATH. Hence MATH. CASE: MATH, MATH blocks MATH. From the definition of blocking, we have that MATH. Hence MATH and the MATH-rule guarantees MATH. CASE: REF is satisfied because, for each MATH, either: CASE: MATH is a MATH-neighbour of MATH, so MATH is a MATH-neighbour of MATH and MATH. CASE: MATH and MATH blocks MATH, so MATH and MATH. CASE: MATH and MATH blocks MATH, so MATH. |
cs/0106031 | Let MATH be a tableau for MATH. Using MATH, we guide the application of the non-deterministic MATH-rule such that the algorithm yields a completion tree MATH that is both complete and clash-free. The algorithm starts with the initial tree MATH consisting of a single node MATH, the root, with MATH. MATH is a tableau, hence there is some MATH with MATH. When applying the expansion rules to MATH, the application of the non-deterministic MATH-rule is guided by the labelling in the tableau MATH. We will expand MATH in such a way that the following invariant holds: there exists a function MATH that maps the nodes of MATH to elements of MATH such that MATH . If MATH holds for a completion tree MATH and a rule is applicable to MATH, then it can be applied in a way that maintains MATH. We have to distinguish the different rules: CASE: If the MATH-rule can be applied to MATH in MATH with MATH, then MATH are added to MATH. Since MATH is a tableau, MATH, and hence the MATH-rule preserves MATH. CASE: If the MATH-rule can be applied to MATH in MATH with MATH, then there is a MATH such that MATH, and the MATH-rule can add MATH to MATH. Hence the MATH-rule can be applied in a way that preserves MATH. CASE: If the MATH-rule can be applied to MATH in MATH with MATH, then MATH and there is some MATH with MATH and MATH. The MATH-rule creates a new successor MATH of MATH and we extend MATH by setting MATH, that is, MATH is the extension of MATH that maps MATH to MATH. It is easy to see that the extended completion tree together with the function MATH satisfy MATH. CASE: If the MATH-rule can be applied to MATH in MATH with MATH and MATH is a MATH-neighbour of MATH, then MATH, and thus MATH. The MATH-rule adds MATH to MATH and thus preserves MATH. The deletion of nodes can never violate MATH. CASE: If the MATH-rule can be applied to MATH in MATH with MATH, MATH, and MATH a MATH-neighbour of MATH, then MATH, and thus MATH. The MATH-rule adds MATH to MATH and thus preserves MATH. The deletion of nodes can never violate MATH. From this claim, the lemma can be derived as follows. It is obvious that the initial tree satisfies MATH: since MATH is a tableau for MATH, there exists an element MATH with MATH and hence the function MATH that maps MATH to MATH satisfies the required properties. The claim states that, whenever a rule is applicable, it can be applied in a way that preserves MATH. Obviously, no completion tree that satisfies MATH contains a clash as this would contradict REF . Moreover, from REF , we have that the expansion process terminates and thus must eventually yield a complete and clash-free completion tree. |
cs/0106031 | REF is an immediate consequence of REF , and REF. Moreover, since MATH is closed under negation, subsumption of concepts MATH can be reduced to the (un-)satisfiability of MATH. |
cs/0106031 | For each node MATH of the completion tree, MATH only contains two kinds of concepts: the concept that triggered the generation of the node MATH, denoted by MATH, and concepts which were propagated down the completion tree by the first alternative of the MATH- or MATH-rules. Moreover, MATH holds for any node in the completion tree. Firstly, consider the elements of MATH for MATH. Let MATH denote the concept that caused the generation of the node MATH. Then MATH contains only concepts which have been inserted using the MATH- or the MATH-rule. Let MATH. Then either MATH and the MATH-rule makes sure that MATH, or MATH is already of the from MATH and has been inserted into MATH by an application of the MATH-rule to MATH. In both cases, it follows that the MATH- or the MATH-rule yield MATH. Hence we have MATH and, since we have MATH choices for MATH, MATH . Secondly, consider MATH. Again, the MATH- and the MATH-rules yield MATH which implies MATH . Summing up, within MATH nodes, there must be at least two nodes MATH which satisfy MATH . This implies that one of these nodes is blocked by the other. |
cs/0106031 | Let MATH be a completion tree generated for MATH by the MATH-algorithm. For every node MATH of MATH we define MATH. If MATH is a predecessor of MATH in MATH, then this implies MATH. If not MATH and MATH, then this implies MATH. Furthermore, for MATH (but possibly MATH), MATH and MATH implies MATH. The only way that the maximal length of concepts does not decrease is along a pure MATH-path with MATH. However, the MATH- and the MATH-rule must be applied before the MATH-rule may generate a new successor. Together with REF , this guarantees that these pure MATH-paths have a length of at most MATH. Summing up, we can have a path of length at most MATH before decreasing the maximal length of the concept in the node labels (or blocking occurs), which can happen at most MATH times and thus yields an upper bound of MATH on the length of paths in a completion tree. |
cs/0106031 | Let MATH be the MATH-concept to be tested for satisfiability. We can assume MATH to be in NNF because every MATH-concept can be turned into NNF in linear time. Let MATH. For each node MATH of the completion tree, the labels MATH and MATH can be stored using MATH bits for each set. Starting from the initial tree consisting of only a single node MATH with MATH, the expansion rules, as given in REF , are applied. If a clash is generated, then the algorithm fails and returns ``MATH is unsatisfiable". Otherwise, the completion tree is generated in a depth-first way: the algorithm keeps track of exactly one path of the completion tree by memorizing, for each node MATH, which of the MATH-concepts in MATH successors have yet to be generated. This can be done using additional MATH bits for each node. The ``deletion" of all successors in the MATH- or the MATH-rule of a node MATH is then simply realized by setting all these additional bits to ``has yet to be generated". There are three possible results of an investigation of a child of MATH: CASE: A clash is detected. This stops the algorithm with ``MATH is unsatisfiable". CASE: The MATH- or the MATH-rule leads to an increase of MATH. This causes reconsideration of all children of MATH, re-using the space used for former children of MATH. CASE: Neither of these first two cases happens. We can then forget about this subtree and start the investigation of another child of MATH. If all children have been investigated, we consider MATH's predecessor. Proceeding like this, the algorithm can be implemented using MATH bits for each node, where the MATH bits are used to store the two labels of the node, while MATH bits are used to keep track of the successors already generated. Since we reuse the memory for the successors, we only have to store one path of the completion tree at a time. From REF , the length of this path is bounded by MATH. Summing up, we can test for the existence of a completion tree using MATH bits. Unfortunately, due to the MATH-rule, the MATH-algorithm is a non-deterministic algorithm. However, NAME 's theorem CITE tells us that there is a deterministic implementation of this algorithm using at most MATH bits, which is still a polynomial bound. |
cs/0106031 | The set MATH can be viewed as a directed graph MATH with vertices MATH and MATH. The strongly connected components of MATH can be calculated in quadratic time. For every non-trivial strongly connected component MATH, select an arbitrary MATH such that MATH if MATH. For every MATH, replace MATH in MATH and MATH by MATH. The results of this replacement are called MATH and MATH, respectively. It is obvious that these can be obtained from MATH in polynomial time and that MATH is cycle-free. For every MATH with MATH, it easy to see that, for every strongly connected component MATH of MATH, MATH holds for every MATH and, if MATH, then MATH is transitive for every MATH. Hence, MATH is satisfiable with respect to MATH iff MATH is satisfiable with respect to MATH . |
cs/0106031 | If MATH then MATH and hence MATH. If MATH then MATH for every MATH with MATH. Hence MATH. |
cs/0106031 | Like in the proof of REF , it is easy to see that the smallest set MATH that contains MATH and is closed under sub-concepts and MATH contains MATH concepts. For MATH, we additionally have to add concepts MATH to MATH if MATH with MATH and MATH, and again close MATH under sub-concepts and MATH to obtain MATH. This may yield at most two concepts for every concept in MATH and every role MATH that occurs in MATH because, for every such concept MATH, it suffices to add MATH and MATH. Since it is a sub-concept of MATH, MATH does not need to be reconsidered in the closure process. |
cs/0106031 | For the only-if-direction, let MATH be a MATH-concept and MATH a set of role axioms. Assume that MATH is a (MATH-)model of MATH with respect to MATH. Let MATH bet the set of freshly introduced concept names from REF . We will construct a MATH-model MATH for MATH and MATH from MATH by setting MATH for every MATH, and maintaining the interpretation of all other concept and role names. For all MATH, MATH. This claim is proved by induction on the structure of concepts. For the base case MATH, MATH holds, and hence MATH. For all other cases, except for MATH and MATH, the claim follows immediately by induction because MATH for every MATH, since MATH and because of REF . For MATH, MATH and by construction of MATH, MATH. For MATH, MATH and MATH which finishes the proof of the claim. In particular, since MATH, also MATH. It remains to show that MATH holds. For the first set of axioms, this holds because MATH since MATH because of REF , and MATH due to REF . For the second set of axioms, let MATH be a role with MATH and MATH. Then MATH unless there is a MATH and a MATH with MATH and MATH. This implies the existence of an element MATH with MATH and MATH. Since MATH and MATH, transitivity of MATH implies MATH. Thus MATH and MATH because MATH. This implies MATH, which is a contradiction because MATH and MATH. Summing up, MATH is contained in the interpretation of every conjunct that appears on the right-hand side of the axioms, and hence MATH . This holds for every MATH and hence MATH. Thus, we have shown that MATH is satisfiable with respect to MATH. For the if-direction, let MATH be an interpretation with MATH and MATH. From MATH we construct an interpretation MATH such that MATH and MATH. To achieve the latter, we define MATH as follows: MATH . Since MATH is cycle-free, MATH is well-defined for every MATH and it is obvious that, for every MATH with MATH, MATH is transitive. First, we check that MATH indeed satisfies MATH. MATH for every MATH. If MATH, this is immediate from the construction. If MATH, then the proof is more complicated. It is by induction on the number MATH, where the case for MATH does not make use of the induction hypothesis. CASE: If MATH and MATH, then MATH due to REF because MATH. CASE: If MATH and MATH, then MATH since MATH, and hence MATH. CASE: If MATH and MATH, then MATH. For every MATH with MATH, MATH because MATH is cycle-free. Since MATH holds by the definition of MATH, induction yields MATH. Also, since MATH, MATH and hence MATH. If MATH, then MATH or there exists a role MATH with MATH and a path MATH such that MATH, and MATH for MATH. Again, then proof is by induction on MATH and, if MATH holds, then we do not need to make use of the induction hypothesis. CASE: If MATH and MATH, then MATH and thus MATH implies MATH. CASE: If MATH and MATH, then MATH. If MATH, then we are done. Otherwise, there exists a path MATH with MATH, and MATH. Also, by definition, MATH. CASE: If MATH and MATH, then either MATH or MATH for some MATH with MATH. In the latter case, MATH and, by induction, either MATH, or there exists a role MATH with MATH and a path MATH with MATH, and MATH for MATH. Since MATH, also MATH. For a simple role MATH, MATH. The proof is by induction on the number