paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/0007015 | REF by induction on MATH. The base case MATH trivially follows from the domain of w variables, which have non-negative values. The same observation concerning the domain of w variables simplifies the proof obligation to the case MATH. It is useful also to observe base cases for MATH and MATH, since by the end of round ... |
cs/0007015 | REF by induction on MATH, for MATH. Note that round MATH occurs in a based computation, since within two rounds following MATH the computation is based. The base case for induction is shown for MATH and MATH, since the main induction step relies on two previous rounds of a based computation. For MATH, since every clock... |
cs/0007015 | In any computation, either some process executes a double-reset or no process does so. In the latter case, the core system component stabilizes within MATH rounds, and the repair timer concurrently reaches the timer-final condition within MATH rounds by REF . This demonstrates MATH stabilization time if no double-reset... |
cs/0007030 | The identity function from MATH to itself trivially is a step refinement from MATH to itself. Hence MATH is reflexive. Transitivity follows from the observation that if MATH is a step refinement from MATH to MATH and MATH is a step refinement from MATH to MATH, then the function composition MATH is a step refinement fr... |
cs/0007030 | For REF , suppose MATH. By induction on MATH we prove MATH . If MATH then both MATH and MATH are MATH. Clearly, MATH. For the induction step, suppose MATH. For reasons of symmetry we may assume, without loss of generality, that MATH. Let MATH be the largest index with MATH and MATH. (By monotonicity, MATH can only be r... |
cs/0007030 | REF follows from the definitions. REF follow immediately from REF and the definitions. |
cs/0007030 | Suppose MATH is a step refinement from MATH to MATH. Let MATH be an execution of MATH. Inductively, we define an execution MATH of MATH and an index relation MATH such that MATH and MATH are MATH-related via MATH. To start with, define MATH and declare MATH to be an element of MATH. Now suppose MATH and MATH is a nonfi... |
cs/0007030 | The relation MATH is a step refinement from MATH to MATH. |
cs/0007030 | Together with an arbitrary norm function, any step refinement (viewed as a relation) is a normed forward simulation. |
cs/0007030 | Let MATH be a function such that MATH and MATH implies CASE: If MATH then MATH. CASE: If MATH then MATH. CASE: If MATH then MATH. The existence of MATH, which chooses between a left move REF of MATH, a common move REF of MATH and MATH, or a right move REF of MATH, is guaranteed by the fact that MATH is a normed forward... |
cs/0007030 | Immediate from the definitions and REF . |
cs/0007030 | REF follows by REF . The proof of REF is routine. |
cs/0007030 | If MATH is infinite then let MATH. If MATH is finite then let MATH be the finite prefix of MATH up to and including the first state whose index is related by MATH to the final index of MATH. Inductively we define a sequence MATH of pairs in MATH. The first element of the sequence is MATH. If MATH is an element of the s... |
cs/0007030 | For reflexivity, observe that the identity function from MATH to itself is a branching forward simulation from MATH to itself. For transitivity, suppose MATH and MATH are branching forward simulations from MATH to MATH and from MATH to MATH, respectively. We claim that MATH is a branching forward simulation from MATH t... |
cs/0007030 | The relation MATH is a branching forward simulation from MATH to MATH. |
cs/0007030 | Trivial. |
cs/0007030 | Similar to the proof of REF . |
cs/0007030 | REF follows immediately by REF and the totality of MATH. In order to prove REF , suppose that MATH is image-finite. Let MATH be an execution of MATH. We have to establish the existence of an execution MATH of MATH with MATH. If MATH is finite then this follows by REF and the totality of MATH. So assume that MATH is inf... |
cs/0007030 | For REF , suppose that MATH is deterministic and that MATH is a normed backward simulation from MATH to MATH. Suppose that MATH is a reachable state of MATH. We will prove that MATH contains exactly one element. Since any normed backward simulation that is functional on the reachable states trivially induces a step ref... |
