paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/0102007 | For each vertex MATH of MATH, let MATH be the deepest allocation node of MATH in MATH. In MATH time, we can compute MATH for all vertices MATH of MATH. For a set MATH, if a pole of MATH is in MATH, then MATH is the root of MATH; otherwise, MATH is the least common ancestor of all MATH with MATH. So, MATH can be computed in MATH time. Let MATH be the deepest one among all MATH with MATH. We can compute MATH in MATH time. Since MATH is the least common ancestor of all MATH with MATH, it can be computed in MATH time. Thus, in MATH total time, we can compute MATH and MATH for all MATH in MATH and all MATH in MATH. Afterwards, in MATH total time, we can compute MATH and MATH for all nodes MATH of MATH. |
cs/0102007 | REF are straightforward. REF holds since each MATH is processed only when the node MATH with MATH is processed. Each MATH is processed once when the node MATH with MATH is processed and once when the node MATH with MATH is processed. When MATH is processed, we perform some of Operations REF through REF on it. Since an operation takes MATH time, REF holds. |
cs/0102008 | CASE: MATH. By MATH, we know MATH. Since MATH contains MATH unbeatable bids, the lemma is proved. CASE: MATH. Let MATH consist of the nonzero bids in MATH. It suffices to show that MATH wins no more than MATH bids in MATH on average. By REF , MATH has an optimal algorithm MATH with MATH. Clearly, for each bid MATH, if MATH is the largest index with MATH, then MATH wins MATH bids in MATH on average. Hence, the unit price for MATH to win a bid in MATH is greater than MATH. By MATH and MATH, the expected number of bids in MATH that MATH wins is an integral multiple of MATH. Since the budget of MATH is MATH, the expected number of bids in MATH that MATH wins is at most MATH. |
cs/0102008 | Let MATH be the MATH-set that satisfies Property P. By Property P REF, the MATH-element set MATH is well defined. By Property P REF, MATH. Therefore, there exists a positive number MATH such that MATH satisfies MATH and MATH. Since each bid in MATH is greater than MATH, MATH wins all MATH bids in MATH. By Property P REF, the expected number of bids in MATH that MATH wins with MATH is at least MATH. Thus, MATH. |
cs/0102008 | Let MATH. For each MATH, let MATH. Define MATH, for any integer MATH. Let MATH. Clearly, MATH holds for each MATH. By MATH, we know MATH. Let MATH . CASE: Clearly, the inequality MATH holds for each MATH. By averaging this inequality over all MATH values of MATH, we have MATH. One can easily verify that the statement holds with MATH. CASE: If MATH (respectively, MATH), then one can easily verify that the statement holds with MATH (respectively, MATH). If MATH, then MATH, and thus the statement holds with MATH. If MATH, then MATH, and thus the statement holds with MATH. It remains to prove the statement for the case MATH. Clearly, MATH and MATH. If MATH, then the statement holds with MATH. If MATH, then the statement holds with MATH. Now we assume MATH and MATH. If MATH is even, then clearly MATH. One can verify that the statement holds with MATH. If MATH is odd, then clearly MATH. If MATH, then let MATH; otherwise, let MATH. Clearly, MATH and MATH. One can verify that the statement holds with MATH. CASE: If MATH, then the statement holds with MATH. If MATH, then the statement holds with MATH. If MATH, then the statement holds with MATH. It remains to consider the case that both MATH and MATH hold. If MATH and MATH, then, by MATH, the statement holds with MATH. When either MATH or MATH holds, we show MATH, which implies that the statement holds with MATH. If MATH, then MATH holds trivially. If MATH, then, by MATH, we know MATH. By MATH and MATH, we have MATH, and thus MATH. CASE: If there is a MATH such that MATH, then the statement holds with MATH. If there is a MATH such that MATH, then, by MATH, the statement holds with MATH. If no such MATH or MATH exists, then we have MATH, a contradiction. CASE: By MATH and REF , there is a MATH-set MATH with MATH. We show that the statement holds with MATH. By the proof for REF , MATH is a MATH-set not containing REF. Thus MATH is a MATH-set. Clearly, MATH, MATH, and MATH. |
cs/0102008 | CASE: Assume for a contradiction that MATH holds for some MATH. By MATH, we know MATH. It suffices to show MATH as follows. If MATH is even, then, by MATH, we know MATH. By REF , MATH, and MATH, we have MATH. If MATH is odd, then, by MATH, we know MATH. By REF , MATH, and MATH, we have MATH. CASE: By REF , and REF , we have MATH. Therefore, MATH. By MATH, MATH, and MATH, we have MATH, and thus MATH. |
cs/0102008 | Let MATH be the number of bids in MATH that are less than MATH. If MATH, then the expected number of bids that MATH wins with MATH is at least MATH, ensuring MATH. The rest of the proof assumes MATH. By MATH, we have MATH. By MATH, we have MATH. By MATH, we know MATH, which implies MATH. Let MATH. Let MATH. Clearly, MATH. Let MATH, where MATH is a number such that MATH and MATH. Since MATH, MATH, and the expected number of bids that MATH wins with MATH is at least MATH, the lemma is proved. |
cs/0102008 | By MATH, we know MATH and MATH. By REF and MATH, it suffices to show a MATH-set with at most MATH elements. If MATH, then, by MATH, MATH is a required MATH-set, where MATH. The rest of the proof assumes MATH. Since MATH is an integral multiple of MATH, we know MATH. By MATH and REF , we know MATH, where MATH and MATH. By REF , we know that MATH is a MATH-set with MATH. It remains to consider the case MATH. By MATH, we have MATH, and thus MATH. By MATH, we know MATH. It follows that MATH, where MATH. By REF , MATH is a MATH-set with MATH. |
cs/0102008 | For notational brevity, the proof omits the subscript of MATH. By MATH, we know MATH, MATH, MATH, and MATH. We first show MATH as follows. Let MATH. Clearly, MATH, and thus MATH is an integral multiple of MATH. Therefore, it suffices to show MATH. By MATH and REF , it suffices to show a MATH-set or a MATH-set that satisfies Property P for each of the following cases. CASE: MATH. Let MATH. By REF , we know that MATH is a MATH-set with MATH, proving Property P REF. Being a MATH-set, MATH satisfies Property P REF and MATH. By MATH, MATH, and MATH, we know MATH. Therefore, MATH satisfies Property P REF MATH. Let MATH. By REF , MATH, and MATH, we have MATH. By REF , we have MATH, proving Property P REF. By REF , we have MATH, proving Property P REF. By MATH, MATH, and REF , we know MATH, proving Property P REF MATH. By MATH, we have MATH. By MATH and REF , we have MATH. By MATH, we have MATH. It follows from MATH and MATH that MATH and MATH. By MATH, we know MATH. By REF , and REF , we have MATH. Therefore MATH. By MATH and MATH, we have MATH, and thus MATH. It follows from MATH, REF , and REF that MATH. We prove the statement for the following two sub-cases. CASE: MATH. Let MATH. By REF , we know that MATH is a MATH-set with MATH, satisfying Property P REF. Being a MATH-set, MATH satisfies Property P REF and MATH. By MATH, MATH, and MATH, we know MATH, satisfying Property P REF. CASE: MATH. Let MATH. Let MATH. By the proof of REF , we know MATH. Therefore, MATH. By MATH, we know MATH. By MATH and REF , one can verify that each of MATH and MATH satisfies Properties P REF and P REF. It remains to show that either MATH or MATH satisfies Property P REF as follows. If MATH, then, by MATH and REF , we know MATH. Thus MATH satisfies Property P REF. Now we assume MATH. By MATH, we know MATH, and thus MATH. It follows from MATH and REF that MATH. Thus MATH satisfies Property P REF. |
cs/0102008 | Clearly, MATH holds trivially. By REF , and REF, we know that MATH holds for any bid set MATH of MATH. Therefore, we have REF , and thus the equality MATH. |
cs/0102008 | Straightforward. |
cs/0102009 | Straightforward. |
cs/0102009 | The first statement follows from the fact that MATH has a maximum legal matching that contains a legal pair between MATH and MATH. The second statement follows from the fact that a maximum legal matching can be obtained by iteratively applying any applicable rule below: CASE: If there are one unpaired type MATH pendant block and one unpaired type MATH pendant block, then we pair a type MATH pendant block and a type MATH one. CASE: If there is no unpaired type MATH (respectively, MATH) pendant block and there are one unpaired type MATH (respectively, MATH) pendant block and one unpaired type MATH pendant block, then we pair a type MATH (respectively, MATH) pendant block with a type MATH one. CASE: If all unpaired pendant blocks are type MATH, then we pair two such blocks. |
cs/0102009 | REF is straightforward. To prove REF , it suffices to show MATH and MATH. Let MATH be an optimal biconnector of MATH. To prove MATH, note that MATH is empty. Thus, every block in MATH contains an endpoint of an edge in MATH. Since all the edges in MATH are legal, MATH contains at least MATH edges. To prove MATH, we need such a MATH that the non-block connected components of MATH are all contained in the same connected component of MATH. If a given MATH has not yet satisfied this property, then let MATH and MATH be two non-block connected components of MATH that are contained in two different connected components MATH and MATH of MATH, respectively. Let MATH and MATH be two edges in MATH. Such MATH and MATH exist because MATH and MATH are not biconnected in MATH, but MATH and MATH are biconnected in MATH. Next, let MATH and MATH. Then, MATH remains an optimal biconnector of MATH. Also, MATH connects MATH and MATH, which include MATH and MATH. By repeating this endpoint switching process, we can construct a desired MATH. With such a MATH, we proceed to prove MATH. Since this claim trivially holds if MATH is componentwise biconnected, we focus on the case where MATH is not componentwise biconnected. Then, MATH is maximized by some MATH that is in a non-block connected component MATH. Let MATH be the connected component of MATH containing MATH. Since MATH is componentwise biconnected, MATH is connected. Then, because MATH has MATH connected components, MATH, proving our claim. |
cs/0102009 | By REF , MATH. For ease of understanding, the proof for MATH is delayed to REF. |