of sub-roles of MATH. If MATH is simple and has no sub-roles, then MATH holds by definition of MATH. If MATH is simple, then every role MATH with MATH and MATH has less sub-roles than MATH because MATH is cycle free, and must be simple because otherwise MATH would not be simple. Hence, the induction hypothesis is applicable to each such MATH, which yields MATH. Also, since MATH, MATH holds by REF and hence MATH. MATH for every MATH. The proof is by induction on the value MATH of concepts in MATH, where the function MATH is defined by MATH where the definition of the norm MATH of a MATH-concept is similar to the definition for MATH extended to universal and existential restrictions: MATH . The purpose of this (seemingly rather strange) definition of MATH is to reduce the case for an existential restriction to its dual universal restriction. Except for existential, universal, and number restrictions, all cases are straightforward. CASE: If MATH, then MATH. If MATH then, since MATH, also MATH. By induction, MATH and MATH, and hence MATH. If MATH and MATH, then, by REF , there are two possibilities. CASE: If MATH, then MATH holds because MATH. By induction, MATH, and hence MATH. CASE: There is a role MATH with MATH and a path MATH with MATH, and MATH for MATH. Since MATH and MATH, we have MATH for every MATH, and MATH in particular. Since MATH, it follows that MATH and, by induction, MATH. In any case, we have shown that MATH and, since MATH has been chosen arbitrarily with MATH, MATH holds. CASE: If MATH, then MATH and MATH where MATH follows by induction since MATH. CASE: If MATH, then MATH and MATH is simple. By REF , MATH holds. Also, by induction, MATH and hence MATH. This finishes the proof of REF , which yields MATH. Since we have already shown that MATH, we have proved satisfiability of MATH with respect to MATH. |
cs/0106031 | The lower bound is immediate from REF , since MATH is strictly contained in MATH. For the upper bound, the reduction from REF extended to MATH by setting MATH for every individual MATH yields a reduction from MATH to MATH, for which the corresponding problems are solvable in NAME if unary coding of numbers in the input is assumed. |
cs/0106031 | For the if-direction, the construction of a model of MATH from a tableau for MATH is similar to the one presented in the proof of REF where the interpretation of the roles is defined in the same manner as it was done in the proof of REF . To be more precise, if MATH is a tableau for MATH with respect to MATH and MATH, a model MATH of MATH can be defined as follows: MATH . Like in the proof of REF , it is easy to see that MATH and that MATH iff MATH or there exists a role MATH with MATH and a path MATH such that MATH, MATH, MATH, and MATH for MATH. Moreover, if MATH is simple, then MATH. It remains to show that MATH. This is done by proving that MATH implies MATH for each MATH and MATH. Since MATH, we then have MATH and hence MATH is a model of MATH. The proof is by induction on the norm MATH of concepts as defined in the proof of REF . The two base cases of the induction are MATH or MATH for MATH. If MATH, then, by definition, MATH. If MATH, then, by REF , MATH and hence MATH. For the induction step, we have to distinguish several cases: CASE: The cases MATH, MATH, and MATH are exactly as for MATH in the proof of REF CASE: MATH. Let MATH with MATH, let MATH be an arbitrary individual such that MATH. There are two possibilities: CASE: MATH. Then REF implies MATH and, by induction, MATH. CASE: MATH. Due to REF , this can only be the case if there is a role MATH with MATH and a path MATH with MATH. Then REF implies MATH for all MATH and particularly MATH. Due to REF , MATH also holds. Again, by induction, this implies MATH. In both cases, we have MATH and, since MATH has been chosen arbitrarily, MATH holds. CASE: MATH. For a MATH with MATH, we have MATH by REF . Hence there are MATH individuals MATH such that MATH for MATH, MATH, and MATH for all MATH. By induction, we have MATH and, since MATH, also MATH. CASE: MATH. For this case, it is crucial that MATH is a simple role because this implies MATH. Let MATH be an individual with MATH. Due to REF , we have MATH or MATH for each MATH with MATH. Moreover, MATH holds due to REF . We show that MATH: assume MATH. This implies the existence of some MATH with MATH and MATH but MATH (because MATH). By REF , this implies MATH, which, by induction, yields MATH, in contradiction to MATH. For the only-if-direction, we have to show that satisfiability of MATH with respect to MATH implies the existence of a tableau MATH for MATH with respect to MATH. Let MATH be a model of MATH with MATH. A tableau MATH for MATH can be defined by: MATH . It remains to demonstrate that MATH is a tableau for MATH: CASE: Except for REF , all conditions are satisfied as a direct consequence of the definition of the semantics of MATH-concepts. CASE: For REF , if MATH and MATH for MATH with MATH and MATH, then MATH unless there is some MATH such that MATH and MATH. In this case, since MATH, MATH, and MATH, it holds that MATH. Hence MATH and MATH - in contradiction to the assumption. MATH therefore satisfies REF . CASE: REF is satisfied because MATH and set-inclusion is a transitive property. |
cs/0106031 | Let MATH, MATH, and MATH the maximum MATH that occurs in a concept of the form MATH. Termination is a consequence of the following properties of the expansion rules: CASE: The expansion rules never remove nodes from the tree or concepts from node labels. NAME labels can only be changed by the MATH-rule which either expands them or sets them to MATH; in the latter case, the node below the MATH-labelled edge is blocked and this block is never broken. CASE: Each successor of a node MATH is the result of the application of the MATH-rule or the MATH-rule to MATH. (Note that the MATH-rule, does not move nodes in the tree.) For a node MATH, each concept in MATH can trigger the generation of successors at most once. For the MATH-rule, if a successor MATH of MATH was generated for a concept MATH and later MATH is set to MATH by the MATH-rule, then there is some MATH-neighbour MATH of MATH with MATH. For the MATH-rule, if MATH were generated by the MATH-rule for MATH, then MATH holds for all MATH. This implies that there are always MATH-neighbours MATH of MATH with MATH and MATH for all MATH, since the MATH-rule never merges two nodes MATH with MATH and, whenever an application of the MATH-rule sets MATH to MATH, there is some MATH-neighbour MATH of MATH which ``inherits" both MATH and all inequalities from MATH. Since MATH contains a total of at most MATH concept of the form MATH and MATH, the out-degree of the tree is bounded by MATH. CASE: Nodes are labelled with non-empty subsets of MATH and edges with subsets of MATH, so there are at most MATH different possible labellings for a pair of nodes and an edge. Therefore, if a path is of length MATH, then, from the pair-wise blocking condition, there must be two nodes MATH on this path such that MATH is directly blocked by MATH. Since a path on which nodes are blocked cannot become longer, paths are of length at most MATH. |
cs/0106031 | Let MATH be a tableau for MATH with respect to MATH. We use this tableau to guide the application of the non-deterministic rules. To do this, we will inductively define a function MATH, mapping the nodes MATH of the tree MATH to MATH such that, for each MATH: MATH . Let MATH be a completion-tree and MATH a function that satisfies MATH. If a rule is applicable to MATH, then the rule is applicable to MATH in a way that yields a completion-tree MATH and an extension of MATH that satisfies MATH. Let MATH be a completion-tree and MATH a function that satisfies MATH. We have to consider the various rules. CASE: For the MATH-, MATH-, and MATH-rule, this is analogous to the proof of REF for MATH. CASE: The MATH-rule: If MATH, then MATH, and if MATH is a MATH-neighbour of MATH, then also MATH due to MATH. REF implies MATH and hence the MATH-rule can be applied without violating MATH. CASE: The MATH-rule: If MATH, then MATH, and if there is some MATH with MATH and MATH is a MATH-neighbour of MATH, then also MATH due to MATH. REF implies MATH and hence the MATH-rule can be applied without violating MATH. CASE: The MATH-rule: If MATH, then MATH, and, if there is a MATH-neighbour MATH of MATH, then MATH due to MATH. REF implies MATH. Hence the MATH-rule can add an appropriate concept MATH to MATH such that MATH holds. CASE: The MATH-rule: If MATH, then MATH and REF implies MATH. Hence there are elements MATH such that MATH, MATH, and MATH for MATH. The MATH-rule generates MATH new nodes MATH. By extending MATH, one obtains a function MATH that satisfies MATH for the extended tree. CASE: The MATH-rule: If MATH, then MATH and REF implies MATH. If the MATH-rule is applicable, we have MATH, which implies that there are at least MATH-neighbours MATH of MATH such that MATH. Thus, there must be two nodes MATH such that MATH (because otherwise MATH would hold). Since MATH, we have that MATH cannot hold because of MATH, and MATH can be chosen such that MATH is a successor of MATH because MATH has at most one predecessor. Hence the MATH-rule can be applied without violating MATH. Why does this claim yield the completeness of the tableau algorithm? For the initial completion-tree consisting of a single node MATH with MATH and MATH, the function MATH for some MATH with MATH satisfies MATH. Such a MATH exists due to REF . Whenever a rule is applicable to MATH, it can be applied in a way that maintains MATH, and, since the algorithm terminates, we have that any sequence of rule applications must terminate. REF implies that any tree MATH generated by these rule-applications must be clash-free as there are only two possibilities for a clash, and it is easy to see that neither of these can hold in MATH: CASE: MATH cannot contain a node MATH such that MATH because MATH and hence REF would be violated for MATH. CASE: MATH cannot contain a node MATH with MATH and MATH-neighbours MATH of MATH with MATH and MATH for MATH because MATH, and, since MATH implies MATH, MATH, in contradiction to REF . |
cs/0106031 | Let MATH be a complete and clash-free completion tree. A path is a sequence of pairs of nodes of MATH of the form MATH. We define auxiliary functions MATH by setting, for such a path MATH, MATH and MATH. With MATH we denote the path MATH. The set MATH is defined inductively as follows: CASE: For the root node MATH of MATH, MATH, and CASE: For a path MATH and a node MATH in MATH: CASE: if MATH is a successor of MATH and MATH is not blocked, then MATH, or CASE: if, for some node MATH in MATH, MATH is a successor of MATH and MATH blocks MATH, then MATH. Please note that, due to the construction of MATH, for MATH with MATH, we have that MATH is not blocked, MATH is blocked iff MATH, and MATH is never indirectly blocked - it is either directly blocked or unblocked. Furthermore, the blocking condition implies MATH. Now we can define a tableau MATH with: MATH . MATH is a tableau for MATH with respect to MATH. We show that MATH satisfies all the properties from REF . CASE: MATH because MATH, hence REF holds. CASE: REF holds because MATH is clash-free. CASE: REF hold because MATH is complete. CASE: REF : assume MATH and let MATH. In MATH, there is a MATH-neighbour MATH of MATH with MATH because MATH is not blocked and the MATH-rule is not applicable. There are two possibilities: CASE: MATH is a successor of MATH in MATH. If MATH is not blocked, then MATH and MATH as well as MATH. If MATH is blocked by some node MATH in MATH, then MATH, MATH and, since MATH, MATH. CASE: MATH is a predecessor of MATH. Again, there are two possibilities: CASE: MATH is of the form MATH with MATH. CASE: MATH is of the form MATH with MATH. Since MATH only has one predecessor in MATH, the node MATH cannot be the predecessor of MATH. Then it must be the predecessor of MATH in MATH, MATH, and MATH blocks MATH, all due to the construction of MATH. Together with the definition of the blocking condition, this implies MATH as well as MATH, due to the pair-wise blocking condition. In both cases, MATH and