cs/0007030 | REF follows by REF . The proof of REF is routine. |
cs/0007030 | Similar to the proof of REF . |
cs/0007030 | The relation MATH is a branching backward simulation from MATH to MATH. |
cs/0007030 | Clearly, MATH is a forest. The function MATH which maps each finite execution of MATH to its last state is a step refinement from MATH to MATH, and the relation MATH, together with an arbitrary norm function, is a normed forward simulation from MATH to MATH. |
cs/0007030 | Take MATH. By REF , MATH is a forest and MATH. Since MATH, also MATH by soundness of history relations. Next apply the partial completeness result for backward simulations REF to conclude MATH. |
cs/0007030 | Straightforward from the definitions. |
cs/0007030 | Forward implication follows by REF . For backward implication, suppose MATH. Then MATH by the definition of history relations, and MATH because any step refinement is a normed forward simulation. Now MATH follows by the fact that MATH is a preorder. |
cs/0007030 | Suppose that MATH is a normed history relation from MATH to MATH, and MATH is a normed history relation from MATH to MATH. Because MATH and MATH have fin, both MATH and MATH are finite. Since MATH is a step refinement, it maps start states of MATH to start states of MATH. Using the fact that MATH is a forward simulatio... |
cs/0007030 | Analogous to that of REF , using REF . |
cs/0007030 | By REF , there exists an automaton MATH with MATH. Next, REF yields the required automaton MATH with MATH, which proves REF . The proof of REF is similar, but uses REF . |
cs/0007039 | REF were introduced and shown to be derived in a preferential relation in CITE. For REF, suppose MATH. Applying S, we get MATH and, by Right weakening, we conclude REF. For REF, suppose MATH. Then, by REF, we get MATH and, by Left Logical Equivalence, we conclude REF. |
cs/0007039 | We must prove that MATH satisfies NAME, Left Logical Equivalence, Right Weakening and Consistency Preservation with respect to MATH. The rest of the rules are already satisfied since they do not involve an underlying consequence relation. First notice that Consistency Preservation is immediate by definition of MATH. Fo... |
cs/0007039 | REF were introduced and showed to be derived in a preferential relation in CITE. For REF, suppose MATH. Applying S, we get MATH, and, by Right weakening, we conclude REF. For REF, suppose MATH. Then, by REF, we get MATH, and, by Left Logical Equivalence, we conclude REF. |
cs/0007039 | We must prove that MATH satisfies NAME, Left Logical Equivalence, Right Weakening and Consistency Preservation with respect to MATH. The rest of the properties are already satisfied since MATH is a rational inference relation. First notice that Consistency Preservation is immediate by definition of MATH. For NAME, supp... |
cs/0007039 | REF is immediate from defining condition (MATH). For REF , suppose that MATH. We must show MATH. Since MATH, we have MATH, by Right Weakening. Applying Or, we get MATH. By hypothesis and And, we get MATH. The left to right direction of REF is straightforward. For the right to left direction, suppose MATH. Then, by Comp... |
cs/0007039 | For REF , if MATH we get immediately MATH, by REF . If not, that is MATH, then, by And and Right Monotonicity, we have MATH. Again, by REF , we have MATH. For REF , assume MATH. If MATH. Then we have MATH, so MATH, and hence, by Dominance, MATH, as desired. If not then there must be MATH such that MATH and MATH. Theref... |
cs/0007039 | We shall try not to overlap with the proof of NAME and NAME proof of REF (see proof of REF). Therefore we do not cover the case where the second half of condition REF applies. The list of rules we verify is NAME, Left Logical Equivalence, And, Cut, Cautious Monotony, Or and Rational Monotony. Right Weakening follows fr... |
cs/0007039 | Let MATH and MATH be the ranked consequence operators induced by MATH and MATH respectively, where the latter is the closure of the former under arbitrary unions and intersections. Without loss of generality we can assume that the sets belonging in MATH carry the same indices in MATH. We must prove that MATH is equal t... |
cs/0007039 | Let MATH be a ranked consequence operator. Denote MATH with MATH, and MATH with MATH. We should verify that MATH satisfies NAME, Transitivity, and Conjunctiveness. For NAME, suppose MATH. We have MATH, so if MATH than MATH, for all MATH. Hence MATH. For Transitivity, suppose MATH and MATH. Pick a MATH such that MATH. W... |