cs/0102009 | CASE: Let MATH be the connected component of MATH that is neither an isolated edge nor a block. REF applies to the case where MATH contains at least two vertices in MATH and at least two in MATH. Thus, we may assume without loss of generality that MATH contains exactly one vertex MATH and MATH vertices MATH with MATH. Note that MATH. Because MATH and MATH, there is an isolated vertex MATH or there is a nonsingular block in MATH containing two vertices MATH. In the former case, MATH is an optimal biconnector; in the latter case, MATH is an optimal biconnector. CASE: Since MATH, we may assume without loss of generality that all the pendant blocks are type MATH. Note that MATH, MATH and MATH. Let MATH be the connected components of MATH that are neither isolated edges nor blocks. Since each MATH has more than two vertices, MATH has a vertex MATH. Let MATH be the pendant blocks of MATH. Each MATH contains a noncut vertex MATH. The set MATH is a biconnector. By REF , this biconnector is optimal. CASE: By REF , MATH. To prove the upper bound, let MATH be a legal edge of MATH. Let MATH. We first show how to choose MATH so that MATH. Since MATH, by REF , we can find a legal pair MATH and MATH in different connected components with MATH. By REF , let MATH be a binding edge for MATH and MATH. Note that MATH, MATH, MATH, MATH and MATH. Thus, MATH. This process reduces MATH and MATH by MATH each. We iterate this process until either REF MATH or REF MATH and MATH. In the latter case, we use REF to complete the proof. In the former case, note that we add an edge to combine two non-singular non-biconnected connected components into a connected component. This new connected component is neither an isolated edge nor a block. Thus, MATH; that is, MATH and MATH in the resulting MATH. We then use REF to complete the proof of this case. CASE: Let MATH be the isolated edge. Let MATH and MATH be two isolated vertices. Then, MATH is an optimal biconnector of MATH. CASE: Let MATH be a connected component that is a nonsingular block in MATH. MATH has a vertex MATH and a vertex MATH. Let MATH be the isolated edge of MATH. Then, MATH is an optimal biconnector of MATH. CASE: This case is straightforward. |
cs/0102009 | The proof is straightforward and similar to that for similar constructs CITE. |
cs/0102009 | The proof is straightforward and similar to those for similar constructs CITE. |
cs/0102009 | The proof follows from REF , the inequality MATH, and basic counting arguments for trees. |
cs/0102009 | By REF , we divide the proof into the following five cases. The first case is discussed in REF ; the other cases are proved in REF - REF, respectively. CASE: MATH. CASE: MATH and MATH. CASE: MATH, MATH, and MATH has two critical vertices. CASE: MATH, MATH, and MATH has no massive vertex and at most one critical vertex. CASE: MATH, MATH, and MATH has exactly one massive vertex. |
cs/0102009 | Straightforward. |
cs/0102009 | Let MATH. Since MATH, MATH by REF . It suffices to construct a biconnector MATH of MATH edges for MATH. Let MATH be the pendant blocks of MATH. Since MATH, MATH and we may assume MATH without loss of generality. Then, MATH has a cut edge MATH for each MATH, where MATH. Since MATH and MATH, there is some MATH. Let MATH be the connected component of MATH containing MATH. Let MATH be the set of legal edges MATH for all MATH and MATH for all MATH. It is straightforward to prove that MATH is as desired by means of REF . |
cs/0102009 | REF follows REF follows from basic counting arguments for trees. REF follows from the first two and REF . |
cs/0102009 | We add to MATH a binding edge for each legal pair in the maximum legal matching of REF . By REF , the resulting graph is biconnected. We add MATH edges, which by REF is optimal. |
cs/0102009 | Let MATH be the vertex in MATH of degree at least three. There are two cases: CASE: MATH is a MATH-vertex. Then, MATH. CASE: MATH is a MATH-vertex. Since MATH is not massive, MATH and MATH. In either case, let MATH be a maximum legal matching of MATH; next, let MATH be a set of legal pairs formed by pairing each pendant block not yet matched in MATH with one already matched. Then, MATH is a set of the smallest number of legal pairs of MATH such that each element in MATH is in a pair. We add to MATH a binding edge for each pair in MATH. Since MATH, we add MATH edges. Since MATH in REF and MATH in REF , these edges form a desired biconnector by REF . |
cs/0102009 | Let MATH be the current root of MATH. There are three cases. CASE: MATH is not critical, and either MATH is of degree two and no MATH-branch is a chain or MATH is of degree at least three. We set MATH. CASE: MATH is not critical, MATH is of degree two, and a MATH-branch is a chain. Note that MATH has a vertex MATH of degree at least three. We set MATH. CASE: MATH is critical. Since MATH, MATH is of degree three or more. We set MATH. |
cs/0102009 | There are two cases. CASE: The degree of MATH is two and no MATH-branch is a chain. Let MATH be a MATH-branch. CASE: The degree of MATH is at least three. Since this is Case SREF, some descendant of MATH has degree at least three. Let MATH be the MATH-branch containing that descendant. Let MATH be the set of leaves in MATH. Let MATH. By REF , there exist a legal pair MATH and MATH with MATH. Then, MATH contains MATH as desired. Furthermore, in REF contains a vertex of degree at least three in MATH and another in MATH; in REF itself is of degree at least three, and MATH contain a vertex of degree at least three in MATH. In both cases, MATH is as desired. |
cs/0102009 | We use REF to reroot MATH, use REF to pick a legal pair MATH and MATH, and then add a binding edge for this pair to MATH. By REF , MATH. By REF , MATH. Hence MATH. There are two cases. CASE: MATH has no critical vertex. Then, by REF , MATH. CASE: MATH has a critical vertex. Then, MATH is the critical vertex and MATH. By REF , and REF , MATH. Hence MATH remains to be a critical vertex. In either case, MATH. Then, MATH. Also, MATH has no massive vertex and thus satisfies Case SREF, SREF, SREF or SREF. |
cs/0102009 | For Case SREF, we use REF . For Case SREF, we add one edge to MATH at a time using REF until the resulting graph MATH does not satisfy Case SREF. By REF , MATH satisfies Case SREF, SREF, SREF or SREF. Thus, we apply REF , or REF to MATH accordingly. By REF , the number of edges added is MATH. |
cs/0102009 | CASE: This statement follows from the definition of Case SREF. CASE: Let MATH be the number of MATH-chains. Then, MATH and MATH. So MATH. Let MATH. Because MATH is massive, MATH. Note that MATH. Thus MATH. CASE: Let MATH be a MATH-chain. Let MATH be the leaf of MATH in MATH. Because MATH, MATH contains a leaf MATH that forms a legal pair with MATH. Let MATH be the MATH-branch that contains MATH. Then, MATH, MATH, MATH and MATH are as desired. |
cs/0102009 | Let MATH, MATH, MATH and MATH be as stated in REF . The added edge is a binding edge for MATH and MATH. By REF , the b-vertices and c-vertices on MATH form a new block MATH in MATH. MATH may or may not be a leaf in MATH; in either case, MATH. Note that MATH contains MATH. Thus, by REF , MATH remains a cut vertex in MATH with MATH while MATH for all vertices MATH. Consequently, MATH. |
cs/0102009 | We add one edge to MATH at a time using REF until the resulting graph MATH satisfies Case SREF, SREF, SREF or SREF. Thus, we apply REF or REF accordingly. By REF , MATH edges are added. To implement this proof in linear time, we first define a data structure as follows. Let MATH be the set of leaves of MATH that are in the MATH-chains. We set up a counter for the number of these leaves. We also set up three doubly linked lists containing those of them that are types MATH, MATH, and MATH, respectively. We set up a counter for the number of MATH-branches that are not chains. For each such branch, we set up a doubly linked list for the leaves of MATH in it. We also set up three doubly linked lists for the leaves in these branches that are types MATH, MATH, and MATH, respectively. Given MATH, we can set up these linked lists and counters in linear time. We next use this data structure to find a legal pair MATH and MATH by means of REF . Since MATH by REF , there are two cases. CASE: Some MATH and MATH form a legal pair. This is our desired pair. Note the MATH-chains containing MATH and MATH in MATH are contracted into a new chain in MATH consisting of a single leaf of type MATH. CASE: MATH contains only type MATH or MATH leaves. Select any MATH. Since MATH, some MATH forms a desire legal pair with MATH. Note that MATH and MATH are no longer pendant blocks in MATH and the newly created block is not a pendant block of MATH, either. The MATH-branch containing MATH becomes a chain if in MATH it contains exactly two pendant blocks. It takes MATH time to decide which of these two cases holds. In either case, the selection of MATH and MATH takes in MATH time using the linked lists. Once MATH and MATH are found, we can find a binding edge in MATH time in a straightforward manner. After the edge is added to MATH, we can update the data structure in MATH time for MATH. Then, we use REF and the counters to check whether MATH satisfies REF in MATH time. We repeat this process until MATH does not satisfies Case SREF. At this point, we complete the reduction. Since we iteratively add at most MATH edges in Case SREF, the reduction takes linear time. |
cs/0102009 | CASE: We implement the proof REF using the following steps. CASE: Use REF of MATH to find MATH. CASE: Use REF of MATH and REF to decide which case of the proof of REF holds. CASE: For REF of the proof of REF , set MATH and MATH is unchanged. CASE: For REF of the proof of REF , first use REF of MATH to find the nearest desired descendant MATH of MATH and then reroot MATH at MATH and update it accordingly. CASE: For REF of the proof of REF , if MATH, then recompute MATH from MATH to root at MATH; otherwise, MATH, and MATH is unchanged. Since REF take MATH time, the the time complexity of each case of this statement is bounded by that of REF MATH is critical. REF runs with MATH in MATH time. CASE: MATH is not critical but MATH is. REF runs with MATH in MATH time. CASE: Neither MATH nor MATH is critical. Then, REF or REF is performed. REF takes MATH time. For REF , the search for MATH takes MATH time per vertex on MATH. Since the internal vertices of MATH all have degree two, updating REF of MATH along this path takes MATH time per vertex. REF of MATH outside this path and the other two items remain the same. Thus, this case takes MATH total time as desired. CASE: We implement the proof of REF using the following steps. CASE: Use REF of MATH to decide which case of the proof of REF holds. CASE: Use REF of MATH and REF to find all possible pairs of types MATH and MATH such that MATH has a maximum matching that contains a legal pair between type MATH and type MATH. CASE: For each such pair of MATH and MATH, perform the following computation until MATH and MATH are found. CASE: For REF of the proof of REF , MATH and MATH are in the two branches of the root of MATH separately. Use REF of MATH at the root to decide whether the desired MATH and MATH exist. If they exist, use REF of MATH to search for them. CASE: For REF of the proof of REF , MATH and MATH are in two separate branches of the root, of which one is not a chain. The remaining computation is similar to that of REF . By REF , some pair MATH and MATH yields the desired MATH and MATH. REF take MATH time. There are MATH possible pairs of MATH and MATH. For each such pair, checking the existence of MATH and MATH takes MATH time. If they exist, searching for them takes MATH time per vertex on the path MATH. |