MATH. CASE: REF : assume MATH and MATH. If MATH, then MATH is a MATH-successor of MATH, and thus MATH because the MATH-rule is not applicable. Since MATH, we have MATH. If MATH, then MATH is a MATH-successor of MATH, MATH a MATH-neighbour of MATH and thus MATH because MATH is not indirectly blocked and the MATH-rule is not applicable. CASE: REF : assume MATH and MATH for some MATH with MATH. If MATH, then MATH is a MATH-successor of MATH and thus MATH because otherwise the MATH-rule would be applicable. From MATH, it follows that MATH. If MATH, then MATH is a MATH-successor of MATH, and hence MATH is a MATH-neighbour of MATH. Because MATH is not indirectly blocked, this implies MATH. CASE: REF : assume MATH and MATH. If MATH, then MATH is a MATH-successor of MATH and thus MATH because the MATH-rule is not applicable. Since MATH, we have MATH. If MATH, then MATH is a MATH-successor of MATH, MATH is a MATH-neighbour of MATH, and thus MATH because MATH is not indirectly blocked and the MATH-rule is not applicable. CASE: REF is satisfied due to the symmetric definition of MATH. CASE: REF is satisfied due to the definition of MATH-successors that takes into account the role hierarchy MATH. CASE: REF : assume MATH. Completeness of MATH implies that there exist MATH distinct individuals MATH in MATH such that each MATH is a MATH-neighbour of MATH and MATH. We claim that, for each of these individuals, there is a path MATH such that MATH, MATH, and MATH for all MATH. Obviously, this implies MATH. For each MATH, there are three possibilities: CASE: MATH is a MATH-successor of MATH and MATH is not blocked in MATH. Then MATH is a path with the desired properties. CASE: MATH is a MATH-successor of MATH and MATH is blocked in MATH by some node MATH. Then MATH is the path with the desired properties. Since the same MATH may block several of the MATH-s, it is indeed necessary to include the blocking nodes explicitly into the path construction to make these paths distinguishable. CASE: MATH is a MATH-successor of MATH. Since MATH is a tree, there may be at most one such MATH. This implies that MATH is of the form MATH with MATH. The path MATH has the desired properties and, obviously, MATH is distinct from all other paths MATH. CASE: Assume REF is violated. Hence there is some MATH with MATH and MATH. We show that this implies MATH, in contradiction to either clash-freeness or completeness of MATH. Define MATH and MATH. Due to the assumption, we have MATH. We distinguish two cases: CASE: MATH contains only paths of the form MATH. We claim that the function MATH is injective on MATH. Assume that there are two paths MATH with MATH and MATH. Then MATH is of the form MATH and MATH is of the form MATH with MATH. If MATH is not blocked in MATH, then MATH, contradicting MATH. If MATH is blocked in MATH, then both MATH and MATH block MATH, which implies MATH, again a contradiction. Since MATH is injective on MATH, it holds that MATH. Also for each MATH, MATH is a MATH-successor of MATH, and MATH. This implies MATH. CASE: MATH contains a path MATH where MATH is of the form MATH. Obviously, MATH may only contain one such path. As in the previous case, MATH is an injective function on the set MATH, each MATH is a MATH-successor of MATH and MATH for each MATH. To show that indeed MATH holds, we have to prove the existence of a further MATH-neighbour MATH of MATH with MATH and MATH. We distinguish two cases: CASE: MATH. Hence MATH is not blocked. This implies that MATH is a MATH-successor of MATH in MATH. Since MATH contains only successors of MATH, we have that MATH and, by construction, MATH is a MATH-neighbour of MATH with MATH. CASE: MATH. This implies that MATH is blocked in MATH by MATH and that MATH is a MATH-successor of MATH in MATH. The definition of pair-wise blocking implies that MATH is a MATH-successor of some node MATH in MATH with MATH. Again, since MATH contains only successors of MATH, we have that MATH and, by construction, MATH is a MATH-neighbour of MATH with MATH. |
cs/0106031 | Let MATH be a MATH-concept and MATH a role hierarchy. Let MATH, MATH, and MATH be the maximum MATH that occurs in a qualifying number restriction in MATH. If we set MATH, the it holds that MATH, MATH, and MATH. In the proof of REF , we have shown that paths in a completion tree for MATH become no longer than MATH and that the out-degree of a completion tree is bounded by MATH. Hence, the MATH-algorithm will construct a tree with no more than MATH nodes. Each node of this tree is labelled with a subset of MATH and each edge is labelled with a subset of MATH. Since every application of a rule either adds a node to the tree, a concept or role to one of the labels, or sets the label of an edge to MATH (in which case the corresponding successor is blocked forever), the MATH-algorithm runs in REF-NExpTime. |
cs/0106031 | Nodes are only generated when initializing the tree (with a single constant) and by the MATH-rule and no constants are added to a MATH once MATH has been generated (but some may be removed by application of the MATH-rule). When triggered by the formula MATH, the MATH-rule initializes MATH such that it contains MATH and another constant for every variable in MATH and MATH. Hence, MATH . The set MATH is a subset of MATH, for which MATH holds because there are at most MATH formulas in MATH, each of which has at most MATH free variables. There are at most MATH distinct sequences of length MATH with constants from MATH. Let MATH be MATH distinct nodes. For every MATH, we will construct an injective mapping MATH such that, if a constant MATH is shared between two nodes MATH, then MATH. Let MATH denote the nodes of a subtree of MATH that contains every node MATH and that is rooted at MATH. By induction over the distance to MATH, we define an injective mapping MATH for every MATH as follows. For MATH we pick an arbitrary injective function from MATH to MATH. For a node MATH let MATH be the predecessor of MATH in MATH and MATH the corresponding function, which has already been defined because MATH has a smaller distance to MATH than MATH. For MATH we choose an arbitrary injective function such that MATH for all MATH. All mappings MATH are injective. For any constant MATH the set MATH induces a subtree of MATH. If MATH are neighbours, the definition above ensures MATH. By induction over the length of the shortest connecting path we obtain the same for arbitrary MATH. For every node MATH there is a MATH such that MATH and we set MATH. There are at most MATH distinct subsets of MATH. Hence, there must be two nodes MATH such that MATH and, without loss of generality, MATH. This implies that MATH is blocked by MATH via MATH. Note that for MATH to be well-defined, MATH must be injective. By construction, MATH preserves shared constants. Since MATH, MATH holds. |
cs/0106031 | For any completion tree MATH generated by the tableau algorithm, we define MATH by MATH . The lexicographic order MATH on MATH is well-founded, that is, it has no infinite decreasing chains. Any rule application decreases MATH with respect to MATH for at least one node MATH, and never increases MATH with respect to MATH for an existing node MATH. However it may create new successors, one at a time. Since MATH is well-founded, there can only be a finite number of applications of rules to every node in MATH and hence a finite number of successors and an infinite sequence of rule applications would generate a tree of infinite depth. Yet, as a corollary of REF , we have that the depth of MATH is bounded by MATH. For assume that there is a path of length MATH in MATH with deepest node MATH. By the time MATH has been created (by an application of the MATH-rule to its predecessor MATH), the path from the root of MATH to MATH contains at least MATH nodes, and hence a blocked node. This implies that MATH is blocked too, and the MATH-rule cannot be applied to create MATH. |
cs/0106031 | Since MATH is satisfiable, there is a model MATH of MATH. We will use MATH to guide the application of the non-deterministic MATH-rule. For this we incremently define a function MATH such that MATH. We refer to this property by MATH. The set MATH can contain atomic formulas MATH, where MATH occurs at some positions of MATH. The constant MATH is not mapped to an element of MATH by MATH. We deal with this as described just after REF by setting MATH . If, for a completion tree MATH, there exists a function MATH, such that MATH holds and a rule is applicable to MATH, then it can be applied in a way that maintains MATH. CASE: For the MATH- and the MATH-rule this is obvious. CASE: If the MATH-rule is applicable to MATH with MATH, then, since MATH, MATH must hold. Hence, for every node MATH that shares MATH with MATH, MATH and the rule can be applied without violating MATH. CASE: If the MATH-rule is applicable to MATH with MATH and MATH with MATH for all atoms MATH, then, from the definition of MATH, there is a tuple MATH, such that MATH and MATH. From MATH we get that MATH and since every MATH appears exactly once in MATH, also MATH. Hence, we have MATH which, by REF , implies MATH and hence MATH can be added to MATH without violating MATH. CASE: If the MATH-rule is applicable to MATH with MATH, then this implies MATH . Hence, there are sequences MATH such that MATH. If we define MATH such that MATH and MATH, then MATH. Note, that this might involve setting MATH for some MATH. With this construction the resulting extended completion-tree MATH and extended function MATH again satisfy MATH. CASE: If the MATH-rule is applicable to MATH with MATH and a neighbour MATH with MATH, then it adds MATH to MATH. From MATH we get that MATH, and since MATH, this implies MATH. Hence, adding MATH to MATH does not violate MATH. CASE: If the MATH-rule is applicable to a node MATH with a universally quantified formula MATH and a neighbour MATH which shares MATH with MATH, MATH yields MATH. Hence, adding MATH to MATH does not violate MATH. A completion-tree MATH for which a function MATH exists such that MATH holds is clash free. Assume that MATH contains a clash, namely, there is a node MATH such that either MATH - implying MATH - , or that there is a sequence MATH, and an atomic formula MATH such that MATH. From MATH, MATH would follow, also a contradiction. These claims yield REF as follows. Let MATH be a tableau for MATH. Since MATH, MATH is satisfied for the initial tree together with the function MATH mapping MATH to an arbitrary element of the universe of MATH. By REF , any sequence of applications is finite, and from REF we get that there is a sequence of rule-applications that maintains MATH. By REF , this sequence results in a tableau. This completes the proof of REF . |
