paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0009250 | First let MATH be the unit vector basis for MATH and set MATH. The tree MATH is clearly a MATH-block basis tree on MATH isomorphic to MATH, so that MATH. By REF the block basis index is strictly greater than the order of any block basis tree on the space, so that MATH. But now, by REF , the block basis index is of the ... |
math/0009250 | The argument follows the same lines as the proof that MATH is isomorphic to MATH in CITE. Note that for MATH the node basis of MATH is a family of indicator functions with nested or disjoint supports, and the nested functions are at most MATH sets deep. The required map MATH is found by sending the MATH element of the ... |
math/0009250 | By REF and the proof of REF we have MATH . To complete the proof we show that for each MATH there does not exist a MATH-block basis tree on MATH of order MATH, and hence MATH. Then, since MATH for any space MATH with basis MATH, and MATH where MATH is any admissible enumeration of the node basis for MATH, it follows th... |
math/0009250 | Find MATH so that MATH, let MATH, and find MATH such that MATH. Now, MATH as required. |
math/0009250 | We may write MATH as MATH where MATH. We shall prove the result using induction on MATH. Let MATH and let MATH be the map MATH, restricted from MATH to MATH, let MATH be the NAME functions on MATH and let MATH be the extension of these to MATH with MATH. Let MATH be a tree with constant REF and order MATH on MATH; we c... |
math/0009250 | From REF we know that the MATH-index of MATH is either MATH or MATH and hence MATH or MATH. For each MATH we shall construct a MATH-tree on MATH of order MATH so that MATH by REF and the result follows. This is clear for MATH since MATH embeds isometrically into MATH for each MATH, which immediately yields a MATH-REF-t... |
math/0009250 | Again, from REF we know that MATH is either MATH or MATH. To demonstrate that it is the former we show that for each MATH there does not exist a MATH-tree on MATH of order MATH. We prove this by induction on MATH based on the following lemmas. The idea of the proof is that if we do have a MATH-tree of order MATH on MAT... |
math/0009250 | Fix MATH and let MATH be as in the statement of the lemma. Suppose that MATH for each MATH. Then MATH for each MATH. We may assume that each MATH has finite support with respect to the unit vector basis of MATH, and let MATH. Thus MATH embeds into MATH with constant MATH via the map MATH, and hence MATH has a lower MAT... |
math/0009250 | Choose MATH, set MATH, and let MATH be a tree on MATH as above. If there exist MATH and MATH such that MATH then set MATH and MATH, and we have MATH while MATH as required. Otherwise set MATH, where MATH. Then MATH is a MATH-tree on MATH of order MATH, and from the previous lemma can find MATH and MATH such that MATH. ... |
math/0009250 | We shall prove the result by induction on MATH. Let MATH and let MATH be a tree on MATH of order MATH satisfying the hypotheses of the lemma. We may assume that MATH consists of finitely supported vectors with respect to the basis of MATH, and that MATH is isomorphic to the minimal tree MATH. We write MATH where MATH i... |
nlin/0009006 | of REF : We shall adopt identifications, MATH . For MATH, the left hand side of the first relation should read as MATH. The six elements in REF are interpreted as MATH respectively. This immediately leads to the theorem. |
nlin/0009006 | The lemma is equivalent to MATH . Since REF is linear in MATH, it suffices to show the equality of the coefficients of them in the both sides. First consider the coefficient of MATH, We need to show the equality, MATH . To verify this, we prepare a matrix MATH . Denote by MATH the minor, the determinant of a matrix obt... |
nlin/0009006 | Firstly we substitute MATH to REF and obtain MATH . The application of NAME 's formula yields MATH in the form, MATH . We use REF to the second term in the right-hand side to obtain, MATH . It is now obvious that repeated applications of REF to the last term results REF . |
nlin/0009034 | Due to the fact that MATH take values in the corresponding orthogonal group we find that from REF it follows MATH, MATH and analogous relations for the vector MATH. As a result we get that MATH . Let us now insert REF into REF and take the limit of the r.h.side of REF for MATH. This immediately gives REF . In order tha... |
quant-ph/0009004 | We use a lemma from CITE. If MATH and MATH are two quantum states and MATH then the total variational distance between the probability distributions generated by the same measurement on MATH and MATH is at most MATH. We also use a lemma from CITE. Let MATH. There are subspaces MATH, MATH such that MATH and CASE: If MAT... |