cs/0007039 | We shall give an alternative proof with a straightforward verification of the rules of rational inference. We shall show that a ranked consequence operator MATH based on MATH induced by a chain of sets MATH satisfies NAME, Left Logical Equivalence, Right Weakening, And, Cut, Cautious Monotonicity, Or and Rational Monot... |
cs/0007039 | Denote the comparative rational inference relation by MATH. We shall define a chain of sets MATH which generates a ranked consequence operator MATH equal to MATH. Let MATH be the equivalence relation induced by MATH (an expectation ordering is clearly a preorder). The equivalence classes will be denoted by MATH (where ... |
cs/0007044 | Let MATH be a nonhomogeneous NAME process with intensity function MATH, which implies MATH. Now, the chance that no new tuple was inserted during MATH is the same as the chance that the process MATH has no arrivals during MATH, that is, MATH. The chance that a new tuple was inserted during MATH is just the complement o... |
cs/0007044 | Let MATH be a randomly chosen tuple, and assume that no corresponding tuple to MATH in MATH is deleted. The proof is identical to that of REF , replacing MATH with MATH and MATH by MATH. |
cs/0007044 | Considering all MATH, and using the well-known fact that if MATH for MATH are independent, then MATH we conclude that the remaining lifetime of MATH (denoted MATH) has a nonhomogeneous exponential distribution with intensity function MATH. The probability of a given tuple MATH surviving through time MATH is thus MATH a... |
cs/0007044 | MATH . |
cs/0007044 | Let MATH denote the lifetime of a tuple inserted into MATH at time MATH. Similarly to the proof of REF , we know that MATH. The probability such a tuple survives through time MATH is the random quantity MATH . Considering all the possible elements of the vector MATH, we then obtain MATH . Assume now that MATH has fixed... |
cs/0007044 | Let MATH be the number of insertion events in MATH, and let their times be MATH. Suppose that MATH and that insertion event MATH happens at time MATH. Event MATH inserts a random number of tuples MATH, each of which has probability MATH of surviving through time MATH. Therefore, the expected number of tuples surviving ... |
cs/0007044 | Each tuple in MATH has a survival probability of MATH, which yields that MATH. By the definitions of MATH and MATH, one has that MATH so therefore MATH . |
cs/0007044 | Let MATH denote a generic random variable with cumulative distribution MATH. The probability of MATH surviving throughout MATH is then MATH and therefore the expected number of tuples in MATH that survive through time MATH is MATH . We now consider a tuple MATH inserted at some time MATH. The probability that such a tu... |
cs/0007044 | Let MATH, MATH, and MATH be the times until the next insertion, modification, and deletion in MATH, respectively. From REF, we have that MATH. Now, for each MATH, there are MATH tuples whose deletion would cause a deletion in MATH. The time until deletion of any such MATH is distributed like MATH. The deletion processe... |
cs/0007044 | Consider a continuous-time NAME chain on the same state space MATH, and with the same instantaneous transition probabilities MATH, where MATH. However, in the new chain, the holding time in each state MATH is simply a homogeneous exponential random variable with arrival rate MATH. We call this system the linear-time ch... |
cs/0007044 | We first compute the expected number of surviving tuples MATH whose values MATH migrate to MATH. Given a value MATH, there are MATH tuples at time MATH such that MATH. Using the previous results, the expected number of these tuples surviving through time MATH is MATH, and the probability of each surviving tuple MATH ha... |
cs/0007044 | Let MATH, which is a monotonically nondecreasing function. From REF , MATH for all MATH. We have MATH. By applying the monotonic function MATH to both sides of the inequality MATH, one has that MATH for all MATH. Substituting in the definitions of MATH and MATH, one then obtains MATH for all MATH, and therefore MATH. |
cs/0007044 | In this case, we note that the random variable MATH is identical to MATH (using the notation of REF), and is independent of MATH. The number MATH of modification events in MATH has a NAME distribution with mean MATH, and hence variance MATH. Therefore we have, for any MATH, MATH . |