cs/0106031 | Let MATH a tableau for MATH. For every direct blocking situation we fix a mapping MATH verifying this blocking. Using an unraveling construction, we will construct a model MATH for MATH of width at most MATH from MATH. First, we ``unravel" blocking situations in MATH by successively replacing every blocked node with a copy of the subtree of MATH rooted at the blocking node. Formally, this is achieved by the following path construction. We define MATH . Since from now on we only deal with nodes from MATH, every blocking is direct and we will no longer explicitly mention this fact. The set MATH is inductively defined by CASE: MATH for the root MATH of MATH, CASE: if MATH, the node MATH is a successor of MATH and MATH is not blocked, then MATH, CASE: if MATH, MATH is a successor of MATH blocked by the node MATH, then MATH. The set MATH forms a tree, with MATH being a successor of MATH if MATH is obtained from MATH by concatenating one element MATH at the end. We define the auxiliary functions MATH by setting MATH and MATH for every path MATH. Intuitively, for every node MATH of MATH, the paths MATH with MATH stand for distinct copies of MATH created by the unraveling. The universe of MATH consists of (classes of) constants labelling nodes in MATH paired with the paths at whose MATH they appear to distinguish constants occurring at different copies of a node of MATH. Formally, we define MATH . Constants appearing at consecutive nodes of MATH stand for the same element and the same holds for constants related by a mapping MATH verifying a block. Hence, to obtain the universe of MATH, we factorize MATH as follows. Let MATH be the smallest symmetric relation on MATH satisfying CASE: MATH if MATH is a successor of MATH in MATH, MATH is an unblocked successor of MATH, and MATH, CASE: MATH if MATH is a successor of MATH in MATH, MATH is a blocked successor of MATH, MATH, and MATH for the function MATH that verifies that MATH is blocked by MATH. With MATH we denote the reflexive, transitive closure of MATH and with MATH the class of MATH, that is, the set MATH. Since MATH iff MATH are neighbours in MATH, for every MATH, the set MATH is a subtree of MATH. The classes of MATH will be the elements of the universe of MATH. First we need to prove some technicalities for this construction. Let MATH and MATH. Then MATH iff MATH. Assume the claim does not hold and let MATH with MATH. By definition of MATH, MATH must hold. Hence, there must be a path MATH such that MATH, MATH, and MATH. Without loss of generality, assume we have picked MATH such that this path has minimal length MATH. Such a minimal path must be of length MATH, for if we assume a path of length MATH, there must be MATH such that MATH, because the relation MATH is defined along paths in the tree MATH. If MATH then we can shorten the path between position MATH and MATH and obtain a shorter path. If MATH, then the path MATH is also a shorter path with the same properties. Hence, a minimal path must be of the form MATH. If MATH is not blocked, by the definition of MATH, MATH must hold. Hence, since MATH, MATH must be blocked by MATH. From the definition of MATH we have MATH and MATH for the function MATH verifying that MATH is blocked by MATH. Since MATH must be injective, this is a contradiction. Since the set MATH is a tree, and as a consequence of REF , we get the following. Let MATH with MATH, MATH. If, for MATH, MATH then MATH. If MATH then there must be a path MATH such that MATH, MATH, MATH, and MATH. Since MATH is only defined along paths in the tree MATH, there must be a step from MATH to MATH (or, dually, from MATH to MATH) in this path, more precisely, there must be a MATH such that MATH and MATH holds. Hence, we have the situation MATH . REF implies MATH and MATH and hence MATH. Using REF , we can show that the blocking condition and the MATH- and MATH-rule work as desired. Let MATH, MATH, MATH non-empty tuples, and MATH. CASE: For every atom MATH, MATH iff MATH. CASE: For every universally quantified MATH, MATH iff MATH. Since both propositions are symmetric, we only need to prove one direction. If MATH with MATH and MATH, then MATH and, as an intersection of subtrees of MATH, MATH is itself a subtree of MATH. Hence, in MATH there is a path MATH for which there exist tuples of constants MATH with MATH, MATH, MATH, MATH, and MATH. Since MATH are non-empty, so are the MATH. From REF , we get that for any two neighbours MATH in MATH, MATH implies MATH. By two similar inductions on MATH with MATH we show that if MATH then MATH and if MATH then MATH. For MATH in both cases nothing has to be shown. Now assume that the we have shown these properties up to MATH. Without loss of generality, assume MATH is a successor of MATH in the tree MATH. The other case is handled dually. There are two possibilities: CASE: MATH is not blocked. Then MATH and by the definition of MATH, MATH is a successor of MATH in MATH and MATH holds. If MATH then MATH holds by induction and due to the MATH-rule, this implies MATH. The MATH-rule is applicable because, for the the non-empty tuple MATH, MATH holds. If MATH then by induction MATH and due to the MATH-rule this implies MATH. CASE: MATH is blocked by MATH (with function MATH) and MATH is a successor of MATH in MATH. Then, by definition of MATH, we have MATH and MATH. If MATH then MATH holds by induction and due to the MATH-rule this implies MATH. The MATH-rule is applicable because, for the non-empty tuple MATH, MATH holds. The node MATH blocks MATH, which implies MATH . If MATH then by induction MATH and due to the MATH-rule this implies MATH. Since MATH blocks MATH, MATH holds. We now define the structure MATH over the universe MATH. For a relation MATH of arity MATH, MATH is defined to be the set of tuples MATH for which there exists a path MATH and constants MATH such that MATH for all MATH, and MATH. It remains to show that this construction yields MATH. This is a consequence of the following claim. For every path MATH and MATH, if MATH, then MATH. We show this claim by induction on the structure of MATH. If MATH, then the claim holds immediately by construction of MATH. Assume MATH, but MATH. Then, by the definition of MATH, there must be a path MATH and constants MATH such that MATH and MATH. From REF we have that MATH implies MATH and, since MATH contains no occurrence of MATH, MATH. Hence MATH contains the clash MATH, a contradiction to the fact that MATH is clash-free. Thus, MATH. Assume MATH but MATH. From REF we get that this implies MATH and hence MATH contains the clash MATH. Again, this is a contradiction to the fact that MATH is clash-free and MATH must hold. For positive Boolean combinations the claim is immediate due to the MATH- and MATH-rule. Let MATH and MATH arbitrarily chosen with MATH . We need to show that also MATH holds. In order to bring completeness of MATH and the MATH-rule into play, we show that information about the fact that REF holds is present at a single node in MATH where it triggers the MATH-rule. We rely on the fact that universal quantifiers must be guarded. Every MATH coexists with every other variable MATH in at least one atom MATH and with every element MATH in at least one atom MATH. For any two distinct variables MATH, MATH holds and this can only be the case if there is a path MATH and constants MATH such that MATH and MATH. Similarly, for every element MATH and every element MATH there exists a path MATH and constants MATH such that MATH and MATH. For every MATH and MATH, MATH and MATH are subtrees of MATH. The tree MATH overlaps with the tree MATH at MATH and with the tree MATH at MATH. From this it follows CITE that there exists a common path MATH . Thus, there are tuples MATH, MATH such that MATH . We now show that the preconditions of the MATH-rule are satisfied at MATH for the formula MATH and the tuple MATH. First, due to REF , it holds that MATH because MATH and MATH. For every MATH, MATH holds as follows: from REF we get MATH . Since MATH is an atom, this implies the existence of a path MATH and tuples MATH with MATH . Clearly, MATH and, since MATH, it holds that MATH. Thus, by REF it holds that MATH. Since this is true for every atom MATH, the preconditions of the MATH-rule are satisfied and the completeness of MATH yields MATH. By induction, MATH holds and together with REF this implies MATH. Since MATH have been chose arbitrarily, MATH holds. If MATH, there are two possibilities: CASE: there are MATH with MATH. Then, by induction, we have MATH and hence MATH. CASE: there are no such MATH, then there is a successor MATH of MATH and MATH with MATH. The node MATH can be blocked or not. If MATH is not blocked, then MATH and by induction MATH . From the definition of MATH we have, MATH and hence MATH. If MATH is blocked by a node MATH (with function MATH) then MATH. From the blocking condition, we have that MATH is unblocked and MATH. Hence, by induction MATH and, by definition of MATH, we have that MATH and hence, MATH. As a special instance of REF we get that MATH. From REF , we get that, for every node MATH, MATH and hence the tree MATH together with the function MATH with MATH provides a tree decomposition of MATH of width MATH. This completes the proof of REF . |
cs/0106031 | Let MATH be satisfiable. Then, from REF we get that there is a tableau MATH for MATH. By REF , MATH induces a model for MATH of tree width at most MATH. Note that we have never relied on REF to obtain any of the results in this thesis and hence have indeed given an alternative proof for the generalised tree model property of MATH. For MATH and MATH, observe that the embedding of these logics into MATH may increase the width of the sentence but not by more than a recursive amount. |
cs/0106034 | To see that MATH can be expressed in the powerset algebra, we begin by noting that for any relation type MATH one can write a powerset algebra expression MATH yielding the collection of all relations of type MATH on MATH. For example, MATH is MATH, and MATH is MATH. Hence, if the type of MATH is MATH for MATH, then MATH yields the collection of all potential solutions. Now it suffices to observe that one can write nested relational algebra expressions that apply MATH or MATH to each database in this collection separately. Explicit forms of such expressions have been given by CITE. After that, the actual solutions can be selected by an equality selection. |
cs/0106034 | The if-direction is clear. For the only-if direction, we work by induction on the nesting depth of equations. The base case - expressions that do not contain any equations at all - is trivial. For the inductive step, consider a top-level equation MATH occurring in MATH. The natural strategy to evaluate this equation runs in polynomial space, so in particular, for each expansion of each database MATH over MATH with a candidate solution MATH over MATH, the natural evaluation of MATH and MATH on MATH runs in polynomial space. In this way we consider every possible database MATH over MATH, because the restriction of MATH to MATH is a possible MATH, and MATH itself then is a possible expansion of MATH. Hence, the natural evaluation strategies of MATH and MATH in general run in polynomial space. Formally, we must note here that MATH and MATH might not actually mention certain relation names in MATH or MATH, and that there is still the formal possibility that their natural evaluation might not run in polynomial space on databases over schemas not containing these names. However, using REF , it can be seen that this is impossible. By induction, we can therefore conclude that all equations occurring nested inside a top-level equation are sparse. The top-level equation itself must also be sparse. In proof, if one of the MATH would be of non-flat type, even one candidate solution can already be of exponential size. Indeed, even in the simplest case where MATH would be of type MATH, on a domain with MATH elements, a possible candidate value for MATH is the collection of all subsets of that domain, which is of size MATH. So, every MATH is of flat type. Furthermore, since we store the solution set as an intermediate result, it must be of at most polynomial size on all databases over MATH. Since the individual solutions are flat databases and thus of polynomial size, the cardinality of the solution set must therefore be at most polynomial. |