cs/0102009 | Given MATH in REF as input, the reduction algorithm is as follows: CASE: Construct MATH. CASE: repeat CASE: Use REF to reroot MATH. CASE: Use REF to find a legal pair MATH and MATH. CASE: Add a binding edge MATH for MATH and MATH into MATH. CASE: Use REF to update MATH while rerooting it at the new b-vertex MATH resulting from the insertion of MATH. until MATH does not satisfy Case SREF. Since REF takes MATH time, it suffices to prove that REF takes MATH time. By REF , each iteration of REF reduces MATH by two. Since MATH, the repeat loop has less than MATH iterations. Then, since the until condition can be checked in MATH time per iteration using REF of MATH, the until step takes MATH total time. Similarly, REF takes MATH time per iteration and MATH total time in a straightforward manner. We next show that REF also take MATH total time. For a given iteration, let MATH and MATH denote MATH before and after MATH is inserted, respectively. CASE: We show that each case in the proof of REF takes MATH total time as follows. CASE: This case takes MATH time per iteration and thus MATH total time. CASE: By REF , this case can only happen once in the above augmentation algorithm. Hence, this case takes MATH total time. CASE: This case takes MATH time per edge on MATH for an iteration. Note that the degree of a vertex in MATH never increases by edge insertion. Then, since MATH is rooted at MATH with MATH connecting two leaves of MATH, each edge on MATH is traversed only once to reroot MATH for this case throughout all the iterations. Therefore, this case takes MATH total time. CASE: This step takes MATH time per iteration. Since there are MATH iterations, by REF , this step takes MATH total time. CASE: We bound the time for updating each item of MATH as follows. REF of MATH. Notice that MATH passes through the root of MATH. Also, MATH is rooted at MATH. These properties make it straightforward to update this item in MATH time per iteration. Since there are MATH iterations, by REF , this step takes MATH total time. REF of MATH. By REF , MATH. Thus it takes MATH time to update this item per iteration and MATH total time. REF of MATH. Let MATH be a c-vertex in MATH. If MATH, it has the same degree in MATH and MATH and is not relocated in this item. If MATH, its degree reduces at most REF in MATH and can be relocated in MATH time. Therefore, this item can be updated in MATH time per iteration, that is, MATH total time as shown for REF . |
cs/0102010 | Straightforward. |
cs/0102010 | Straightforward based on the assumption that MATH has no duplicates. |
cs/0102010 | CASE: To prove by contradiction, suppose that MATH contains a cycle MATH. By the construction of MATH, MATH must be of the form MATH where MATH; MATH; MATH; and MATH. By definition, if MATH is a path in MATH, then MATH and MATH overlap by MATH in any valid permutation of MATH. Thus, for MATH, the existence of the subpath MATH of MATH in MATH means that MATH overlaps with MATH and MATH and MATH overlaps with MATH and MATH. To enable both MATH and MATH overlap with MATH, MATH must be in the middle of MATH and MATH for MATH. Consequently, for MATH, MATH is in the middle of MATH and MATH, which is impossible. CASE: For any diameter MATH of MATH, we show that every subtree MATH hanged on MATH must be a dangler. First, MATH must be hanged on MATH at a node in MATH. Otherwise, if MATH is hanged on MATH at a node MATH, MATH has degree greater than MATH, contradicting REF . Then, MATH has more than one node because the root of MATH is a node in MATH and must be of degree MATH. If MATH cannot have more than MATH nodes, REF follows. To prove by contradiction, suppose that MATH has more than two nodes. Without lost of generality, assume that MATH is hanged on MATH at a node MATH and the root of MATH is a node MATH. Note that MATH has another neighbour, say MATH, from MATH. If MATH contains more than two nodes, MATH must has a child, say MATH, from MATH and MATH must has a child, say MATH, from MATH. Thus, MATH must have a root-to-leaf path of length more than MATH. Then, the two paths from MATH to both ends of MATH must be of length more than MATH. Otherwise, MATH cannot be a diameter of MATH. From those observations, MATH has the pattern shown in REF . According to the pattern, MATH and MATH overlap with MATH. Therefore, in any valid permutation, one of MATH and MATH, say MATH, must be in the middle of the other two fragments and MATH can only overlap with MATH. However, according to the pattern in REF , for MATH, MATH overlaps with another fragment MATH, reaching a contradiction. |
cs/0102010 | For each MATH, if MATH contains only one element, then the lemma follows. Otherwise, MATH is of degree at least MATH. Then, MATH must be on the diameter MATH. Let MATH and MATH be elements in MATH which are the two neighbours of MATH on MATH. The remaining nodes in MATH must be located in the danglers hanged on MATH at MATH. By dangler-first search, all the elements in MATH must form a consecutive subsequence in MATH. By symmetry, for each MATH, the elements in MATH must form a consecutive subsequence in MATH. |
cs/0102010 | Suppose the lengths from MATH and MATH are plotted on the same line according to the order given by MATH, MATH and MATH, respectively. Consider the stripes formed from MATH and MATH. By REF , for each MATH, MATH overlaps with MATH for all MATH. By symmetry, for each MATH, MATH overlaps with MATH for all MATH. Then, by the definition of the EDD problem, MATH is a valid permutation. |
cs/0102010 | The only-if part follows from REF . The if part follows from REF . |
cs/0102010 | First, by REF , NAME is correct. As for its time complexity, REF requires MATH time as MATH contains MATH edges and we can find each edge in MATH time. REF checks whether MATH satisfies the two properties in REF . For REF , we can determine whether a graph is a tree in MATH time. For REF , we can compute a diameter of a tree in linear time first, then, we verify whether MATH satisfies REF by detecting whether the subtrees hanged on the diameter are danglers. Thus, REF requires MATH time. REF finds MATH using dangler-first search. Since the search scans every node in MATH once, it runs in MATH time. REF finds the induced permutation MATH of MATH in MATH time. In summary, a valid permutation of MATH can be computed in MATH time. |
cs/0102010 | The two directions are proved as follows. CASE: Let MATH be a hamiltonian path in MATH. Let MATH and MATH be permutations of MATH and MATH as shown in REF . It is easy to check that MATH is a valid permutation to MATH. CASE: Let MATH be a valid permutation of MATH. The remainder of this proof shows that the ordering of the lengths in MATH defines a hamiltonian path in MATH. Assume the lengths from MATH are plotted on a line according to the order given by MATH and similarly, the lengths from MATH are also plotted on this line according to MATH. For each MATH, the line fragment corresponds to MATH is called MATH. For each MATH, the line fragment corresponds to MATH, is called MATH. For every MATH, since MATH, MATH overlaps with two consecutive line fragments from MATH; in addition, the overlapping regions between MATH and these two line fragments must be of length MATH and MATH, respectively. Observe that MATH and MATH for all MATH. One of these two fragments, which overlaps with MATH, must be MATH. The other line fragment can be MATH for any MATH with MATH, that is, MATH. Let MATH. From the above argument, we know that, for every two consecutive line fragments MATH and MATH, there exists a fragment MATH (where MATH is either MATH or MATH) which overlaps with both MATH and MATH. The above argument also implies that MATH. Thus, MATH forms a path in MATH. As MATH contains all the MATH nodes of MATH, this path is a hamiltonian path. |
cs/0102013 | It is assumed that MATH, since there is nothing to show in the case MATH. It is also assumed that the values of MATH are even (odd cases can be dealt with a similar argument). Given a protocol MATH of a MATH-message MATH-restricted quantum single-prover interactive proof system, we construct a MATH-message MATH-restricted quantum prover MATH such that the probability of acceptance by MATH is exactly equal to the one by MATH on every input. We construct MATH by showing, for every input MATH, how to construct each MATH based on the original MATH. In the following proof, each MATH and MATH will be abbreviated as MATH and MATH, respectively. Let MATH be the NAME space corresponding to the private qubits of MATH. Let each MATH, for MATH, denote a state of the original MATH-message MATH-restricted quantum interactive proof system defined in a recursive manner by MATH . Here MATH is the initial state in which all the qubits are the MATH-states. Notice that MATH for each MATH, since each MATH acts only on the qubits in MATH. From REF , there exist states MATH such that MATH for each MATH. Thus we have MATH for each MATH. Therefore, by REF , there exists a unitary transformation MATH acting on MATH such that MATH for each MATH. Having defined MATH, MATH, and MATH for each MATH, compare the state just before the final measurement is performed in the original protocol and that in the constructed protocol. Let MATH and MATH. These MATH and MATH are exactly the states we want to compare. Noticing that MATH, we have MATH, since MATH acts only on MATH. This implies that the verifier MATH cannot distinguish MATH from MATH at all. Hence, for every input MATH, the probability of accepting MATH in the protocol MATH is exactly equal to the one in the original protocol MATH. Thus we have the assertion. |