quant-ph/0009004 | Let MATH be the minimal deterministic automaton of a language MATH. If it contains at least one of "forbidden constructions" of REF , then MATH cannot be recognized by a REF-way QFA. We now show that, if MATH does not contain any of the two ``forbidden constructions" and does not contain ``two cycles in a row" construc... |
quant-ph/0009004 | We have a QFA MATH which accepts MATH with probability MATH and a QFA MATH which accepts MATH with probability MATH. We will make a QFA MATH which will work like this: CASE: Runs MATH with probability MATH, CASE: Runs MATH with probability MATH, CASE: Accepts input with probability MATH. CASE: MATH and MATH input is ac... |
quant-ph/0009004 | Follows immediately from REF . |
quant-ph/0009004 | For a contradiction, assume that MATH is a QFA that recognizes MATH with a probability MATH. We construct REF words such that MATH gives a wrong answer on at least one of them. Let MATH be a superposition of QFA corresponding to the state MATH (the superposition after reading some word MATH that leads to MATH). Similar... |
quant-ph/0009004 | The minimal automaton of MATH has a structure similar to REF , with some more states. Similarly to REF , the states of the minimal automaton of MATH can be partitioned into REF levels: CASE: The starting state (nothing read so far). CASE: The states after reading MATH, MATH or MATH. CASE: The states after reading MATH,... |
quant-ph/0009007 | (Existence) We presuppose the canonical isomorphism between MATH and MATH (see CITE). Every state MATH on MATH gives rise to a function MATH via the equation MATH. Conversely, a function MATH gives rise to a state MATH on MATH just in case MATH and the map MATH is a positive definite kernel CITE. Define MATH by MATH wh... |
quant-ph/0009007 | Since MATH is faithful, we have MATH . Using the NAME density theorem, and the fact that multiplication is jointly continuous (in the strong operator topology) on bounded sets, it follows that MATH is contained in the strong-operator closure of MATH. The conclusion then follows immediately. |
quant-ph/0009007 | We show first that MATH is cyclic for MATH. By the GNS construction, MATH is cyclic for MATH. Since MATH is generated by products of NAME operators, it follows that the set MATH is a total set in MATH. Let MATH . Then it will suffice for our conclusion to show that MATH is a total set in MATH. Let MATH. That is, MATH, ... |
quant-ph/0009007 | Let MATH be the GNS representation of MATH induced by MATH, and let MATH. By REF , it will suffice to find a NAME operator MATH for MATH such that MATH. Since MATH is type II, there is a projection MATH such that MATH is equivalent to MATH CITE. That is, there is a partial isometry MATH such that MATH and MATH. [Note t... |
quant-ph/0009009 | Following NAME 's diagonalization proof CITE let us consider the following family of unary predicates over MATH depending on the parameter MATH : MATH . Clearly: MATH and so: MATH . Anyway: MATH implying the formula REF . |
quant-ph/0009032 | By definition, MATH, and by REF , MATH. Write MATH as a telescoping sum. Then MATH, and the theorem follows. |
quant-ph/0009032 | For any oracle MATH, we will think of MATH as an infinite bit-string where MATH for all MATH. Operator MATH defined by REF is then given by MATH . Let MATH denote the identity operator. For every MATH, let MATH denote the projection operator onto the subspace querying the MATH-th oracle bit. Let MATH. By REF . For ever... |
quant-ph/0009038 | As given in Ref. CITE. |
quant-ph/0009038 | The proof of REF is given in CITE and CITE. We stress that MATH direction holds in any OL. |
quant-ph/0009038 | REF fail in lattice OREF, so they imply the orthomodular law. For the converse with REF we start with MATH. It is easy to show that MATH and MATH. By applying the NAME theorem (which we shall subsequently refer to as F-H) CITE to our starting expression we obtain: MATH. The first conjunction is by orthomodularity equal... |
quant-ph/0009038 | To obtain REF from REF we apply REF CASE : MATH. Reversing the steps yields REF from REF . To get REF we first note that one can easily derive MATH from REF . Then one gets REF by applying REF . To arrive at REF starting from REF we apply REF and reduce the right hand side of REF to MATH what yields REF . |
quant-ph/0009038 | REF follows from REF and by REF an ortholattice that admits a strong set of classical states is orthomodular. Let now MATH and MATH be any two lattice elements. Assume, for state MATH, that MATH. Since the lattice admits a strong set of classical states, this implies MATH, so MATH. But MATH for any state, so MATH. Henc... |