cs/0106034 | The crux is a representation of nested relational databases by ``pseudo-flat" ones, also used by CITE. Given a nested relational database MATH, we define its extended domain, denoted by MATH, as the union of its finite domain of atomic values with the set of all relations occurring (possibly deeply nested) in MATH. We regard the relations in the extended domain as if they were atomic values. Now for any nested relational database schema MATH we can construct a flat one MATH, together with a mapping MATH from the set of databases over MATH onto the set of databases over MATH, expressible in the nested relational algebra. The details of this mapping need not concern us here. Important is that we can furthermore construct a converse mapping MATH from the set of databases over MATH to the set of databases over MATH, also expressible in the nested relational algebra, with the following properties for each database MATH over MATH: REF the finite domain of MATH equals MATH; and REF MATH. Note that while MATH is expressed in the nested relational algebra, the result MATH is not really a flat database, because of the nested relations in the extended domain. However, it is ``pseudo-flat," in the sense that we regard these relations as if they were atomic values. For any relation name MATH of MATH, we denote the nested relational algebra expression defining the MATH-component of the mapping MATH by MATH. Likewise, we denote the expression defining the MATH-component by MATH. Given this representation, the proof is straightforward. Let MATH be an NP property of databases over MATH, closed under isomorphism. Define the property MATH of databases over MATH as follows: MATH satisfies MATH if MATH satisfies MATH. Then MATH is in NP, and is also closed under isomorphism. Hence, Fagin's theorem gives us a MATH sentence MATH over MATH expressing MATH. By the equivalence of relational algebra and first-order logic, there is a flat relational algebra expression MATH over MATH such that the first-order logic sentence MATH is equivalent to MATH. Now modify MATH as follows: for every relation name MATH of MATH, replace every occurrence of MATH in MATH by MATH. Likewise, replace every occurrence of MATH in MATH by MATH. Denote the resulting nested relational algebra expression by MATH. We now have, for any database MATH over MATH, that MATH satisfies MATH if and only if MATH is true on MATH. The condition MATH can easily be written as an equation (compare REF ). |
cs/0106034 | The usual proof of Fagin's theorem immediately yields the case where MATH is flat. Indeed, in that proof, to express an NP property decided by some polynomial-time bounded non-deterministic NAME machine MATH, one writes a MATH sentence MATH where MATH stands for an order on the domain; MATH, , MATH encode (using the order in MATH) a computation of MATH; and MATH checks whether the computation is accepting. As we are dealing with a NAME property, MATH has only polynomially many accepting computations. Hence, the equation MATH would be sparse were it not for MATH, as there are exponentially many possible orders on a finite domain. On ordered databases, however, there is no need for MATH and we obtain a genuinely sparse equation. This is for flat databases; for general nested relational databases we use the same representation technique as in the proof of REF . |
cs/0106034 | Given a binary relation MATH and a natural number MATH, we define the relation MATH as MATH (MATH times MATH), where MATH is the classical composition operator of binary relations: MATH. Further, define MATH as MATH, and define MATH as MATH. Note that MATH equals the transitive closure of MATH, and that for MATH, MATH. Now consider the following REF-ary relation MATH: MATH . We show next that there is an equation whose only solution, given MATH, is MATH. This proves the Proposition, because all we then have to do is unnest the solution set and project on the middle two columns to get the transitive closure. (The only exception is when MATH is empty, in which case the transitive closure of MATH is MATH itself, but this can also easily be tested in the nested relational algebra.) The desired equation expresses the conjunction of the following conditions on relation variable MATH: CASE: For any pair MATH, we denote the relation MATH by MATH, and denote further MATH . Then for every MATH, we must have CASE: MATH; CASE: MATH; CASE: MATH; and CASE: MATH. CASE: Furthermore, every pair MATH in the latter difference belongs to MATH, with MATH (and thus also MATH). CASE: MATH, and for every MATH, we have MATH. CASE: For every MATH such that MATH, there exists a pair MATH with MATH. CASE: For every MATH such that MATH, there exists a pair MATH with MATH. The conjunction of the above conditions expresses that MATH equals MATH. Indeed, by REF we know that MATH. By induction and by REF we know that MATH and hence MATH. Moreover, for every MATH we have MATH. On the other hand, if MATH then MATH, in which case MATH by REF , or MATH, in which case we know by induction and by REF that MATH for some MATH. This proves MATH. |
cs/0106034 | Suppose we want to express nesting of a flat binary relation MATH. The first application of the powerset operator is to the result of a flat relational algebra expression MATH applied to MATH. Let us focus on the case where MATH is the identity relation on a finite domain of MATH elements. A straightforward argument by structural induction shows that, on identity relations, every relational algebra expression is equivalent to a finite disjunction of equality types. Here, an equality type is a maximally consistent conjunction of equalities MATH and non-equalities MATH over the variables MATH, where MATH is the output arity of MATH. We thus see that either MATH is empty on all such MATH (this is when the disjunction is empty), or MATH is of size at least MATH when MATH is at least MATH. In the empty case, the powerset operator is useless, and we continue to the next application of powerset. Otherwise, the powerset operator explodes and the overall expression is not sparse. |
cs/0106034 | Let MATH be a relation name of type MATH, and let MATH and MATH be relation variables of type MATH. We can write a relational algebra expression MATH such that on any database MATH over MATH, MATH is empty if and only if MATH is a singleton MATH with MATH, and MATH. An example of a MATH that works is: (MATH stands for symmetric difference) MATH . Hence, the expression MATH is a sparse equation expression equivalent to MATH. The construction for general nesting operations is analogous. |
cs/0106034 | Suppose, to the contrary, that we have a sparse equation expression to express the parity of the cardinality of a finite domain MATH. We may assume that the input schema is empty, that is, an input database consists of MATH and nothing else. Consider an innermost equation MATH occurring in our expression. It may be nested inside other equations MATH, enumerated from the inside to the outside. By REF , for each MATH, MATH can be written in the form MATH, for some MATH, where MATH is a flat relational algebra expression over the flat schema MATH possibly expanded with certain free variables MATH, where MATH and MATH. By our assumption there are at most polynomially many solutions to REF. For each solution MATH of MATH there are at most polynomially many solutions MATH of MATH and so on. A sequence MATH of databases is a solution vector for MATH, if MATH is a solution for MATH, and each MATH, MATH is a solution for MATH, given MATH. Now let MATH be a solution vector for MATH, given an input MATH of size MATH. Then for every permutation MATH of MATH, MATH is also a solution vector for MATH. The number of different such MATH is precisely MATH, where MATH is the group of automorphisms of MATH. Since all equations are supposed to be sparse, this number is at most MATH for some fixed MATH, or, equivalently, MATH. Putting MATH, this implies MATH for sufficiently large MATH. We thus need to know more about large permutation groups. The following crucial lemma will give us the information we need. The group of permutations of a finite set MATH is denoted by MATH, and its alternating subgroup of even permutations by MATH. If MATH is a subgroup of MATH, a fixed set for MATH is a subset MATH, such that every MATH maps MATH to MATH. The action of MATH on a fixed set MATH (as a subgroup of MATH) is denoted by MATH. Let MATH be a fixed natural number. Let MATH be a subgroup of MATH, MATH, MATH sufficiently large. Then MATH implies the existence of a fixed set MATH with MATH, such that MATH contains MATH. For background on finite permutation groups, we refer to NAME 's book CITE, but here are a few preliminaries. An orbit of a permutation group MATH on a set MATH is a set of the form MATH, for some MATH. We call MATH transitive if MATH is one single orbit of MATH. Further, MATH is called primitive if it is transitive and has no nontrivial blocks. Here, a block of MATH is a subset MATH such that for all MATH, the set MATH is either equal to MATH, or disjoint from it. Trivial blocks are MATH, MATH, and the singletons. If MATH is not primitive but transitive, there is always a complete block system which partions MATH in equal-sized non-trivial blocks. We recall: NAME 's REF Let MATH be primitive on MATH, not containing MATH. Let MATH. Then MATH. For the proof of the Lemma, first assume that MATH is transitive. There are two possibilities: CASE: MATH is imprimitive with, say, MATH blocks of size MATH (MATH, MATH, MATH). Then MATH, which in turn is at most MATH for MATH sufficiently large. Thus, by what is given about MATH, MATH . However, this is impossible for MATH sufficiently large. CASE: MATH is primitive. Then, unless MATH contains MATH (in which case the Lemma is proved), by NAME 's theorem, MATH . Again, this is impossible for MATH sufficiently large. Now assume MATH is intransitive. Let MATH be an orbit of MATH of maximal size; let MATH be such that the size of MATH equals MATH. We have MATH. Suppose MATH. Then MATH reaches its maximum at MATH. Hence, MATH and thus MATH which is impossible for large enough MATH. So, MATH, or in other words, the size of MATH is at least MATH. We have MATH, so MATH. By REF is transitive on MATH. We can now apply the same arguments as in the case ``MATH is transitive" above, for MATH instead of MATH, and get that MATH must contain MATH. Indeed, in the right-hand sides of REF , MATH now becomes MATH, which has for effect that the upper bounds become smaller. Hence, as these inequalities already were impossible, they now become even more impossible. The extra factor of MATH in the left-hand sides does not have a significant influence. Invoking this Lemma for MATH, we get a fixed set MATH of size at least MATH such that any even permutation of MATH can be extended to an automorphism of MATH. Now let MATH be one of the relation variables of an equation MATH, of arity, say, MATH, and consider any MATH-tuple MATH whose components are either the symbol MATH or are in MATH. Let MATH be the number of components that are the symbol MATH. Further, let MATH be an equality type of MATH-tuples. Denote by MATH the set of MATH-tuples over MATH of equality type MATH such that, if we replace the MATH-components of MATH by the components of the MATH-tuple (from left to right), we get a tuple in MATH. MATH is either empty, or consists of all MATH-tuples over MATH of equality type MATH. Suppose to the contrary that MATH is neither empty nor full. Take MATH in MATH, and take MATH (of arity MATH and of type MATH) not in MATH. Take two arbitrary elements from MATH that neither appear in MATH nor in MATH, and remove them from MATH, resulting in MATH. Take a third tuple MATH (of the right arity and type) over MATH, and disjoint from MATH and MATH. If MATH is in MATH, initialize the set MATH to MATH; otherwise, put MATH. Now complete MATH to a maximal set of pairwise disjoint MATH-tuples over MATH of equality type MATH. There are at least MATH tuples in MATH. Assume at least half of MATH is outside MATH; denote the set of these by MATH. (The case where at least half of MATH is in MATH is symmetric.) Fix a MATH. For each tuple MATH in MATH, we consider the permutation MATH that transposes MATH and MATH and leaves everything else fixed. If MATH happens to be odd, we make it even by adding the transposition of the two dummy elements we took out of MATH (when we defined MATH). Then each set MATH contains MATH, but does not contain any other tuples from MATH. Thus, we produce in this way at least MATH different sets of MATH-tuples over MATH. Since they are even, each MATH can be extended to an automorphism. Hence, each of the MATH sets must be the MATH of some MATH. However, there are less than MATH different possibilities for MATH, while MATH is larger than that for MATH sufficiently large. So we get to the desired contradiction. We call MATH a MATH-ary pattern. If MATH is nonempty (and thus full), we say that the pattern is instantiated in MATH. Note also that the extreme cases, where MATH consists exclusively of stars or where MATH has no star at all, are also allowed and make sense. By the above, we thus see that any solution vector MATH can be generated by the following non-deterministic procedure: CASE: Initialize all relations of MATH to empty. CASE: Choose at most MATH different elements from MATH, playing the role of the elements outside MATH. CASE: For every relation variable MATH (of arity MATH, say), run through all MATH-ary patterns, and for each of them, non-deterministically instantiate it in MATH, or not. Since MATH and the number of relation variables are fixed, the number of possible patterns is also fixed. Hence, we can write an expression in the nested relational algebra that, given MATH, constructs the set of all possible vectors MATH of the above non-deterministic procedure. This set is a superset of the actual set of solution vectors for MATH. REF can now be replaced by a nested relational algebra expression which REF constructs the set of solution candidates MATH, REF projects out the relations for MATH and REF selects those relations which fulfil MATH. The latter is an easy task for the nested relational algebra CITE. Hence, we can get rid of MATH. Repeating this process, we can get rid of all equations, so that in the end we are left with a standard nested relational algebra expression for the parity query. But this is well known to be impossible CITE. |