cs/0102013 | Since MATH gives the minimum dimension of MATH such that there is a state MATH satisfying MATH, it is sufficient to show that MATH. This can be proved as follows: MATH . The first inequality directly comes from the definition of the entanglement measure. To prove the second and third inequalities, let MATH be the decomposition of MATH with respect to the orthonormal bases MATH, MATH, and MATH of MATH, MATH, and MATH, respectively, and let MATH be that of MATH with respect to the orthonormal bases MATH and MATH of MATH and MATH, respectively. The second and third inequalities are the consequences of the equality MATH and the equality MATH respectively, where MATH and MATH are the NAME decompositions of MATH and MATH, respectively. The fourth inequality is from the definition of the entanglement measure, which ensures that MATH holds. |
cs/0102013 | It is assumed that MATH, since there is nothing to show in the case MATH. It is also assumed that the values of MATH are even, and thus MATH (odd cases can be dealt with a similar argument). Given a protocol MATH of a MATH-message MATH-restricted MATH-prior-entangled quantum MATH-prover interactive proof system, we first show that MATH can be replaced by a MATH-message MATH-restricted MATH-prior-entangled quantum prover MATH such that the probability of acceptance by MATH is exactly equal to the one by MATH on every input. Having shown this, we repeat the same process for each of provers to construct a protocol MATH from MATH and so on, and finally we obtain a protocol MATH in which all of MATH are MATH-message MATH-restricted MATH-prior-entangled quantum provers. We construct MATH by showing, for every input MATH, how to construct each MATH based on the original MATH. In the following proof, each MATH and MATH will be abbreviated as MATH and MATH, respectively. Let each MATH, for MATH, denote a state of the original MATH-message MATH-restricted MATH-prior-entangled quantum MATH-prover interactive proof system defined in a recursive manner by MATH . Here MATH is the initial state in which the first MATH qubits in each MATH may be entangled with private qubits of other provers than MATH. All the qubits other than these prior-entangled qubits are the MATH-states in the state MATH. Note that MATH, for each MATH. We define each MATH recursively. To define MATH, consider the states MATH and MATH. Let MATH. Since all of the last MATH qubits in MATH in the state MATH are the MATH-states and MATH, by REF , there exists a state MATH in MATH such that MATH and all but the first MATH qubits in MATH are the MATH-states in the state MATH. Furthermore we have MATH . Therefore, by REF , there exists a unitary transformation MATH acting on MATH such that MATH and MATH is of the form MATH, where MATH is a unitary transformation acting on qubits in MATH and the first MATH qubits of MATH, and MATH is the MATH-dimensional identity matrix. Assume that MATH, MATH, and MATH have been defined for each MATH, MATH, to satisfy CASE: MATH, MATH. CASE: All but the first MATH qubits in MATH are the MATH-states in the state MATH. CASE: All but the first MATH qubits in MATH are the MATH-states in the state MATH. Notice that MATH, MATH, and MATH defined above satisfy such conditions. Define MATH, MATH, and MATH in the following way to satisfy the above four conditions for MATH. Let MATH and define MATH. Then all but the first MATH qubits in MATH are the MATH-states in the state MATH, since none of MATH acts on the space MATH and MATH satisfies the fourth condition. Since MATH, by REF , there exists a unitary transformation MATH acting on MATH such that MATH. Thus we have MATH . Hence, by REF , there exists a state MATH such that MATH and all but the first MATH qubits in MATH are the MATH-states in the state MATH. From REF , we have MATH since MATH and MATH act only on MATH. Therefore, by REF , there exists a unitary transformation MATH acting on MATH such that MATH. It follows that MATH is of the form MATH, where MATH is a unitary transformation acting on qubits in MATH and the first MATH qubits of MATH, because all of the last MATH qubits in MATH are the MATH-states in both of the states MATH and MATH. One can see that MATH, and MATH satisfy the four conditions above by their construction. Having defined MATH for each MATH, compare the state just before the final measurement is performed in the original protocol and that in the modified protocol applying MATH's instead of MATH's. For MATH, let MATH and MATH. These MATH and MATH are exactly the states we want to compare. Noticing that MATH, we have MATH, since none of MATH acts on MATH. Thus we have MATH which implies that the verifier MATH cannot distinguish MATH from MATH at all. Hence, for every input MATH, the probability of accepting MATH in the protocol MATH is exactly equal to the one in the original protocol MATH, and MATH uses only MATH qubits in his private space. In the protocol MATH, each MATH is described as MATH, where MATH is a unitary transformation acting on qubits in MATH and the first MATH qubits of MATH. Consequently, by constructing a MATH-message MATH-restricted quantum prover MATH from each MATH, for every input MATH, the probability of accepting MATH in the protocol MATH is exactly equal to the one in the original protocol MATH. Now we repeat the above process for each of provers, and finally we obtain a protocol MATH in which all MATH provers are MATH-message MATH-restricted quantum provers. It is obvious that, for every input MATH, the probability of accepting MATH in the protocol MATH is exactly equal to the one in the original protocol MATH, and we have the assertion. |
cs/0102013 | For convenience, we assume that the values of MATH are even (odd cases can be dealt with a similar argument). Let MATH be a language in MATH. Then, from REF together with REF , there exist polynomially bounded functions MATH and a MATH-message MATH-restricted quantum verifier MATH for a quantum MATH-prover interactive proof system such that, for every input MATH, REF if MATH is in MATH, there exists a set of MATH quantum provers MATH of MATH-message MATH-restricted MATH-prior-entangled such that MATH accepts MATH with certainty, and REF if MATH is not in MATH, for all sets of MATH quantum provers MATH of MATH-message MATH-restricted MATH-prior-entangled, MATH accepts MATH with probability at most MATH. For an input MATH of length MATH, consider a classical simulation of this quantum MATH-prover interactive proof system by a non-deterministic NAME machine. Let MATH be arbitrary fixed polynomial. First, for the initial state MATH, an approximation MATH of MATH can be guessed in time non-deterministic exponential in MATH with accuracy of MATH. Next, since each MATH applied in the original proof system is polynomial-time uniformly generated and MATH and MATH are polynomially bounded functions, it is routine to show that an approximation MATH of a matrix description of MATH can be computed in time exponential in MATH with accuracy of MATH. Finally, since MATH and MATH are polynomially bounded functions, for each operation MATH of the MATH-th prover applied in the original proof system, an approximation MATH of a matrix description of MATH can be guessed in time non-deterministic exponential in MATH with accuracy of MATH. Thus, for the quantum state MATH which is the state just before the final measurement in the proof system, the approximation MATH of MATH can be computed in time non-deterministic exponential in MATH with accuracy of MATH for any fixed polynomial MATH by appropriately choosing MATH. Now, after having computed MATH, a measurement of the output qubit is simulated by summing up squares of the computed amplitudes in the accepting states. The input MATH is accepted if and only if this sum, the computed probability that the measurement results in MATH, is more than MATH. From the property of the original proof system, this computed probability is more than MATH if MATH is in MATH, while it is less than MATH if MATH is not in MATH. Thus, taking MATH and MATH, the input MATH is accepted if and only if MATH is in MATH and the whole computation is done in time non-deterministic exponential in MATH. |
cs/0102013 | For simplicity, we assume that the values of MATH are even, and thus MATH (odd cases can be proved with a similar argument). Let MATH be a language in MATH. Then, from REF together with REF , there exist polynomially bounded functions MATH and a MATH-message MATH-restricted quantum verifier MATH for a quantum MATH-prover interactive proof system such that, for every input MATH of length MATH, REF if MATH is in MATH, there exists a set of MATH-message MATH-restricted quantum provers MATH without prior entanglement such that MATH accepts MATH with probability at least MATH, and REF if MATH is not in MATH, for all sets of MATH-message MATH-restricted quantum provers MATH without prior entanglement, MATH accepts MATH with probability at most MATH. We construct a MATH-oracle-call verifier MATH of a quantum oracle circuit as follows. Let us consider that quantum registers (collections of qubits upon which various transformations are performed) MATH, MATH, and MATH, for MATH, are prepared among the private qubits of the verifier MATH, and quantum registers MATH and MATH are prepared among the qubits for oracle calls. MATH consists of MATH qubits, each MATH and MATH consist of MATH qubits, and each MATH and MATH consist of MATH qubits. Let MATH, each MATH, and each MATH denote the NAME spaces corresponding to the registers MATH, MATH, and MATH, respectively. Take the NAME space MATH corresponding to the qubits private to the verifier MATH as MATH. Accordingly, the number of private qubits of MATH is MATH. Let MATH and MATH denote the NAME spaces corresponding to the registers MATH and MATH, respectively. Take the NAME space MATH corresponding to the qubits for oracle calls as MATH. Accordingly, the number of qubits for oracle calls is MATH. Consider each MATH, the MATH-th quantum circuit of the verifier MATH of the original quantum MATH-prover interactive proof system, which acts on MATH. For each MATH, let MATH be just the same unitary transformation as MATH and MATH acts on MATH, corresponding to that MATH acts on MATH. Define the verifier MATH of the corresponding quantum oracle circuit in the following way: CASE: At the first transformation of MATH, MATH first applies MATH, and then swaps the contents of MATH for those of MATH. CASE: At the MATH-th transformation of MATH for each MATH, MATH first swaps the contents of MATH and MATH for those of MATH and MATH, respectively, then applies MATH, and finally swaps the contents of MATH and MATH for those of MATH and MATH. CASE: At the MATH-th transformation of MATH for each MATH, MATH first swaps the contents of MATH and MATH for those of MATH and MATH, respectively, then swaps the contents of MATH and MATH for those of MATH and MATH. CASE: In the case the input MATH of length MATH is in MATH: In the original MATH-message quantum MATH-prover interactive proof system, there exist MATH-message MATH-restricted prior-unentangled quantum provers MATH that cause MATH to accept MATH with probability at least MATH. Hence, if we let MATH for each MATH be just the same unitary transformation as MATH (MATH acts on MATH corresponding to that MATH acts on MATH), it is obvious that the probability of accepting MATH by MATH with access to MATH is exactly equal to the one the original MATH accepts it, which is at least MATH. CASE: In the case the input MATH of length MATH is not in MATH: Suppose that there were an oracle MATH that makes the verifier MATH accept MATH with probability more than MATH. Consider MATH-message MATH-restricted prior-unentangled provers MATH of the original MATH-message quantum MATH-prover interactive proof system such that, for each MATH, MATH is just the same transformation as MATH (MATH acts on MATH corresponding to that MATH acts on MATH). By their construction, it is obvious that the probability with which these provers MATH can convince the verifier MATH is exactly equal to the one with which the oracle MATH can, which is more than MATH. This contradicts the assumption. |