quant-ph/0009038 | CITE By REF , to any two orthogonal atoms MATH and MATH there correspond orthogonal one-dimensional subspaces (vectors) MATH and MATH from MATH such that MATH and MATH. The unitary orthoautomorphism MATH maps into the unitary operator MATH so as to give MATH for some MATH. From this and from the unitarity of MATH we ge... |
quant-ph/0009038 | We need only to use pure states defined by unit vectors: If MATH and MATH are closed subspaces of NAME space, MATH such that MATH is not contained in MATH, there is a unit vector MATH of MATH belonging to MATH. If for each MATH in the lattice of all closed subspaces of MATH, MATH, we define MATH as the square of the no... |
quant-ph/0009038 | The proof is similar to that in CITE. By REF we have MATH etc., because MATH, that is, MATH in any ortholattice. Assuming MATH we get MATH. Hence, MATH. Therefore, MATH. Thus, by REF for strong quantum states, we obtain: MATH, , MATH, and MATH, wherefrom we get MATH. By symmetry, we get MATH. Thus MATH. MATH-GO is orth... |
quant-ph/0009038 | We use induction on MATH. The basis is simply the definition of MATH. Suppose MATH. Multiplying both sides by MATH, we have MATH . F-H was used in the last two steps, whose details we leave to the reader. |
quant-ph/0009038 | Lattice OREF violates all of the above equations as well as MATH-Go. Thus for the proof we can presuppose that any OL in which they hold is an OML. REF follows from definitions, replacing variables with their orthocomplements in MATH-Go. Assuming REF , we make use of MATH to obtain the equivalent equation MATH, so MATH... |
quant-ph/0009038 | These obviously follow from REF and (for MATH) the fact that MATH. |
quant-ph/0009038 | For REF , MATH. For REF , we have MATH . In the third step we used REF ; in the fourth MATH and MATH; in the fifth MATH and MATH. Rearranging the left-hand side, this proof also gives us MATH and thus MATH. The other direction of the inequality follows from MATH. |
quant-ph/0009038 | We have already proved the converses in REF . From REF , we have MATH since MATH when MATH for MATH. By rearranging the left-hand side we also have MATH and MATH. Thus MATH, which is the REFGO law by REF . In the penultimate step we used REF . The proof for REF is similar, and from REF we obtain REF . |
quant-ph/0009038 | We illustrate the proof by showing that REF-variable equation MATH holds in any REFGO. The essential identities we use are MATH which hold in any OML. Starting with REF , we have MATH where in the penultimate step we used REF [or more generally REF ] and in the last REF . Substituting MATH for MATH and using REF we obt... |
quant-ph/0009038 | Substitute MATH for MATH in REF-Go. |
quant-ph/0009038 | For REF : MATH. From hypotheses, MATH commutes with MATH and MATH. Using F-H twice, MATH. For REF : From hypotheses, MATH. For REF : From REF , MATH[from hypothesis] MATH. Thus MATH, which by REF is MATH. |
quant-ph/0009038 | Substituting MATH for MATH,, MATH for MATH, MATH for MATH, , MATH for MATH, and MATH for MATH, we satisfy the hypotheses of REF and obtain REF . Conversely, suppose the hypotheses of REF hold. From the hypotheses and REF , we obtain MATH. Thus MATH. Applying REF to the right-hand side, we obtain MATH. Then REF gives us... |
quant-ph/0009038 | For REF : This is the same as REF for MATH. For REF : Using REF we express the REFGO law as MATH . We define MATH, MATH, and MATH. In REF we substitute MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, and... |
quant-ph/0009038 | To show this equation holds in REFGO, we start with REF that occurs in the proof of NAME 's REF , rewriting it as: MATH . We substitute MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, and MATH for MATH. With these substitutions, the hypotheses of REF are satisfied. This results in REF , showi... |
quant-ph/0009038 | If MATH has REF or more variables, we replace it with a chained identity per REF , otherwise we replace it with the extended definition we mention after REF . The proof is then obvious. (In many cases the equation may also hold for smaller MATH or even in OML or OL, for example, when MATH.) |
quant-ph/0009038 | We illustrate the case MATH. In any OML we have MATH. Thus MATH. |
quant-ph/0009038 | In MATH, set MATH. On the other hand, lattice LREF (REF b) is a REFOA because it is an OML in which REF holds, but it is not a REFOA because it violates REF . |
quant-ph/0009038 | For REF - REF : We omit the easy proofs. For REF : If MATH then MATH using REF , so MATH (via F-H) MATH. Conversely, if MATH, then using REF , MATH. |