cs/0106044 | Denote: MATH . Then, MATH . |
cs/0106044 | Assume that the algorithm MATH, when trained on a sample MATH, produces a hypothesis that minimizes the empirical error over MATH. Denote MATH when MATH is sampled according to a distribution that supports only examples with label in MATH. Let MATH be a sample set of size MATH, according to MATH, and MATH the hypothesis produced by MATH. Then, for all MATH, MATH . In the limit, as MATH . In particular this holds if MATH is a hypothesis produced by MATH when trained on MATH, that is sampled according to MATH. |
cs/0106046 | Consider a semi-algebraic set MATH in MATH, and let MATH be the MATH-Whitney stratification of MATH compatible with MATH. Let MATH and define the MATH-map MATH . We will need NAME 's First Isotopy Lemma CITE. Applied to the MATH-map MATH and the MATH-Whitney stratification MATH, this lemma can be stated as follows: For any MATH, CASE: If MATH is proper, that is, MATH is compact, and CASE: if for each stratum MATH, the restriction MATH has no critical points (this will be explained later), then for any MATH, there exists a homeomorphism MATH . Moreover, this homeomorphism satisfies the following two properties: CASE: For each MATH, MATH, and CASE: MATH is a homeomorphism for every connected component MATH of a stratum of MATH. This statement of NAME 's First Isotopy Lemma is a specialized form of REF. Remark that REF is automatically satisfied. Indeed, the inverse image by MATH of any interval MATH is equal to MATH which is closed and bounded in MATH. If REF is satisfied for MATH (so MATH), then every MATH is a cone radius of MATH in MATH. Take an arbitrary real number MATH. The lemma gives a homeomorphism MATH . By REF , we obtain a homeomorphism MATH which equals the restriction MATH. Since the cylinder MATH is clearly homeomorphic to the cone MATH, for example, by the homeomorphism MATH we obtain a homeomorphism MATH . It is easily verified that MATH is a homeomorphism for each connected component MATH of a stratum of MATH. Since MATH also satisfies REF , we have that MATH is a homeomorphism for each connected component MATH of a stratum of MATH. This implies that the restriction MATH is a homeomorphism from MATH to MATH, because MATH is the union of connected components of strata of MATH. The homeomorphism MATH can easily be extended to the point MATH, hence MATH is indeed a cone radius as desired. Let MATH be a schema containing a relation name MATH of arity MATH, and let MATH be a database over MATH. By REF , we can define the following cone radius query MATH where the interval MATH satisfies REF for the map MATH and semi-algebraic set MATH. Let us express this query in FO. We define the critical point query as MATH . The critical points of MATH restricted to a stratum MATH, are the points MATH where the differential map MATH is not surjective. A point MATH is a critical point of MATH if and only if the tangent space of MATH in MATH is orthogonal to the vector MATH. We compute the differential MATH as follows: Locally around MATH, we may assume that the projection on the first MATH coordinates MATH, is a homeomorphism, where MATH is the dimension of MATH in MATH. By definition of the differential, MATH where MATH. By the MATH . Inverse Function Theorem, we may assume that MATH, where MATH are MATH-maps, and hence MATH . An elementary calculation shows that the differential of MATH in MATH is the vector MATH . Since MATH is an isomorphism between the tangent space MATH of MATH in the projection MATH, and the tangent space MATH of MATH in MATH, any tangent vector MATH is of the form MATH. More specifically, any tangent vector MATH can be written as MATH . Hence, the product MATH is equal to MATH. This implies that the differential map MATH is not surjective if and only if MATH for all tangent vectors MATH. This proves the Claim. The proof of the theorem now continues as follows. The tangent space query MATH is expressible in FO CITE. Because the orthogonality of two vectors can be easily expressed in FO, the formula MATH expresses MATH correctly by REF . Here, MATH denotes an NAME expressing MATH for MATH, and MATH is an FO formula expressing MATH. A critical value of MATH is the image by MATH of a critical point. The query which returns the set of critical values is expressible in FO by the formula MATH . We now observe that MATH is finite for each MATH. Indeed, the set of critical points MATH is semi-algebraic and hence admits a MATH-cell decomposition MATH such that MATH is MATH CITE. NAME 's Theorem for MATH-maps CITE implies that each MATH attains only a finite number of values. Hence the image by MATH of the set of critical points is finite. This implies that either there are no critical values, or there exists a minimal critical value. By REF , any value smaller than this minimal value is a cone radius. We therefore conclude that the query expressed in FO as MATH is a cone radius query, as desired. |
cs/0106050 | The result is shown in CITE for standard selection rules. It easily extends to our generalisation of selection rules by noting that MATH and MATH strongly terminate iff they universally terminate with respect to the set of standard selection rules. The only-if part is immediate. The if-part follows by noting that a derivation via an arbitrary selection rule is a (prefix of a) derivation via a standard selection rule. |
cs/0106050 | Since MATH and MATH are permutation well moded, every query MATH in a derivation of MATH and MATH is permutation well moded CITE, that is, there exists a permutation MATH of MATH which is well moded. By REF , the leftmost atom in MATH is ground in its input positions and hence bounded with respect to MATH. Consider the selection rule which always selects this ``leftmost" (modulo the permutation) atom. This selection rule is local and delay-safe, and it is a standard selection rule (since there is always a selected atom). Therefore, local delay termination implies MATH-termination. Concerning the second claim, since fair-boundedness is a complete characterisation of MATH-termination, we have the conclusion. |
cs/0106050 | Since MATH and MATH are permutation well moded, every query MATH in a derivation of MATH and MATH is permutation well moded CITE, and so MATH contains an atom that is ground in its input position. The selection rule MATH that always selects this atom together with all program clauses is an input-consuming selection rule, and also a standard selection rule. Therefore, input termination implies universal termination with respect to MATH and hence MATH-termination. Concerning the second claim, by REF , MATH and MATH input terminate. As shown above, this implies that they MATH-terminate. Since fair-boundedness is a complete characterisation of MATH-termination, we have the conclusion. |
cs/0106050 | The conclusion follows by showing that any derivation of MATH and any permutation simply moded query MATH via a local delay-safe selection rule (with respect to MATH) is also a derivation via an input-consuming selection rule. So, let MATH be a local delay-safe selection rule and MATH a permutation simply-well moded query such that MATH selects atom MATH. Then by REF , for each MATH, head of a renaming of a clause from MATH, if MATH and MATH unify, then MATH is an instance of MATH, that is, MATH for some substitution MATH such that MATH. By CITE, this implies that the resolvent of MATH and any clause in MATH is again permutation simply moded. Moreover, by applying the unification algorithm CITE, it is readily checked that, if MATH and MATH unify, then MATH is an mgu. Permutation simply-modedness implies that MATH and MATH are variable-disjoint. Moreover, MATH and MATH are variable-disjoint. This implies that MATH, and so the derivation step is input-consuming. By repeatedly applying this argument to all queries in the NAME of MATH and MATH via MATH, it follows that the derivation is via some input-consuming selection rule. |
cs/0106052 | For all proofs we refer to CITE. |
cs/0106053 | Proof is done similarly to CITE. Replace each call to adorned predicate MATH in the body of clauses originating from MATH by the corresponding call to MATH. Call the obtained program MATH. Since MATH has a clause for every adornment in MATH, every path in the NAME of MATH with respect to MATH has a corresponding path in the NAME of MATH with respect to MATH. Thus, MATH. Similarly, every path in the NAME of MATH with respect to MATH has a corresponding path in the NAME of MATH with respect to MATH. Indeed, every call to MATH on the path of the NAME of MATH with respect to MATH is followed by calls to all adorned versions of MATH via rules originating from MATH, and subsequently to the rules of those predicates. However, in MATH these rules are directly defining MATH. Thus, MATH. We conclude, that MATH. MATH . |
cs/0106053 | Assume that MATH terminates with respect to MATH and does not terminate with respect to MATH. Let MATH be a clause in MATH, such that MATH can be unified with MATH and the application of this rule starts an infinite branch in the NAME. In the resolution with respect to MATH the only clauses to be applied to resolve with MATH are those of MATH. This will reduce the query to queries of the form MATH, where MATH are adorned versions of the predicate MATH of MATH and MATH are arguments of MATH. First of all, we prove that for some adornment of MATH there exists an adorned variant of MATH exists in MATH. For the sake of contradiction assume that this does not hold, that is, there is no possible adornment of atoms of predicates mutually recursive with MATH that is consistent with comparisons of the maximal prefix of the clause and with one of the possible adornments for MATH. This verbal description above can be rewritten in the following way: MATH where MATH are all possible adornments for MATH, MATH is a conjunction of comparisons of the maximal prefix of the clause and MATH are all possible adornments for the atoms of predicates mutually recursive with MATH and appearing in the body of the clause. Disjunction of these conjunctions is, on the one hand, MATH and on the other hand, MATH (since MATH is complete). Thus, MATH. Observe, that comparisons of the maximal prefix are not affected by the transformation. Thus, the body of the clause above starts with a sequence of inconsistent comparisons. Since comparisons are part of the maximal prefix, inconsistency will be discovered before any atom other than a comparison is reached. Thus, any application of this clause will cause a failure and it cannot start an infinite branch of the NAME. Thus, the adorned version of MATH exists in MATH. Let MATH be this adorned version, where MATH denotes either an adorned version of MATH, if MATH was adorned, and is identical to MATH otherwise. Let MATH be the first (while going from left to right) adorned atom in the clause body. Since transformation does not affect the preceding body atoms the corresponding queries are identical with respect to MATH and with respect to MATH. Let MATH be a query corresponding to MATH, and let MATH be a query corresponding to MATH. We have to prove that MATH terminates. Assume that MATH does not terminate. Let MATH be a clause in MATH, such that MATH is resolved with on the infinite branch of the NAME. By the previous claim there is an adorned version of this clause that belongs to MATH. If there is no adorned version of this clause with the adornment of MATH then by reasoning similar to above one can conclude that this adornment is inconsistent with the comparisons of the maximal prefix. Since those are not affected by the transformation and are identical in the clause of MATH and in the corresponding clause of MATH. Thus, any application of this clause will cause a failure and it cannot continue an infinite branch of the NAME. This means, that there exists an adorned variant of the clause, such that its head can be unified with MATH. Since computed answer substitutions of queries with respect to MATH and of MATH are identical (follows from the fact that unfolding preserves computed answer substitutions CITE) the same reasoning can be done for any of the subsequent calls and clauses, that is, we will mimic the resolution started by MATH with respect to MATH by a resolution that is started by MATH with respect to MATH. Since any resolution of MATH with respect to MATH is finite, contradiction to the assumption is obtained. Since computed answers are preserved the same claim can be proved also for the queries, originating from other atoms that MATH, thus, completing the proof. MATH . |