cs/0102013 | Given a classical MATH-prover interactive protocol, consider such a quantum MATH-prover protocol without prior entanglement that a quantum verifier performs measurements in MATH basis on every qubit of his part at every time he sends questions to quantum provers and at every time he receives responses from them, and for the rest part of computation the quantum verifier behaves in the same manner as the classical verifier does. Such a protocol can be simulated without intermediate measurements by only using unitary transformations CITE. Furthermore, since there is no prior entanglement among private qubits of the quantum provers, such a quantum protocol makes no difference from a classical protocol in which a classical verifier chooses a set of MATH classical provers probabilistically at the beginning of the protocol. Therefore, in such a quantum MATH-prover protocol, for every input, the quantum provers can be only as powerful as the classical provers, that is, the quantum provers can behave just in the same way as the classical provers do, while no set of MATH quantum provers can convince the quantum verifier with probability more than the maximum probability with which a set of MATH classical provers can convince the classical verifier. Now we explain in more detail. Let MATH be a language in MATH, then MATH has a one-round two-prover interactive proof system. Let MATH be the classical verifier of this one-round two-prover interactive proof system. We construct a two-message quantum two-prover interactive proof system by just simulating this classical protocol. Assume that, just after the classical verifier MATH has sent questions to the provers MATH and MATH, the contents of MATH's private tape, the question to MATH, and the question to MATH are MATH, MATH, and MATH, respectively, with probability MATH. Our two-message quantum verifier MATH prepares the quantum registers MATH, MATH, MATH, MATH, and MATH among his private qubits. MATH first stores MATH, MATH, and MATH in MATH, MATH, and MATH, respectively, then copies the contents of each MATH to the message qubits shared with a quantum prover MATH. That is, MATH prepares the superposition MATH where, for each MATH, MATH denotes the quantum register that consists of the message qubits between MATH and MATH, and MATH denotes the quantum register that consists of MATH's private qubits. Next the quantum provers MATH and MATH apply some unitary transformations on their qubits. Now the state becomes MATH where each MATH denotes the transition amplitude and each MATH is a unit vector in the private space of MATH. Finally, MATH copies the contents of the message qubits shared with the quantum prover MATH to MATH to have the following state MATH and does just the same computation as the classical verifier MATH using MATH, MATH and MATH. MATH accepts the input if and only if MATH accepts it. CASE: In the case the input MATH of length MATH is in MATH: The quantum provers have only to answer in just the same way as the classical provers do, and MATH accepts MATH with probability MATH. CASE: In the case the input MATH of length MATH is not in MATH: Since no quantum interference occurs among the computational paths with different REF-tuple MATH, and from the fact that any pair of classical provers cannot convince the classical verifier with probability more than MATH (actually MATH), it is obvious that, for any pair of quantum provers, MATH accepts MATH with probability at most MATH (actually MATH). |
cs/0102024 | The proof is similar to the proof of REF . Let MATH be an infinite emptiness-testable NAME word-decreasing-self-reducible set having MATH-gaps. If MATH is finite then MATH is not MATH-immune, so we henceforth consider only the case that MATH is infinite. We will show that MATH has an infinite subset in MATH. Let MATH be a constant such that MATH. Let MATH be a deterministic polynomial-time NAME machine witnessing the NAME word-decreasing-self-reducibility of MATH. By REF , there exist a constant MATH and a deterministic MATH-time-bounded NAME machine MATH such that MATH. Let MATH. By REF. Consider the following deterministic NAME machine MATH: CASE: On input MATH simulate the work of MATH while answering the generated queries as follows: (Initialize MATH. The variable MATH will work as a flag to indicate whether the answers to certain oracle queries are correct.) CASE: Every query MATH with MATH is answered according to the outcome of MATH, that is, simulate MATH and answer ``yes" to the query MATH if MATH accepts and answer ``no" otherwise. CASE: Every query MATH with MATH is answered ``no." If MATH, that is if MATH, set MATH, otherwise leave MATH unchanged. (Informally put, we change the value of MATH if we answered ``no" to a query that is of a length at which MATH is not empty.) CASE: Accept if and only if both the simulation of MATH while answering the queries as described above accepts and MATH. From here on the proof proceeds in analogy to the proof of REF . |
cs/0102024 | Regarding REF, let MATH be a set in MATH having NAME. Let MATH be a constant such that for infinitely many MATH it holds that MATH. Since MATH there exist a constant MATH and a MATH-time-bounded NAME machine MATH such that MATH. Suppose that MATH is MATH-hard. Let MATH be a polynomial-time computable such reduction, that is, for all MATH, MATH. Since MATH is positive-Turing self-reducible (even disjunctive-Turing self-reducible) there exists a deterministic polynomial-time machine MATH such that CASE: MATH, CASE: on each input MATH, MATH queries only strings of length strictly less than MATH, and CASE: for all MATH and MATH such that MATH it holds that MATH. We will show that, for every MATH, there exists a MATH set MATH such that, for infinitely many MATH, MATH . We will do so by showing that, for every MATH, there exist a MATH set MATH and an infinite tally set MATH such that for all MATH, MATH . Though the format here is MATH versus MATH rather than MATH versus MATH, it is not hard to see that this suffices. Let MATH be polynomial such that for all MATH and for all MATH, MATH and MATH. Let MATH. Consider the following deterministic NAME machine MATH: CASE: On input MATH, MATH, simulate MATH and each time MATH asks a query MATH to MATH compute MATH and answer the query MATH as follows: CASE: If MATH then answer ``yes" if and only if MATH accepts and ``no" otherwise. CASE: If MATH then answer ``no." CASE: Accept if and only if the simulation of MATH, answering the queries as described above, accepts. It is not hard to see that the above machine MATH runs in time polynomial in MATH. Let MATH. Since MATH is globally positive and the above machine answers queries by exploiting the many-one reduction from MATH to MATH or by answering ``no," it follows that MATH. Define MATH. Note that MATH is infinite. To see this let MATH be such that, for all MATH, MATH. Such a MATH clearly exists, since MATH is a monotonic polynomial of degree greater than zero. Now let MATH be any natural number such that MATH and MATH. Define MATH. Note MATH and MATH. It follows that MATH. Since there are infinitely many MATH satisfying both, MATH and MATH, it follows that MATH is an infinite set. We are now prepared to show that for all MATH, MATH . Let MATH. In light of the definition of MATH, there exists some MATH such that MATH and MATH. Choose such a MATH (which implicitly is MATH). Note that MATH implies MATH. Hence any string MATH satisfying MATH cannot be in MATH. Let MATH be such that MATH and suppose that MATH. Note that MATH. So, since MATH and MATH are monotonic, MATH, and of course MATH. This implies that any string MATH satisfying MATH cannot be in MATH. Now consider the action of MATH. MATH essentially simulates the work of MATH. Note that for all queries MATH generated by MATH, MATH and hence MATH. Furthermore, any query MATH with MATH is correctly answered during the simulation of MATH in our algorithm since MATH. On the other hand, for all queries MATH with MATH (recall that those queries are answered ``no" by MATH during the simulation of MATH) MATH is in the gap associated with MATH (that is, the gap that extends at least down to the length MATH and stretches at least up to the length MATH), in other words, MATH and consequently MATH. This shows that during the run of MATH all queries generated in the simulation of MATH are answered correctly and hence MATH implies MATH. So we showed that, under the assumption of REF, MATH has MATH-easiness bands. The same proof works for any positive-Turing self-reducible MATH set, or indeed, with the obvious minor change in the proof, for any positive-Turing word-decreasing-self-reducible MATH set. This completes the proof of REF . Regarding the proof of REF we note that if MATH is emptiness-testable, then the above-defined set MATH is in MATH. This can be seen easily in light of the definition of MATH, using also REF . Though the set MATH of this proof marks upper ends of bands in contrast with REF which requires the marking of the lower ends, it is not hard to see that this suffices, though due to rounding issues one has to be slightly careful. In particular, if we wish to prove bands of the form MATH-MATH, we use the above proof for the value MATH to get bands of the form MATH-MATH and to get an upper-edge-marking set MATH. The set MATH will also be in MATH, will be infinite, and will serve as the desired lower-edge-of-band marking tally set in the sense of REF . |
cs/0102025 | By simple induction on the structure of LO proofs. |