quant-ph/0009038 | We will work with the dual of REF , MATH . First we show that the REFOA law implies REF . In any OL we have MATH . Substituting MATH for MATH, MATH for MATH, and MATH for MATH; simplifying with REF ; and applying REF to the left-hand side of the conclusion we obtain MATH . We convert the REFOA law MATH to MATH to MATH ... |
quant-ph/0009038 | Using the definitions, REF can be written in the dual form MATH. We substitute MATH for MATH and MATH for MATH throughout; simplifying with REF we obtain MATH. This is easily shown to be equivalent to REF . |
quant-ph/0009038 | The proof is analogous to that for REF . |
quant-ph/0009038 | For REF , assuming MATH we have MATH by REF . Conversely, from REF MATH and MATH and from the hypothesis MATH, so MATH. For REF , the proof is similar, noticing for its converse that MATH. |
quant-ph/0009038 | This equation is obviously a substitution instance of the REFOA law REF . On the one hand, it fails in lattice LREF but holds in lattice MATH (REF a). On the other hand, the REFOA law REF holds in lattice LREF but fails in lattice MATH. |
quant-ph/0009038 | Writing the REFOA law as MATH, we weaken the left-hand side of the inequality with MATH, etc. to obtain MATH . This equation fails in OML LREF but holds in LREF. |
quant-ph/0009038 | It is obvious by definition that MATH is the REFOA law REF . It is also obvious MATH implies MATH and thus the REFOA law: each subexpression of MATH is greater than or equal to the subexpression of MATH that replaces it. To show that MATH holds in MATH, we closely follow the proof of the orthoarguesian equation in CITE... |
quant-ph/0009038 | The proof is analogous to the proof of REF . |
quant-ph/0009038 | In any OML, from REF we have MATH. Using this, we rewrite MATH as MATH. In any OML we also have MATH. Repeatedly applying the commutation law MATH, we prove MATH. Similarly, for any MATH we have MATH. Repeatedly applying the commutation laws MATH and MATH, we can build up MATH for any expression MATH constructed from v... |
quant-ph/0009038 | REF . |
quant-ph/0009038 | For REF : MATH[using MATH, MATH] MATH[using MATH, MATH] MATH[using MATH, MATH] MATH[using MATH, MATH] MATH[since MATH and MATH] MATH. For REF : MATH[using MATH, MATH] MATH[using MATH] MATH. The other direction of the inequality is obvious. For REF : MATH[using MATH, MATH] MATH. |
quant-ph/0009038 | For REF : Let MATH abbreviate MATH. By REF we have MATH. Hence from the first hypothesis MATH, so MATH and by REF MATH. Using REF , we obtain MATH . Since MATH, it follows that MATH, MATH, and MATH. In a similar way we obtain REF from REF respectively. To obtain the MATH-GO law from REF , we substitute MATH for MATH, M... |
quant-ph/0009038 | For REF : By F-H we have MATH. From REF we have MATH. Combining these we have MATH. Substituting MATH for MATH and MATH for MATH and simplifying with REF gives the result. For REF , MATH: Expanding the definition of MATH REF and applying F-H, we have MATH [using REF ] MATH. For REF , MATH: Expanding the definition of M... |
quant-ph/0009038 | Assume that REF holds. Substitute MATH for MATH, MATH for MATH, and MATH for MATH. It is easy to see the hypotheses of REF are satisfied [use REF to establish the third hypothesis]. Using REF , the left-hand side of the conclusion evaluates to MATH. The right-hand side is MATH, which by REF is MATH, establishing the RE... |
quant-ph/0009038 | Assume that REF holds. Since MATH, we also have that REF holds. So by REF we have MATH . By REF we also have MATH, so MATH; applying REF again to the right-hand side we obtain MATH . In REF we substitute MATH for MATH, MATH for MATH, and MATH for MATH. It is easy to see the hypotheses of REF are satisfied, and the conc... |
quant-ph/0009039 | The theorem is a special case of one in CITE, and the reader is referred to that paper for a strictly formal proof. Here we will give a slightly less formal sketch. Let us say that a diagram MATH is accepted by the algorithm if a call scan-MATH occurs. We will first prove that at least one member of each isomorphism cl... |
quant-ph/0009039 | Consider a diagram MATH with more than one block. If MATH is not restricted to connected diagrams, MATH for any MATH, so MATH is reducible. Suppose instead that MATH contains only connected diagrams. Choose a longest possible sequence MATH of distinct blocks MATH, where MATH for MATH. Let MATH and MATH be two atoms of ... |
quant-ph/0009091 | For each individual pair MATH, where MATH, by the definitions of MATH and MATH, we have MATH . By the properties of the inner product and that MATH is unitary, the above expressions are simplified to MATH . The effects of MATH and MATH cancel out on most base vectors, except for these MATH such that MATH. Therefore REF... |