hep-th/0007163 | The uniqueness of such a function follows immediately from the NAME - NAME Theorem. To show the existence we shall consider the following function MATH . The set of REF and the constraint MATH (the latter means that MATH is an elliptic function) form the system of MATH equations on the functions MATH . For generic data... |
hep-th/0007163 | Consider a function MATH. It's straightforward to check that the function MATH satisfy all defining properties of the function MATH. The uniqueness of MATH implies that MATH. |
hep-th/0007163 | To obtain these equations one has to divide REF by MATH and compare the residues of the both sides of the obtained equation at the points MATH . |
hep-th/0007163 | Let a function MATH be a solution to the differential equation MATH . We define new variables MATH, MATH by the conditions MATH. In terms of these variables system REF has the following form MATH . The rest of the proof is parallel to the genus one case considered above. |
hep-th/0007163 | The uniqueness of such a function (provided it exists) follows immediately from the NAME - NAME Theorem. To show the existence we write this function down explicitly in terms of the NAME MATH-function. The NAME MATH-function, associated with an algebraic curve MATH of genus MATH is an entire function of MATH complex va... |
hep-th/0007163 | Consider a function MATH. Projections MATH of its zeros onto MATH-plane satisfy matrix REF . The first MATH equations ensure that restrictions REF are satisfied. Therefore the polynomial MATH is of degree MATH and the last MATH equations in system REF state that the points MATH are its zeros. The coefficients of MATH t... |
hep-th/0007194 | It follows easily from REF that MATH; since MATH the result follows. |
hep-th/0007194 | By REF , MATH . It follows that MATH. |
hep-th/0007194 | The vector fields MATH commute with MATH; it follows that MATH . Written out explicitly, this equation says that MATH . Applying the operator MATH, MATH, we obtain MATH . The lemma is an immediate consequence of these formulas. |
hep-th/0007194 | Since MATH, REF implies that MATH . Since MATH by REF, we conclude that MATH. By REF , the restriction of MATH to the small phase space equals MATH. It follows that the restriction of MATH to the small phase space equals MATH; hence the functions MATH form a coordinate system in a neighbourhood of the small phase space... |
hep-th/0007194 | By the genus MATH dilaton equation MATH, we have MATH, and the formula for MATH follows. |
hep-th/0007194 | Let MATH; then MATH and MATH. Thus MATH satisfies the equations MATH for MATH, and MATH, as well as the dilaton equation MATH, and is thus determined by the restrictions of the partial derivatives MATH to the small phase space. But these vanish; we conclude that MATH. |
math-ph/0007003 | This proof is from Ref. CITE. Define MATH by MATH for MATH, and MATH for MATH. Clearly MATH. Computing the strain MATH by REF gives MATH showing that MATH is a smooth isometry with MATH. |
math-ph/0007003 | Setting MATH and MATH in the definition of a flat bilinear form and using the symmetry of MATH gives MATH for every MATH. We will use this result repeatedly in the proof of the lemma. Let MATH and MATH. We need to demonstrate the existence of at least MATH linearly independent vectors MATH such that MATH for all MATH a... |
math-ph/0007003 | CASE: Set MATH. Let MATH for all points in the open subset MATH. Let MATH be the orthogonal complement of MATH in MATH. Evidently the dimension of MATH is MATH. On the other hand, we may express MATH via MATH given MATH, there exist MATH and MATH ,such that MATH . Take smooth local extensions MATH and MATH. Clearly, MA... |
math-ph/0007003 | Given a set MATH, let MATH denote the boundary of MATH, MATH denote the closure of MATH and MATH denote the interior of MATH. First, we note that MATH is closed in MATH implies that MATH is nowhere dense. This can be shown as follows: Since MATH is closed, MATH. Therefore, MATH. However, MATH. Consequently, it follows ... |
math-ph/0007003 | CASE: Let MATH and MATH. Then MATH so that MATH, and MATH for any arbitrary vector MATH. Therefore, MATH . Using REF , we obtain MATH . This implies that MATH. Also, MATH . A similar argument shows that MATH. Therefore, MATH. By the NAME 's Theorem cited above, MATH is an integrable distribution in MATH. By a similar a... |