cs/0102025 | The proof of MATH and MATH is by simple induction. The proof of MATH is by (complete) induction on MATH (see REF ). CASE: If MATH, then, no matter which MATH you choose, MATH; CASE: if MATH is a fact, then MATH means MATH, which in turn implies that there exists MATH such that MATH, therefore MATH; CASE: if MATH, then by inductive hypothesis there exist MATH and MATH such that MATH and MATH. Therefore, if MATH, by MATH we have that MATH and MATH, therefore MATH, that is, MATH as required; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. |
cs/0102025 | Immediate from MATH definition and REF MATH. Continuity. We prove that MATH is finitary. Namely, given an increasing chain of interpretations MATH, MATH is finitary if MATH. We simply need to show that if MATH then there exists MATH such that MATH. The proof is by induction on MATH. CASE: If MATH, then, no matter which MATH you choose, MATH; CASE: if MATH is a fact and MATH, then, by definition of MATH, there exist a fact MATH and a clause MATH, such that MATH and MATH. CASE: MATH implies that MATH, therefore, again by definition of MATH, MATH, that is, MATH as required; CASE: if MATH, then by inductive hypothesis, there exist MATH and MATH such that MATH and MATH. Then, if MATH, by REF MATH we have that MATH and MATH. This implies MATH, that is, MATH as required; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. |
cs/0102025 | CASE: We prove that for every MATH and context MATH, if MATH then MATH. The proof is by (complete) induction on MATH, where MATH is an induction measure defined by MATH, and MATH if and only if REF or (MATH and MATH) (lexicographic ordering). CASE: If MATH, the conclusion is immediate; CASE: if MATH is a fact, then MATH, so that MATH and MATH. By definition of MATH we have that there exist a fact MATH and a clause MATH, such that MATH and MATH. By REF MATH we have that MATH, and then, by inductive hypothesis, MATH, therefore by LO MATH rule, MATH, that is, MATH; CASE: if MATH then by inductive hypothesis MATH and MATH, therefore MATH by LO MATH rule; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. CASE: We prove that for every context MATH if MATH then there exists MATH such that MATH by induction on the structure of the LO proof. CASE: If the proof ends with an application of MATH, then the conclusion is immediate; CASE: if the proof ends with an application of the MATH rule, then MATH, where MATH are atomic formulas, and there exists a clause MATH. For the uniformity of LO proofs, we can suppose MATH to be a fact. By inductive hypothesis, we have that there exists MATH such that MATH, then, by REF MATH, MATH, which, by definition of MATH, in turn implies that MATH, therefore MATH, that is, MATH; CASE: if the proof ends with an application of the MATH rule, then MATH and, by inductive hypothesis, there exist MATH and MATH such that MATH and MATH. Then, if MATH we have, by REF MATH, that MATH and MATH, therefore MATH, that is, MATH; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. |
cs/0102025 | CASE: By induction on MATH. CASE: MATH and MATH and MATH for any MATH; CASE: if MATH then MATH for MATH. Since MATH, we have that MATH, therefore MATH, so that MATH for all MATH such that MATH, because MATH is upward closed; CASE: if MATH then MATH and MATH and MATH. By inductive hypothesis, MATH and MATH for any MATH such that MATH and MATH. That is, MATH for any MATH. It follows that MATH for all MATH; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. CASE: By induction on MATH. CASE: The MATH-case follows by definition; CASE: if MATH then MATH, that is, there exists MATH such that MATH. Since MATH, it follows that for MATH, MATH and MATH; CASE: if MATH then MATH for MATH. By inductive hypothesis, there exists MATH such that MATH, MATH for MATH, that is, MATH. The thesis follows noting that MATH; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. CASE: If MATH, then by MATH, MATH. Since MATH then, by REF MATH, MATH. Thus, by MATH, there exists MATH such that MATH. CASE: By induction on MATH. CASE: If MATH, then, no matter which MATH you choose, MATH; CASE: if MATH is a fact, then MATH, that is there exists MATH such that MATH and MATH. Therefore there exists MATH such that MATH and MATH, that is MATH; CASE: if MATH, then by inductive hypothesis there exist MATH and MATH such that MATH and MATH. Therefore, if MATH, by REF MATH, we have that MATH and MATH, therefore MATH, that is, MATH; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. |
cs/0102025 | Let MATH where MATH and MATH then, by REF MATH, MATH for any MATH such that MATH. Thus, for any MATH such that MATH, MATH. Vice versa, if MATH then MATH where MATH and MATH. By REF MATH, there exists MATH such that MATH and MATH, that is, MATH and MATH. |
cs/0102025 | If MATH, then, by definition of MATH, it follows that MATH. This implies that MATH. Thus, by REF it follows that MATH, that is, MATH. |
cs/0102025 | Monotonicity. For any MATH there exists MATH such that MATH. Assume now that MATH. Then, by REF MATH, we have that MATH for MATH. Thus, there exists MATH such that MATH, that is, MATH. Continuity. We show that MATH is finitary. Let MATH be an increasing sequence of interpretations. For any MATH there exists MATH such that MATH. By REF MATH, MATH. By REF MATH, we get that MATH for some MATH, and by REF MATH, MATH for MATH. Thus, MATH, that is, MATH, that is, MATH. |
cs/0102025 | REF show that MATH. REF guarantees that the fixpoint of MATH can always be reached after finitely many steps. |
cs/0102025 | To prove the result we present an encoding of Vector Addition Systems (VAS) as MATH programs. A VAS consists of a transition system defined over MATH variables MATH ranging over positive integers. The transition rules have the form MATH where MATH is an integer constant. Whenever MATH, guards of the form MATH ensure that the variables assume only positive values. Following CITE, the encoding of a VAS in MATH is defined as follows. We associate a propositional symbol MATH to each variable MATH. NAME now becomes a rewriting rule MATH where MATH if MATH (tokens removed from place MATH) and MATH if MATH (tokens added to place MATH). We encode the set of initial markings (that is, assignments for the variables MATH's) MATH using MATH clauses as follows. The MATH clause MATH is such that if MATH is the assignment MATH then MATH for MATH. Based on this idea, if MATH is the program that encodes the VAS MATH it is easy to check that MATH corresponds to the set of reachable markings of MATH (that is, to the closure MATH of the MATH operator MATH with respect to MATH and the initial markings). From classical results on NAME Nets (see for example, the survey CITE), there is no algorithm to compute the set of reachable states of a VAS MATH (=MATH). If not so, we would be able to solve the marking equivalence problem that is known to be undecidable. |
cs/0102025 | CASE: By induction on MATH. CASE: If MATH, then every MATH (with MATH) is solution of MATH, and MATH for every fact MATH; CASE: if MATH, then MATH is the only solution of MATH, and MATH; CASE: if MATH then there exists MATH such that MATH is satisfiable. Then for every solution MATH of MATH there exists a vector MATH such that MATH is satisfiable and MATH. From this we get that for MATH, MATH is a solution for MATH, therefore MATH so that we can conclude MATH; CASE: if MATH then MATH and MATH, MATH. By inductive hypothesis, MATH and MATH for every MATH and MATH solutions of MATH and MATH, respectively. Thus MATH for every MATH which is solution of both MATH and MATH, that is, for every MATH which is solution of MATH; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. CASE: By induction on MATH. CASE: MATH for every MATH, and MATH, where MATH, and every MATH is solution of MATH; CASE: if MATH, MATH, then MATH, where MATH, and MATH is solution of MATH; CASE: if MATH then MATH, therefore there exists MATH such that MATH. Therefore, if MATH is such that MATH, we have that MATH is satisfiable, MATH is solution of MATH and MATH; CASE: if MATH then MATH and MATH. By inductive hypothesis, there exist MATH and MATH such that MATH and MATH, and MATH is a solution of MATH and MATH. Therefore MATH and MATH is a solution of MATH; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. CASE: By simple induction on MATH. CASE: By induction on MATH. CASE: The MATH and MATH-cases follow by definition; CASE: if MATH, then there exists MATH such that MATH is satisfiable. Then there exists MATH such that MATH and MATH; CASE: if MATH, then MATH, and, by inductive hypothesis, there exist MATH and MATH such that MATH and MATH. Then, for MATH, we have, by MATH, MATH and MATH, therefore MATH; CASE: the MATH-case follows by a straightforward application of the inductive hypothesis. |
cs/0102025 | Immediate from MATH definition and REF MATH. Continuity. Let MATH, be an increasing sequence of interpretations. We show that MATH. If MATH, by definition there exists a clause MATH such that MATH and MATH is satisfiable. By REF MATH, there exists MATH such that MATH. This implies that MATH, that is, MATH. |
cs/0102025 | Let MATH, then there exist MATH and a clause MATH, such that MATH is solution of MATH, MATH and MATH. Then there exists MATH solution of MATH such that MATH, and, by REF MATH, MATH. Therefore, by definition of MATH, MATH. Vice versa, let MATH, then there exists MATH such that MATH and MATH. By REF MATH, there exists MATH such that MATH and MATH. Therefore MATH, and MATH. |
cs/0102025 | By REF . |
cs/0102025 | By definitions. |
cs/0102025 | As MATH and MATH, the proposition follows by monotonicity of MATH. |
cs/0102025 | Suppose MATH and MATH. Then there exists MATH such that MATH. By definition of MATH, there exists a clause MATH (the case for unit clauses is trivial) such that MATH and MATH. As MATH, we also have MATH, which implies that there exists MATH such that MATH. Let MATH (it is immediate to prove that such a MATH exists), and let MATH. We have that MATH, therefore MATH (because MATH and MATH is upward-closed). Now, MATH, and, by definition of MATH (MATH), MATH. By repeatedly applying MATH (the proof is by induction on MATH) we get MATH. Therefore MATH. |
cs/0102025 | By a simple induction, using REF , we have that for every MATH there exists MATH such that MATH. Therefore MATH. |
cs/0102030 | As MATH and MATH and MATH, it follows that MATH. Since MATH is satisfiable in MATH, MATH . Thus MATH is satisfiable in MATH. |