cs/0106053 | The proof is immediate by checking the definitions MATH . |
cs/0106056 | If FAREF accepts MATH, then, corresponding to the MATH moves, we can augment the sequence MATH with an access MATH in the interval MATH of each operation execution MATH - or select the single access involved if MATH as in the case of a reset operation execution - to obtain a new sequence MATH that is accepted by REF. By the way FAREF composes FAREF, it accepts MATH, the subsequence of atomic accesses MATH with MATH contained in MATH. Furthermore, letting MATH denote the time of access MATH, we have MATH iff MATH. Defining MATH if MATH, the total order of accesses in MATH, then MATH is a total order that extends the partial order MATH. That is, the sequence of operation executions of MATH, linear ordered by MATH, is accepted by REF. |
cs/0106056 | We show that the set of letters in an entry in the table of REF is a representative set for the state of process MATH, indexing the row, and the state of process MATH, indexing the column. The entries were chosen excluding all states from the representative sets with all outgoing moves consisting of MATH-moves (but the representative sets contain the states the outgoing MATH-moves of the excluded states point to). This gives the most insight into the workings of the protocol by considering only the result of executing MATH-moves from a state if its only outgoing moves are MATH-moves. A MATH-entry indicates an unreachable state pair. (The number ending an entry gives the expected number of accesses to finish the current operation execution of process MATH - and by symmetry, that for an equivalent state pair with respect to MATH. We will use this later.) Thus, every state MATH of the implementation execution corresponds with a set of states MATH of REF. The representative sets are given by the entries of REF . The proof of the claim is contained in the combination of REF . Below we give the inductive argument. The mechanical verification of the subcases has been done by hand, and again by machine. The setting up of the exhaustive list subcases and subsequent verification by a computer program is the essennce of a finite-state proof. In this particular case, exceptionally, the finite state machines involved (and the table of representative sets) have been minimized so that ``mechanical" verification by hand by the reader is still feasible. Induction is on the length of the sequence of accesses: CASE: Initially, after an empty sequence of accesses, FAREF is in the state MATH. CASE: Induction: Every non-reachable state has a MATH-entry in the table of REF . Consider an arbitrary atomic transition from a reachable state MATH to a state MATH, that is, using a single arc in the state chart in REF for either process MATH or MATH. This way, either MATH or MATH but not both. Then, for every FAREF state MATH, REF , according to the table of REF , there is an FAREF state MATH according to REF , such that FAREF can move from MATH to MATH by executing: either the access corresponding to the transition in the state chart in REF , if that access belongs to MATH, or no access otherwise (there is a sequence of MATH-moves from MATH to MATH). Since every reachable state of the system MATH, with MATH a state of the state chart of REF , has a representative set in FAREF , and every state of of REF is an accepting state, the lemma follows from REF . |
cs/0106056 | By REF the implementation by the state chart in REF correctly implements the specification of two-process test-and-set given by REF . The implementation is linearizable (atomic) by REF . The system makes progress (every operation execution is executed completely except for possibly the last one of each process) since MATH contains only the start and finish accesses of each operation execution performed by the implementation. |
cs/0106056 | In REF every arc is an access. Double circled states are idle states (in between completing an operation execution and starting a new one). Consider process MATH (the case for process MATH is symmetrical). The longest path without completing an operation and without cycling is from state `tstREF': tstREF, free, me, notme, choose, tohe, he, tstREF. This takes REF accesses. Four of these accesses are parts of a potential cycle of length REF. The remainder is REF accesses outside the potential cycle. In state `choose', the outgoing arrow is a random choice only when process MATH is also in the CHOOSE group. If it is, then with MATH probability MATH makes (or has already made) a choice which will cause process MATH to loop back to the `choose' state again. This can happen again and again. The expected number of iterations of loops is MATH. Since a loop has length REF, this gives a total of expected accesses of REF for the loops. Together with REF non-loop accesses the total is at most MATH accesses. Such a computation holds for every state in the state chart of REF , the only loop being the one discussed but the longest possible path is the one starting from `tstREF'. For definiteness, we have in fact computed the expected number of accesses for every accessible state MATH according to the state chart of REF , and added that number to the representative set concerned in the table of REF . Since the expected number of accesses is between REF for all operation executions, the algorithm given by the state chart of REF is wait-free. |
hep-th/0106029 | CASE: The relations MATH can be verified diagrammatically using the identities that hold in ribbon categories. REF shows the calculation for MATH. The other cases are analogous. CASE: This claim can also be verified diagrammatically. It is essentially a consequence of the fact that the maps MATH on the dual state spaces in REF to REF are given by the mirror images of the maps on the original state spaces. |
hep-th/0106029 | Using the standard abbreviations, the given isomorphisms are of the form MATH for all triangles MATH with standard vertex order MATH. For any permutation MATH we define isomorphisms MATH by MATH . Observe that this assignment is compatible with REF . These definitions provide us with isomorphisms MATH and with their dual maps MATH for all triangles MATH with arbitrary vertex order. Furthermore, there are induced linear isomorphisms of the state spaces, MATH and MATH . A convenient abbreviated notation for these maps is MATH, MATH writing MATH etc. Now the following diagram for the traces MATH commutes: To see this, imagine REF drawn for maps MATH etc. and insert the definitions of the linear isomorphisms MATH of REF. Then the isomorphisms in MATH, MATH, appear twice in each ribbon in a way such that they cancel. Let MATH denote the pairing REF applied to the state spaces which use the colouring MATH. We find MATH for all MATH and MATH. As a consequence the following diagram involving the MATH-simplex maps themselves also commutes: Analogous diagrams are available for MATH and REF in the case of opposite orientation. Finally, each tetrahedron occurs precisely twice in the boundaries of MATH-simplices, once with positive and once with negative relative orientation. Therefore the tensor product of all MATH-simplex maps in REF is conjugated by a linear isomorphism MATH which can be obtained from a tensor product of the MATH, MATH . Since MATH is the trace of REF, it agrees with MATH. |
hep-th/0106029 | Consider another set of colours MATH such that each colouring MATH induces a colouring MATH for which MATH are isomorphic in MATH for all triangles MATH. The partition function REF defined using MATH agrees with that one defined using MATH because the weights satisfy MATH and because MATH according to REF . |
hep-th/0106029 | The calculation is described in diagrammatic language and can be translated into equalities for morphisms of the ribbon category MATH as described in REF. First, a number of coupons are moved around in the plane without twisting or braiding any ribbons: Move the coupon MATH to the left and place it above MATH, then move MATH down and to the right and place it below and right of MATH. Move MATH to the right and place it below MATH and below and left of MATH. Rotate the coupon MATH by MATH degrees in order to place its ribbons as depicted in REF . Finally, lift the ribbon labelled MATH out of the plane, move it across the entire diagram, and place it as shown in REF . |
hep-th/0106029 | The proof is again explained diagrammatically: Start with REF . Lift the ribbon MATH out of the plane and move it across the coupon MATH so that MATH now over-crosses MATH, MATH and MATH rather than MATH. Then the coupons can be moved around in the plane without introducing twists or braidings such that the configuration in REF is obtained. |
hep-th/0106029 | Consider REF whose quantum trace agrees with MATH of REF according to REF . The linear isomorphisms MATH of REF can now be used to replace the dashed boxes of REF by morphisms of the state spaces MATH with a different vertex order. The result is very similar to the trace MATH of REF for the MATH-simplex MATH. Observe, however, that in REF the arrows of the ribbons corresponding to the triangles MATH, MATH and MATH are reversed compared with REF . We can reverse these arrows in REF if we label them instead by MATH, MATH and MATH, respectively. Consider the triangle MATH. If MATH is an even permutation of the standard vertex order, then REF contains MATH for some object MATH, that is, upon reversal of the arrows this label changes to MATH. This is the same label as the label arising in MATH. If, however, MATH is an odd permutation of the standard vertex order, then REF contains MATH for some object MATH, that is, upon arrow reversal this becomes MATH. This is in general not identical, but still isomorphic to MATH which arises in MATH in this case. This is the reason why the present lemma holds only for a colouring MATH with isomorphic objects at all triangles. Let MATH denote any triangle in standard vertex order, MATH. Define the colouring MATH by MATH if MATH and by MATH for the other triangles. Then the quantum trace of REF with arrows MATH, MATH and MATH reversed, agrees with the trace of REF for the colouring MATH. The following diagram therefore commutes: Using REF , this implies the commutativity of the diagram claimed in the present lemma. |
hep-th/0106029 | Consider REF whose quantum trace agrees with MATH of REF according to REF . The linear isomorphisms MATH and MATH of REF can now be used to replace the dashed boxes of REF by morphisms of the state spaces MATH with a different vertex order. The result is the trace MATH of REF for the MATH-simplex MATH up to the choice of isomorphic objects for the triangles MATH, MATH and MATH. These isomorphisms arise from double dualization as in REF . The following diagram therefore commutes: Employing REF , this proves the claim. |