quant-ph/0009102 | Let MATH be fixed; putting MATH we find that MATH satisfies REF , since if MATH then MATH is in MATH. As a consequence of uniqueness, we have the desired result. |
quant-ph/0009102 | Using the properties of integration of projection valued measures, it is easy to see that the MATH-timelike component is a c-number if and only if MATH is constant almost everywhere according to MATH. It is constant only on the two-dimensional affine subspaces of MATH parallel to MATH. But considering the transformatio... |
cs/0010005 | We prove the lemma by induction on MATH. Let MATH be a total, associative function. CASE: For MATH, both REF above hold trivially. CASE: Let MATH be such that MATH. Since MATH is total, MATH. Since MATH, either MATH or MATH (or both). Therefore, for MATH, MATH generates one of the sets that satisfies one of REF or REF ... |
cs/0010005 | Let MATH be a nondecreasing, unbounded, total, recursive function. We will construct a MATH-to-REF, total, commutative, associative, recursive function MATH. Our construction uses a downward self-reducible trick that results in a total, single-valued, one-to-one function MATH (recall that MATH is the ``power multiset" ... |
cs/0010005 | Let MATH, MATH, and MATH be as assumed above. Let MATH be a nondeterministic NAME machine that accepts MATH, runs in polynomial time, and on input MATH has no more than MATH accepting paths. We will use MATH to build an associative, one-way function MATH that is strong, total, and MATH-to-one. First, we introduce some ... |
cs/0010005 | If MATH UP MATH P, then MATH is accepted by a nondeterministic NAME machine that runs in polynomial time and has, at most, one accepting path. Taking MATH, by REF there exists a MATH-to-one AOWF. |
cs/0010005 | For the ``only if" direction, suppose that MATH is a language accepted by a nondeterministic NAME machine that runs in polynomial time and, on input MATH, has at most MATH accepting paths (where MATH is a polynomial). We can easily find another polynomial MATH that is nondecreasing and greater than or equal to MATH. By... |
cs/0010005 | Let MATH be an associative function in MATH. We will prove the above lemma by induction over MATH. First, assume that MATH. Clearly, MATH satisfies the conditions of the lemma. Next, suppose that MATH. By the induction hypothesis, there exists MATH such that MATH satisfies one of REF or MATH from REF , that MATH, and t... |
cs/0010005 | Suppose that MATH satisfies REF . We can write REF equivalently as MATH . We will use induction over MATH to prove that MATH satisfies the conditions of the lemma. Suppose that MATH. It follows immediately that, for all MATH, MATH. Next, suppose that MATH. By associativity, MATH for our choice of MATH. Now, MATH . Supp... |
cs/0010005 | Suppose that MATH is a total, associative function that satisfies REF . By REF , there exists MATH where MATH such that for all MATH where MATH, and all MATH, MATH. We will prove, by contradiction, that MATH is not MATH-to-REF, where MATH inverts MATH, defined as MATH. Assume that, for all MATH, MATH is MATH-to-one. Le... |
cs/0010007 | Note that MATH is usually not accounted for in the I/O model, but we will keep track of the internal memory computation done in MATH in our emulation. The idea behind the emulation is as follows. We will mimic the behavior of the I/O algorithm MATH in the cache model, using an array NAME of MATH blocks to play the role... |
cs/0010007 | Any lower bound in the number of block transfers in MATH carries over to MATH. Since the lower bound is the maximum of the lower bound on number of comparisons and the bound in REF , the theorem follows by dividing the sum of the two terms by REF. |
cs/0010007 | The MATH-way mergesort algorithm described in CITE has an I/O complexity of MATH. The processing time involves maintaining a heap of size MATH and MATH per output element. For MATH elements, the number of phases is MATH, so the total processing time is MATH. From REF , and REF , the cost of this algorithm in the cache ... |
cs/0010007 | See REF. |
cs/0010007 | The probability of conflict misses is MATH when MATH is MATH. Therefore the expected total number of conflict misses is MATH for MATH elements. The I/NAME mergesort uses MATH-way merging at each of the MATH levels, hence the second part of the theorem follows. |