cs/0102030 | We assume the congruence axioms hold and prove that, for any MATH, we have MATH. The proof is by induction on the depth of MATH. Suppose, first that the depth of MATH is one. If MATH is a variable not in MATH or a constant, then MATH and the result follows from REF . If MATH, then, for some MATH, MATH. Thus MATH. If the depth of MATH is greater than one, then MATH has the form MATH where MATH have depth less than the depth of MATH. By the inductive hypothesis, for each MATH, , MATH, we have MATH. Therefore, applying REF , we have MATH. |
cs/0102030 | Suppose first that MATH and that MATH. Then MATH and hence, MATH. Next, suppose that for all MATH, MATH. Let MATH. We will show that MATH by induction on the depth of MATH. If MATH is a constant or MATH, then the result follows from the fact that MATH. If MATH, then the result follows from the hypothesis. Finally, if MATH, then, by the inductive hypothesis, MATH for MATH, , MATH. Therefore we have MATH. Thus, by REF , as MATH, MATH. |
cs/0102030 | Observe that, since MATH, the relation MATH is trivial. To prove the opposite relation, suppose that MATH. Then there exists MATH such that MATH. Now, if MATH, then MATH and MATH. On the other hand, if MATH, then MATH so that MATH and hence, as MATH, MATH. |
cs/0102030 | As MATH has no circular subset and MATH is finite, there exists a MATH such that MATH and hence, MATH. As MATH is variable-idempotent, we have MATH . Hence MATH. |
cs/0102030 | We assume that the congruence and identity axioms hold. Let MATH have distinct outer-most symbols so that, by the identity axioms, MATH. By REF , either MATH or, for some MATH, MATH. We consider each case separately. If, for some MATH, MATH, then, as MATH and MATH have distinct outer-most symbols, there exists a MATH such that MATH and MATH have distinct outer-most symbols. Thus, by the identity axioms, MATH. Let MATH. It follows from REF that, as MATH and MATH is satisfiable, MATH and is satisfiable. By REF and the congruence axioms, MATH. However, MATH, so that MATH. Thus, by the congruence axioms, we have MATH, which is a contradiction. Suppose then that MATH. If MATH, then it follows from REF that MATH and, as MATH is satisfiable, MATH is satisfiable. By REF and the congruence axioms, MATH. However, MATH, so that MATH. Thus, by the congruence axioms, we have MATH, which is a contradiction. Hence MATH as required. |
cs/0102030 | We assume that the congruence and identity axioms hold. To prove the result, we suppose that there exists MATH and derive a contradiction. By hypothesis, MATH. Thus, using REF and the congruence axioms, we have, for any MATH, MATH. By REF , for all MATH, MATH so that MATH. By REF , MATH, so that, as MATH is ordered and MATH, MATH. In particular, MATH, so that as MATH and MATH is ordered, we would have MATH, which is a contradiction. |
cs/0102030 | We assume that the congruence and identity axioms hold. Note that, by the hypothesis, MATH and MATH so that, using REF and the congruence axioms, we have MATH and MATH, for all MATH. Let MATH. We prove, by induction on the depth MATH of MATH, that there exists MATH such that MATH. The base case is when MATH so that MATH. Now, for each MATH, MATH and hence, by REF (as MATH), MATH. As a consequence, MATH for all MATH and MATH. By REF , MATH. Thus, we define MATH. For the inductive step, we assume that MATH so that, for some MATH, we have MATH and, for some MATH, MATH and MATH has depth MATH. By REF , either MATH or there exists a MATH such that MATH. First, suppose that MATH. Now, MATH so that, as MATH, by REF , we have MATH. Thus, by REF , there exists MATH such that MATH. Then, using the identity axioms, we have MATH and MATH. By the inductive hypothesis (letting MATH be the empty substitution), we have MATH. However, MATH so that MATH. As MATH and MATH, MATH. Thus, in this case, let MATH. Secondly, suppose that there exists a MATH such that MATH. Then, as MATH, it follows from the identity axioms that MATH and MATH. By the inductive hypothesis, there exists MATH such that MATH. However, MATH so that we must have MATH. As MATH, MATH as required. |
cs/0102030 | Since MATH, there exists MATH with MATH such that MATH and MATH. If MATH, MATH and the result is trivial. Suppose now that MATH. We define MATH . We first show that MATH and MATH. If MATH, then MATH so that MATH. Also, as MATH has no circular subset, MATH has no circular subset and MATH. If MATH, then MATH and MATH. Thus, as MATH has no circular subset, MATH so that MATH. Moreover, neither MATH nor MATH have circular subsets. Hence MATH has no circular subset. Thus MATH. Now, since MATH it follows that MATH. Therefore, it remains to show that, for any equality theory MATH, MATH. To do this, we assume that the congruence axioms hold, and show that MATH. By REF , we have MATH . It therefore follows from REF that MATH. |
cs/0102030 | The proof is by induction on the length of the sequence of MATH-steps transforming MATH to MATH. The base case is the empty sequence. For the inductive step, the sequence has length MATH and there exists MATH such that MATH and MATH has length MATH. By REF , MATH, MATH, MATH and MATH. By the inductive hypothesis, MATH, MATH, MATH and MATH. Hence we have MATH, MATH, and MATH. |
cs/0102030 | To prove the theorem, we construct a MATH-transformation and show that the resulting substitution has the required properties. Suppose that MATH, MATH and, for each MATH, , MATH, MATH where, if MATH, MATH and, for each MATH, , MATH with MATH, we have MATH. It follows from the definition of MATH that, for MATH, , MATH , MATH can be obtained from MATH by two sequences of MATH-steps of lengths MATH and MATH: MATH where, for MATH, , MATH with MATH, MATH . Hence, by REF , MATH. We next show, by induction on MATH, with MATH, that, for each MATH, , MATH and each MATH, , MATH, we have MATH. For the base case when MATH there is nothing to prove. Suppose, therefore, that MATH and that, for each MATH, , MATH and MATH, , MATH, MATH . Now by the definition of MATH where MATH, MATH, we have MATH . Therefore, for each MATH, , MATH and MATH, , MATH, using REF and the inductive hypothesis, we have MATH . Letting MATH we obtain, for each MATH, , MATH, MATH . Thus, by REF , for all MATH, MATH. The result follows by taking MATH. |
cs/0102030 | Since MATH, MATH and MATH, we have that MATH. It follows from REF that MATH can be obtained from MATH by a sequence of MATH-steps so that, by REF , we have REF . To prove REF , we suppose that, for some MATH, there exist MATH, MATH and MATH such that MATH. We need to prove that MATH. It follows from REF , that MATH and MATH. Suppose first that MATH. Then MATH and hence MATH. Therefore, as MATH and MATH, we can conclude MATH. Thus, we now assume that MATH. As MATH, we have MATH, so that MATH and hence, MATH. If MATH we have MATH, so that MATH. On the other hand, if MATH then, by the hypothesis, MATH. Thus, in both cases, as MATH, we obtain MATH and hence MATH. It follows, using REF , that REF holds. |
cs/0102030 | The proof is by induction on MATH. For the base case (when MATH), if MATH, then MATH and MATH. Thus, MATH so that, by REF , MATH. Suppose MATH. Then, if MATH, we have, by REF , MATH. By the induction hypothesis, MATH so that MATH and thus MATH. |
cs/0102030 | By REF , MATH and, for all MATH, we have MATH if MATH. Thus, MATH, for all MATH, so that, by REF , MATH. |
cs/0102030 | Suppose first that MATH. Then MATH . Also, by REF , MATH. Suppose next that MATH. It follows from REF , that MATH . However, as MATH, we have MATH. Thus, as MATH, MATH and hence, by REF , we have also MATH, for all MATH. Therefore, by REF , MATH . |
cs/0102030 | As MATH we have, for all MATH, MATH. Also, as MATH, we can apply REF so that MATH . |
cs/0102030 | Suppose first that MATH. Thus we assume that MATH , where MATH, and that MATH . Suppose MATH. Then we show that MATH. If MATH, then MATH and there is nothing to prove. Also, if MATH then, by REF , MATH so that by REF , MATH. We now assume that MATH and MATH. We first prove that, for each MATH, MATH . The proof is by induction on MATH. By REF , we have that MATH so that REF holds for MATH. Suppose then that MATH and that MATH . Then, to prove REF , we must show that MATH . By REF , there exists MATH . Hence, by the inductive hypothesis, MATH. If MATH, then, by REF , MATH. Suppose now that MATH. Since, by REF , we have that MATH, it follows, using REF , that MATH and MATH. However, by assumption, MATH, so that MATH and MATH. Thus, by REF , there exists MATH . However, MATH and MATH so that, by REF , we have MATH. Since, by REF , MATH, we have also MATH. Moreover, by REF , MATH so that, by the inductive hypothesis, we have that MATH. Thus, by REF , as MATH, MATH. Conversely, we now prove that, for all MATH, MATH . The proof is again by induction on MATH. As before, MATH so that REF holds for MATH. Suppose then that MATH and MATH . Then, to prove REF , we must show that MATH . By REF , there exists MATH . Hence, by the inductive hypothesis, MATH. If MATH then, by REF , we have MATH. Suppose now that MATH. Since, by REF , we have MATH, it follows, using REF , that MATH and MATH. Hence, since MATH, by REF , we have also MATH. Furthermore, MATH so again, by REF , as MATH, MATH. Combining REF we obtain the result that, if MATH is obtained from MATH by a single MATH-step, then MATH. Thus, as MATH was arbitrary, MATH. Suppose now that MATH. If MATH, then MATH. If MATH, we have by the first part of the proof that, for each MATH, , MATH, MATH, and hence the required result. |
cs/0102030 | By REF , there exists MATH such that MATH and, for any MATH, MATH. Moreover, by REF , MATH and MATH. Thus, by REF , MATH. |
cs/0102030 | Since MATH is obtained from MATH by renaming variables and MATH, we have also that MATH. In addition, MATH so that, since MATH and MATH, we have MATH. To prove REF , we have to show that, if MATH then MATH. By the hypothesis, for all MATH we have MATH if and only if MATH. As MATH, we can use the alternative characterisation of MATH given by REF and conclude that, for each MATH, MATH. Therefore MATH. The reverse inclusion follows by symmetry so that MATH. To prove REF , we first show by induction on the depth of MATH that MATH . For the base case, MATH has depth REF. If MATH is a constant or a variable other than MATH or MATH, then MATH. If MATH, then MATH and MATH. Finally, if MATH, then MATH and we have, using the congruence axioms, that MATH. For the inductive step, let MATH. Then MATH. Thus, using the inductive hypothesis, for each MATH, , MATH, MATH. Hence, by the congruence axioms, REF holds. Note that MATH. Thus, it follows from REF that, for each MATH, MATH and hence, using the congruence axioms, MATH. Thus, MATH. Since MATH, the reverse implication follows by symmetry so that MATH. |