math-ph/0007003 | CASE: Let MATH. By REF , MATH. We see that MATH which is open by REF , where, as before, MATH denotes the complement of MATH. Let MATH, where MATH is the open ball with radius MATH. Then MATH is a nonempty open set with MATH for all MATH . If MATH for all MATH, by REF MATH is smooth and integrable on MATH so that we ca... |
math-ph/0007003 | We begin with the following observations which are easily verified using the fact that geodesics are extremal curves for the arc length. CASE: If MATH, the unit speed geodesic through MATH and MATH is unique (except for reparameterization) and is given by MATH . CASE: Every unit speed curve MATH with MATH and MATH has ... |
math-ph/0007003 | We will identify the tangent spaces MATH and MATH with MATH and MATH respectively by the standard exponential map. Since MATH, REF implies that MATH where MATH and MATH . Similarly, MATH where MATH and MATH . Since MATH is an isometric immersion, MATH . This along with MATH and MATH, and using the identification of the... |
math-ph/0007003 | We will first show that for every MATH, where MATH is as defined in REF , there exists a MATH such that MATH. We will prove this proposition by induction. REF implies that this statement is true if MATH. Note that the statement is trivially true for all MATH if MATH. We will assume that this statement is true for MATH ... |
math-ph/0007006 | See NAME 's book CITE for a proof of a more general result. An outline of the proof is as follows: NAME first transforms the equation into another complex MATH-plane by using the NAME transform. And then he compares MATH with the solutions of the sine equation MATH and finally transforms back to the original complex MA... |
math-ph/0007006 | Let MATH be an eigenfunction with eigenvalue MATH, so that MATH where MATH for some MATH, MATH. Write MATH . Fix MATH with MATH. Let MATH. Then MATH and MATH . Thus our ODE becomes MATH . Then we multiply this by MATH, integrate and use integration by parts to get MATH for all MATH, where we note that the line MATH sta... |
math-ph/0007006 | The main idea of the proof is the same as that of the proof of REF . NAME if REF is little different from the equation for REF , NAME regions for REF are the same as for REF (if MATH has the same sign as MATH) or else are rotated by MATH (if MATH has the opposite sign). See Section MATH in CITE for details. But in eith... |
math-ph/0007006 | Let MATH for MATH. Then MATH and MATH . Hence by integration by parts, MATH . Now by the formula MATH and splitting real and imaginary parts of the above, we get the lemma. |
math-ph/0007006 | For any MATH, by REF with MATH and by MATH, we have that MATH and this is negative for MATH with MATH, because then MATH for all MATH and so MATH. This argument also shows that MATH in MATH; see REF . Similarly in MATH, for all MATH we have that MATH. For MATH (so that MATH), we use REF along vertical line segments sta... |
math-ph/0007006 | In the regions MATH and MATH, we use REF with horizontal lines to infinity to get the statements in REF and MATH of this theorem. In the region MATH between MATH and MATH with the real part less than or equal to that of the zero of MATH in the third quadrant (see REF ), we use REF with vertical lines MATH to show that ... |
math-ph/0007006 | On horizontal line segments MATH in MATH, REF becomes MATH . Since MATH in MATH, MATH is a strictly increasing function of MATH on each horizontal line segment in MATH. |
math-ph/0007006 | The proofs of REF give everything except the last statement of the theorem. For that, recall we can take MATH by REF ; this implies MATH is an odd function with respect to reflection in the imaginary axis, so MATH on the whole imaginary axis. Now we use REF to complete the proof. |
math-ph/0007006 | We will first prove this for MATH. Suppose that MATH. Then we could find a vertical line segment MATH in MATH whose end points are MATH and MATH. We apply REF to this line segment to get MATH . This would imply MATH on the curve MATH since MATH in MATH. So then since MATH is analytic, MATH in MATH. This is a contradict... |