hep-th/0106029 | Equip the set of vertices with a different linear order which is induced from the given one by a permutation MATH of the vertices. The partition function using this new order can be expressed in terms of the definitions of REF which use the original order, if MATH is applied both to the vertices and to the colouring, MATH . Here MATH replaces the partition function per colouring in the case of the new vertex order, MATH denotes MATH for a given MATH-simplex MATH, and MATH is the colouring induced by MATH, that is, MATH for all triangles MATH. The permutation MATH replaces triangles MATH by MATH and therefore just permutes the factors of the product in REF. This product can be reorganized so that we obtain MATH where the vertex order of the triangles does not matter because of the reality REF . Any permutation MATH which just permutes the MATH-simplices but does not change the vertex order of these MATH-simplices, permutes the tensor factors in REF and therefore leaves the trace invariant. It is thus sufficient to prove invariance under permutations MATH that change the vertex order for fixed MATH-simplices. Consider a MATH-simplex MATH and and let MATH be an elementary transposition, MATH. The colouring MATH associates with each triangle MATH either the object MATH assigned to some triangle MATH or the dual of that object. Since the set of colours MATH contains for each given object MATH exactly one object that is isomorphic to MATH, there exists a unique colouring MATH such that MATH for all triangles. Moreover since MATH is a transposition, MATH so that MATH induces an involution on the set of colourings MATH. We can now sum over MATH rather than MATH in REF and obtain MATH where we have used REF and where we have written MATH instead of MATH for simplicity. In the preceding lemmas we have constructed linear isomorphisms which form the vertical maps in commutative diagrams of the following form: Here the MATH are suitable state spaces such that the bottom horizontal map is related to the MATH-simplex map MATH by the pairing REF. The colouring MATH is such that MATH for all triangles MATH. Since each tetrahedron occurs twice in the boundary of some MATH-simplices, once with positive and once with negative relative orientation, each state space MATH occurs twice among the MATH, once as MATH and once as the dual state space MATH. In both cases, the corresponding map MATH is the same, either one of the MATH or the identity. Therefore the tensor product of all MATH-simplex maps is conjugated by a linear isomorphism MATH which can be obtained from a tensor product of these MATH, MATH . Observe that here MATH and that the colouring MATH can be replaced by MATH as a consequence of REF . Since MATH is the trace over REF, we find MATH and therefore MATH. |
hep-th/0106064 | The local normality follows directly from the limit construction via a sequence of normal states. Positiveness of MATH is equivalent to the proposition CITE: for any sequence of test functions MATH, MATH being MATH functions defined for MATH and with MATH except for a finite number of MATH's, it holds the inequalities MATH . But, if we define the functions MATH we can write the above sum as MATH which turns out to be non-negative as a result of vacuum positivity. From a similar expression, we can see MATH is faithful on MATH since vacuum is faithful on MATH, note that MATH and MATH . Then MATH . |
hep-th/0106122 | Define the space MATH and a surjection MATH given by MATH. This gives rise to the sequence MATH . We give MATH the tensor product coproduct structure, the subalgebra structures of MATH and MATH and the cross relations induced by the pull-back of the cotriangular structure of MATH. This makes REF into a sequence of cotriangular NAME algebras. (Note that the coquasitriangular structure on MATH is thus trivial on the first component, hence the notation without the prime.) Now consider the injection of cotriangular NAME algebras MATH given by MATH. The surjection MATH inverts MATH. Thus, according to REF there is a braided NAME algebra MATH in the category of left MATH-comodules so that MATH. We can recover MATH (as an algebra) as MATH and observe that on this space the map MATH restricts to the subspace MATH of MATH. Identifying MATH as MATH we obtain precisely a surjection MATH as required. It remains to show that MATH is the spin-statistics reduction MATH of MATH. For this observe that th REF implies MATH. This in turn implies for an element MATH in MATH that MATH by composition with MATH. Thus MATH and it follows that MATH. On the other hand clearly MATH and thus MATH. This completes the proof. |
hep-th/0106194 | CASE: This follows from applying REF which shows that MATH is isomorphic to the semi-group of all vector bundles on REF If we pull a bundle back from MATH to MATH and tensor it with a bundle gerbe module the result is still a bundle gerbe module. CASE: If MATH is a bundle gerbe module for MATH and MATH is a bundle gerbe module for MATH it is straightforward to see that MATH defines a bundle over the fibre product of MATH and MATH which is a bundle gerbe module for MATH. CASE: This follows easily by pull-back. |
hep-th/0106194 | Let MATH be the image of MATH under the map MATH were we contract the two copies of MATH with the inner product. Then MATH. So take MATH. |
hep-th/0106194 | MATH . This is just the reduced version of NAME 's definition of twisted MATH-theory REF. MATH . This is a standard construction. MATH . Notice that the MATH bundle MATH is MATH covariant. It follows that if we pull it back to MATH by a MATH equivariant map MATH that we must get a MATH covariant bundle on MATH. MATH . Let MATH be a MATH covariant MATH bundle. By composing the projections MATH we think of MATH as a bundle on MATH. Both groups MATH and MATH act on MATH and the combined action is an action of the semi-direct product and realises MATH as a bundle over MATH for this semi-direct product. Conversely consider a bundle MATH for the semi-direct product for which the induced MATH bundle is isomorphic (as a MATH bundle) to MATH. Identify this bundle with MATH and hence MATH is a bundle over MATH and, in fact, a MATH covariant MATH bundle. MATH A MATH bundle over MATH is determined by a classifying map MATH. A little thought shows that we can realise this latter space as MATH which fibres over MATH. The composition MATH of MATH with the projection to MATH is the classifying map of the induced MATH bundle which is MATH. This means that we can find a MATH equivariant map MATH covering MATH. Using this we define MATH by MATH. This is well-defined. Moreover if MATH then MATH so that MATH and hence MATH proving equivariance. |
math-ph/0106002 | Let MATH. Note that MATH is an ideal of MATH. Since MATH is simple, by the above discussion either MATH or MATH is invertible, that is, MATH is a skew-field. Now one can argue as in REF . |
math-ph/0106002 | The proof is the same as in the ``even" case, see REF , and REF . |
math-ph/0106002 | As in the ``even" case, see REF . |
math-ph/0106002 | As in the ``even" case, see REF . |
math-ph/0106002 | As in the ``even" case, see REF . |
math-ph/0106002 | We can apply the NAME REF to MATH and REF in order to use the same argument as in REF . |
math-ph/0106002 | By CITE, Corollary to REF , MATH is isomorphic to MATH as a topological NAME superalgebra. Since MATH is open, MATH. The fact that a linearly compact space can be identified with its double dual implies that MATH is isomorphic to MATH. |
math-ph/0106002 | If we choose the MATH-th member MATH REF of the canonical filtration associated to MATH as another fundamental subalgebra, say MATH, then, by definition, the associated canonical filtration is MATH. Hence this change of the fundamental subalgebra does not affect the growth. Now, if MATH is another fundamental subalgebra of MATH and MATH is the associated canonical filtration, then, by NAME 's principle, MATH for sufficiently large MATH, hence MATH for all MATH. Therefore, MATH. Exchanging the roles of MATH and MATH we get the opposite inequality. |
math-ph/0106002 | If MATH contains an open subalgebra MATH such that all MATH are open subspaces of MATH (then this is automatically an algebra filtration), let us denote by MATH the growth of this filtration. Since the growth is invariant under rescaling, we see (using NAME 's principle as in the proof of REF ) that MATH is independent of the choice of MATH. Furthermore, we clearly have: MATH if we choose MATH to be a fundamental subalgebra of MATH. Furthermore, it follows from the classification of simple linearly compact NAME superalgebras CITE that MATH has a NAME filtration MATH for which the associated graded NAME superalgebra MATH has the property that the space MATH generates MATH (in fact, in all cases except for MATH, one has MATH, MATH), hence MATH . On the other hand, for the canonical filtration MATH associated to MATH one has MATH . It follows from REF that MATH. Therefore, by REF , the growth of any algebra filtration of MATH is equal to MATH. |
math-ph/0106002 | Using NAME Lemma (see REF ), one can apply the same argument as in REF . |
math-ph/0106002 | Let MATH be the maximal integer such that MATH. Then there exists MATH such that MATH. The automorphism MATH is well-defined and converges uniformly on MATH, and we have: MATH . By repeating this argument, we obtain MATH in the limit. |
math-ph/0106002 | Let MATH be an even surjective derivation of MATH. It clearly transforms MATH surjectively into itself. Moreover, MATH leaves the radical MATH of MATH invariant, hence it induces a derivation of MATH which is not inner because it is surjective. On the other hand, MATH is a semisimple NAME algebra, so that every derivation is inner. Consequently, MATH is solvable, but this in turn implies (see REF ) that MATH is solvable. |
math-ph/0106002 | Let MATH. Then MATH is a MATH-basis of the free MATH-module MATH and MATH is spanned by the set MATH. We have a canonical surjective map MATH, and since the images of the MATH's are linearly independent in MATH, the kernel of this map is zero. |
math-ph/0106002 | If MATH is a MATH-cocycle on MATH, it can be written in the form MATH, where MATH is an endomorphism of the space MATH. One can easily check that the skew-symmetry of MATH is equivalent to the skew-symmetry of MATH, that the cocycle equation is equivalent to MATH being a derivation and that MATH is a trivial cocycle iff MATH is an inner derivation. |
math-ph/0106002 | The NAME superalgebras MATH, MATH, MATH, MATH, MATH, MATH and MATH are contragredient with a symmetrizable NAME matrix, hence they have an even invariant form CITE. The obvious pairing between the even and the odd part of MATH is an odd invariant form. Define on MATH the NAME bracket MATH . This NAME superalgebra has a bilinear form MATH (where MATH denotes the NAME integral, compare CITE), which is invariant on the derived subalgebra MATH that is, the span of all monomials except the top one. The kernel of the restriction of the form to MATH is MATH. Since MATH, the proposition is proved in this case as well. In order to show that in the remaining cases there is no invariant form on MATH, we take a maximal reductive NAME subalgebra MATH of MATH and show that the MATH-modules MATH and MATH are not isomorphic. |
math-ph/0106002 | In the cases when MATH carries an invariant, supersymmetric, non-degenerate bilinear form, we apply REF and the known description of derivations (see REF ). Notice that in all cases except MATH all derivations are skew-symmetric, whereas for MATH, MATH, only one of the two outer derivations, namely MATH, has this property. In the remaining cases, that is, MATH, MATH, MATH and MATH, we apply REF to compute MATH. |
math-ph/0106002 | Note that all NAME differentials of MATH are exact. Hence one can use the same argument as in CITE. |
math-ph/0106002 | This follows from REF . |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.