cs/0102030 | The proof is by induction on the number MATH of the bindings MATH such that MATH and MATH (the number of unordered bindings). For the base case, when MATH, MATH is ordered and the result holds by taking MATH. For the inductive case, when MATH, let MATH be an unordered binding and define MATH. Then, by REF , we have MATH, MATH, MATH, for all MATH, and, finally, MATH, for any equality theory MATH. In order to apply the inductive hypothesis to MATH, we must show that the number of unordered bindings in MATH is less than MATH. To this end, roughly speaking, we start showing that any ordered binding in MATH is mapped by MATH into another ordered binding in MATH, therefore proving that the number of unordered bindings is not increasing. There are three cases. First, any ordered binding MATH such that MATH is mapped by MATH into the binding MATH which is clearly ordered, since MATH. Second, consider any ordered binding MATH such that MATH. Since MATH, we have MATH. If also MATH then we have MATH and MATH; otherwise MATH so that MATH and, as MATH, MATH. Thus, in either case, such a binding is mapped by MATH into the binding MATH which is ordered since MATH. Third, consider any ordered binding MATH such that MATH and MATH. The ordering relation implies MATH and we also have MATH, since MATH. Hence, we obtain MATH. Now, as MATH, MATH. If MATH, then MATH. On the other hand, if MATH, then MATH so that MATH. Thus, in both cases, as MATH, MATH. and hence, MATH is ordered. Finally, to show that the number of unordered bindings is strictly decreasing, we note that the unordered binding MATH is mapped by MATH into the binding MATH, which is ordered. Therefore, by applying the inductive hypothesis, there exists a substitution MATH such that MATH is ordered, MATH, MATH, for all MATH, and MATH, for any equality theory MATH. Then the required result follows by transitivity. |
cs/0102030 | By REF , we can assume that MATH, MATH and, for any MATH, MATH. By REF , we can assume that MATH are also ordered substitutions so that, by REF , MATH. To prove the result we need to show that, for all MATH, we have both MATH and MATH. We just prove the first of these as the other case is symmetric. Suppose that MATH and that MATH. Then, using the alternative characterisation of MATH for variable-idempotent substitutions given by REF , we just have to show that MATH. By REF (replacing MATH by MATH, MATH by MATH and MATH by MATH), we have that there exists MATH such that MATH. Thus as MATH, MATH, and hence, MATH so that MATH, as required. |
cs/0102030 | We first prove the result under the assumption that MATH. We do this in two parts. In the first, we partition MATH into two substitutions one of which, called MATH, is the same as MATH when MATH and MATH are idempotent. We construct a new substitution MATH which, in the case that MATH and MATH are idempotent, is a most general solution for MATH. Finally we compose MATH with MATH to define a substitution that has the same abstraction as MATH but with a number of useful properties including that of variable-idempotence. In the second part, we use this composed substitution in place of MATH to prove the result. CASE: By REF , we can assume that MATH and that all subsets of MATH are in MATH. Let MATH be defined such that MATH . Then, it follows from the above assumption on subsets of MATH that MATH . Now, suppose MATH. Then MATH for some MATH. Thus, by REF , for some MATH, MATH and, again by REF , MATH so that, by REF , MATH. Therefore, as MATH was an arbitrary variable in MATH, MATH . Let MATH be a most general solution for MATH in MATH so that MATH . By REF , we can assume that MATH . By REF , and REF , we have MATH . Therefore, as MATH (by REF ), we can use REF to obtain the following properties for MATH. MATH . Now we have MATH . Therefore, by REF , MATH . CASE: To prove the result under the assumption that MATH, we define MATH so that MATH . Then, by REF , MATH. We show that MATH. If MATH, there is nothing to prove. Therefore, we assume that there exists MATH so that MATH and, for some MATH, MATH . Note that REF imply that MATH . Let MATH . We show that MATH . By REF , MATH and, by REF , MATH and MATH. Thus, it follows from REF with REF , that it suffices to show that, for each MATH, MATH if and only if there exists MATH such that MATH. First, we suppose that MATH. Thus, there exists MATH such that MATH. Since MATH (by REF ), MATH (by REF ), MATH (by REF ) and MATH (using REF ), we can apply REF (replacing MATH by MATH, MATH by MATH and MATH by MATH) so that there exists MATH such that MATH. We want to show that MATH. Now either MATH or MATH so that, by REF (if MATH) or REF (if MATH), MATH. However, MATH, so that, by REF , MATH. Thus, it remains to prove that MATH. Now, as MATH and MATH, we have MATH. So we must show that MATH. To see this note that, if MATH, then MATH and, by REF , MATH so that MATH. On the other hand, if MATH, then, by REF , MATH so that MATH . Now, as MATH and MATH (by REF ), we can apply REF so that MATH. Hence, MATH. Secondly, suppose there exists MATH such that MATH. Then MATH. We need to show that MATH. By REF , if MATH, then MATH so that MATH. On the other hand, if MATH, then again, by REF , MATH. Moreover, MATH so that, by REF with the congruence axioms, MATH. Hence, since MATH (by REF ) and MATH (by REF ), we can apply REF (replacing MATH by MATH, MATH by the empty substitution and MATH by MATH) and obtain MATH. Therefore, as a consequence of the previous two paragraphs, for each MATH, we have MATH if and only if there exists MATH such that MATH. It therefore follows that REF holds. Let MATH . Note that by REF and the fact that MATH we have MATH . We now consider the two cases MATH and MATH separately. Consider first the case when MATH. Then, by REF , for some MATH, MATH . Thus, by REF , MATH and hence, by REF , MATH. Also, by REF , MATH. Thus as MATH (by REF ) we can use REF to see that, for each MATH, MATH and hence, MATH. Therefore, by REF , MATH. As MATH (by REF ), we can apply REF to both MATH and MATH. Thus, as MATH, MATH and also (using REF ) MATH so that MATH. It therefore follows from REF that MATH and hence from REF , that MATH . Now consider the case when MATH. By REF , and the assumption that MATH, MATH . As a consequence, since MATH was any set in MATH, we have MATH and hence, by REF , MATH . We now drop the assumption that MATH and just assume the hypothesis of the theorem that MATH. Suppose MATH. Then MATH. It follows from REF that MATH is monotonic on its first argument so that MATH . Thus, by REF (replacing MATH by MATH), we obtain the required result MATH . |
cs/0102030 | Suppose MATH. Then MATH means that, for some MATH, there exist sets MATH and MATH such that MATH. Thus MATH. However MATH and MATH. Thus MATH. On the other hand, MATH means that MATH where, for some MATH, MATH, and MATH, we have MATH and MATH. Let MATH be the maximum of MATH and suppose that, for each MATH where MATH and MATH, we define MATH and MATH. Then, MATH. However, for MATH, MATH. Thus MATH. |
cs/0102030 | Let MATH . Then, by REF , MATH . Hence, we have MATH . Now, by REF , MATH . |
cs/0102030 | Let MATH. Then MATH means MATH and MATH. In other words, there exist MATH, , MATH such that MATH and, for each MATH, , MATH, we have MATH. This amounts to saying that there exist MATH, , MATH such that MATH, which is equivalent to MATH. |
cs/0102030 | Let MATH. Then MATH means that there exist MATH, , MATH such that MATH, for each MATH, , MATH, and MATH. Thus MATH and MATH. Hence, MATH. |
cs/0102030 | If MATH the statement is trivial. Suppose MATH. Then, for some MATH, there exists MATH sets MATH such that MATH for each MATH, , MATH, and MATH. Suppose MATH for MATH, MATH. Thus we have MATH, MATH, and MATH. Suppose MATH with MATH and MATH not both empty. Then, for some MATH and MATH, there exist MATH and MATH such that MATH and MATH. Then MATH and MATH . Thus MATH. |
cs/0102030 | Suppose MATH. Then MATH means MATH and MATH. Similarly, MATH means that MATH and MATH. |
cs/0102030 | We let MATH, MATH, MATH, and MATH (possibly subscripted) denote elements of MATH. The subscripts reflect certain properties of the sets. In particular, subscripts MATH indicate sets of variables that definitely have a variable in common with the subscripted set. For example, MATH is a set in MATH that has a common element with MATH and MATH is a set in MATH that has common elements with MATH and MATH. In contrast, the subscript MATH indicates that the subscripted set does not share with one of the sets MATH or MATH. Of course, in the proof, each set is formally defined as needed. Suppose that MATH . The converse then holds by simply exchanging MATH and MATH, and MATH and MATH. There are two cases due to the two components of the definition of MATH in REF . CASE: Assume MATH . Then MATH and MATH. Again there are two possibilities. CASE: Suppose first that MATH . CASE: Suppose now that MATH . Then, there exist MATH such that MATH, where MATH . Therefore, MATH . CASE: Assume MATH . Then there exist MATH such that MATH . Then, by REF , MATH where MATH . CASE: Suppose MATH. Then, by REF , MATH . By REF , MATH and hence, using REF , MATH . However, by REF , MATH so that, by REF , MATH . CASE: Suppose MATH . Then, by REF , MATH . The proof of this subcase is in two parts. In the first part we divide MATH and MATH into a number of subsets. In the second part, these subsets will be reassembled so as to prove the required result. First, by REF , there exist MATH such that MATH . Thus, if either MATH or MATH, it follows that MATH . However, by REF , MATH, so that we have MATH . We now subdivide the sets MATH, MATH, MATH, and MATH further. First note that MATH . Hence, by REF , sets MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH exist such that MATH where MATH and also MATH . We note a few simple but useful consequences of these definitions. First, it follows from REF using REF , and REF , that MATH . Secondly, using REF with REF , we have MATH and then, using this with REF , and REF , it follows that MATH . In the second part of the proof for this subcase, the component subsets of MATH are reassembled in an order that proves the required result. First, let MATH . By REF we have MATH . By REF , MATH. Thus, MATH and, as a consequence of REF , MATH. For similar reasons, MATH. Hence, by REF , MATH . Now, by REF , at least one of the following two inequalities holds: MATH . Hence, by REF , MATH . Similarly, assuming MATH and MATH it follows that MATH . Thus, as one of the inequalities in REF holds, one of REF or REF holds so that MATH . However, since MATH . |
cs/0102030 | We have MATH . |
cs/0102030 | Let MATH be a most general solution for MATH. Then MATH . |
cs/0102030 | We have MATH . |
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