math-ph/0007006 | Suppose MATH is an eigenfunction of REF with eigenvalue MATH with MATH. Since MATH has infinitely many zeros in MATH (by the paragraph shortly before REF ), certainly MATH also has infinitely many zeros in MATH. Proof of REF . Suppose that MATH for some point MATH on the imaginary axis. By REF with MATH, we have that M... |
math-ph/0007006 | We omit the proof because it is very similar to the proof of REF . We use REF instead of REF , and also make use of REF . |
math-ph/0007006 | It is useful to have the following two formulas, which follow from multiplying REF by MATH and separating real and imaginary parts: MATH and MATH . At the end of the proof we will justify the fact that we can differentiate through the integrals that follow. MATH . This is clear by integrating REF , using the zero bound... |
math-ph/0007006 | This is a consequence of REF with MATH, or it can be proved using the subharmonicity of MATH . |
math-ph/0007007 | The first inequality follows from the strict positivity of MATH. The upper bound can be given by choosing the projection onto the ground state of the Hamiltonian MATH, as a test density matrix MATH. Then we apply the bound to the repulsion by attraction REF , together with the observation, that the nucleus has to be at... |
math-ph/0007007 | Applying the diamagnetic inequality CITE to the lower bound in REF , we get MATH . To get the upper bound, we apply NAME 's inequality REF MATH to the upper bound in REF . |
math-ph/0007007 | We follow closely the proof of the analogous REF. Let MATH be a minimizing sequence for MATH with MATH. First note that MATH is bounded above, because MATH is bounded above, and since the other contributions to the energy are bounded relative to MATH. Now we show that the magnetic-kinetic energy is bounded below by REF... |
math-ph/0007007 | This follows immediately from the strict convexity of MATH. |
math-ph/0007007 | (We proceed essentially as in CITE). For any MATH . Especially, for all MATH, MATH . Now if there exists a MATH with MATH and MATH we can choose MATH small enough to conclude that MATH which contradicts the fact that MATH minimizes MATH. |
math-ph/0007007 | For MATH this was shown in CITE, and MATH was computed numerically to be MATH in CITE. Fix MATH. We assume that MATH, and will show that MATH has an eigenvalue strictly below zero. Using MATH with some MATH as a (not normalized) test function we compute MATH where MATH is the potential generated by the charge distribut... |
math-ph/0007007 | This follows from REF and the bound on MATH given in CITE (see REF below). |
math-ph/0007007 | For MATH, that is, the hydrogen atom, this was shown in CITE (and also in CITE). So we can restrict ourselves to considering the case MATH. Let MATH be a normalized ground state for MATH, with angular momentum MATH. Define MATH . The function MATH is continuous and bounded. It achieves its minimal value at MATH, becaus... |
math-ph/0007007 | Mimicking the proof of the last theorem, we see that if MATH has a ground state, the corresponding wave function MATH has MATH. Hence it is given by REF . |
math-ph/0007007 | Fix MATH, and let MATH be the minimizer of the density functional MATH under the constraint MATH. Using MATH as a trial function we get MATH . It is well known that MATH falls off more quickly that the inverse of any polynomial, so MATH is finite CITE. For the converse, we estimate MATH . The observation MATH then lead... |
math-ph/0007007 | For MATH we use again REF , and the analogous inequality in the other direction for MATH, to conclude that MATH where we have also used REF . With the aid of REF one easily sees that MATH . Using again MATH we finally get MATH . In particular, if we choose MATH, we arrive at the desired result, as long as MATH is large... |
math-ph/0007007 | The equation for the kinetic energy is obvious by definition of MATH. We calculate the energy of attraction as MATH where MATH . For the term in square brackets we use REF, MATH to estimate the difference to the expected limit in the NAME: MATH . For the repulsion MATH we calculate MATH where MATH . The inequality is s... |
math-ph/0007007 | Combining all the bounds of the Lemma, and using MATH to simplify, gives MATH . For the upper bound on MATH we specify MATH as MATH, the minimizing wave function for the NAME. See REF or the following REF . It remains MATH, for all MATH, so these wave functions are normalized to MATH. Since the variational principle wi... |
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