cs/0010007 | If MATH and MATH map to the same cache set in MATH cache then their MATH-th level memory block numbers (to be denoted by MATH and MATH) differ by a multiple of MATH. Let MATH. Since MATH (both are powers of two), MATH where MATH. Let MATH be the corresponding sub-blocks of MATH and MATH at the MATH-th level. Then their... |
cs/0010007 | Let the set of blocks of size MATH be MATH (we are assuming that the blocks are aligned). Let the target block in the contiguous area for each block MATH be in the corresponding set MATH where each block MATH is also aligned with a cache line in MATH . Cache. Let block MATH map to MATH, MATH where MATH denote the set o... |
cs/0010007 | The proof of NAME and NAME can be modified to disregard block transfers that merely rearrange data in the external memory. Then it can be applied separately to each cache level, noting that the data transfer in the higher levels do not contribute for any given level. |
cs/0010007 | We perform a MATH-way mergesort using the variation proposed by CITE in the context of parallel disk I/Os. The main idea is to shift each sorted stream cyclically by a random amount MATH for the MATH-th stream. If MATH, then the leading element is in any of the cache sets with equal likelihood. Like CITE, we divide the... |
cs/0010007 | The two input buffers and the output buffer, even if they map to the same cache set can reside simultaneously in the cache. Since at any stage only one REF-merger is active there will be no conflict misses at all and the cache misses will only be in the form of capacity or compulsory misses. |
cs/0010007 | If MATH and MATH are such that MATH we have the total number of conflict misses MATH . Note that the condition is satisfied for MATH for any fixed MATH which is similar to the tall-cache assumption made by NAME. The set associative case is proved by REF . |
cs/0010007 | : We will split up the summation of REF into two parts, namely, MATH and MATH. One can obtain better approximations by refining the partitions, but our objective here is to demonstrate the existence of MATH and MATH and not necessarily obtain the best values. MATH . The first term can be upper bounded by MATH which is ... |
cs/0010008 | The proof is based on the characterisation of polynomial space computable functions by means of ramified reccurrence, reported in CITE. We shall represent ramified functions by MATH-programs. Firstly, let us recall briefly how ramified functions are specified. Let MATH be the set of constructors, and suppose that each ... |
cs/0010008 | REF is a consequence of REF , of REF and of REF is a consequence of REF by observing that for each MATH, we have MATH. |
cs/0010008 | Given a program MATH, the operational semantics of call by value are provided by a relation MATH where MATH is a substitution over MATH. The relation MATH is defined as the union of the family MATH defined below : CASE: MATH, if MATH and MATH. CASE: MATH, if MATH is a MATH-ary constant of MATH. CASE: MATH if MATH CASE:... |
cs/0010008 | The demonstration of the theorem above is tedious. The theorem is a consequence of REF whose proof is detailed in the three next subsections. For each ground substitution MATH and rule MATH, we have MATH by REF below. The result follows from the monotonicity of the interpretation as stated in REF . |
cs/0010008 | Suppose that MATH is a function symbol of rank MATH. Put MATH. By definition, we have MATH. Thus, MATH, since MATH. |
cs/0010008 | Straightforward by induction on MATH. |
cs/0010008 | The proof is by induction on MATH. CASE: Assume MATH. CASE: Suppose that MATH is a constant of MATH. We have MATH, by REF . CASE: Suppose that MATH is a variable of MATH. We have MATH for some MATH satisfying MATH and MATH. By the hypotheses of the lemma, MATH. So, MATH by REF . CASE: Assume MATH. CASE: Suppose that MA... |
cs/0010008 | By induction on MATH. CASE: Assume MATH. CASE: Suppose that MATH is the constant MATH. We obtain MATH, by REF and so REF holds. CASE: Suppose that MATH is a variable. We have MATH for some MATH. By the hypotheses of the lemma we have MATH. Therefore MATH. CASE: Assume MATH. Suppose that MATH. The hypotheses of the lemm... |
cs/0010008 | The proof goes by induction on MATH. However all the cases are the same as in in REF , except the following one. Suppose that MATH where MATH is a function symbol of MATH of rank MATH. We have MATH for each position MATH such that MATH. Now, MATH implies MATH, by REF . So we can apply the lemma hypothesis, and we obtai... |
cs/0010008 | The proof is by induction on MATH. We have MATH. Assume that MATH, MATH. We have MATH, and so MATH by induction hypothesis. Assume that MATH. REF yields that MATH. By REF , MATH. Therefore, MATH. |
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