... proceedings of the theorem prover. It starts applying rules, matching them with the an- tecedent, without making use of the original se- mantic information, and thus resulting in an in- efficient ... rule in (I). ALGORITHMS FOR GENERATION INLAMBEK THEOREM PROVING Erik-Jan van der Linden * Guido Minnen Institute for Language Technology and Artificial Intelligence Tilburg University PO ... generation; theorem proving; bidirection- ality; categorial grammar. 1 INTRODUCTION Algorithms for tactical generation are becoming an increasingly important subject of research in computational linguistics...
... (Dutch). van der Linden, H. 1988b User-documentation for SIMPLEX. Internal report 88 ITI B 34, ITI-TNO, Delft (Dutch). Moottgat, M. 1987a LambekTheorem Proving. In Klein; and van Benthem ... Grammar. To appear in Droste, F., Ed., Main- streams in Linguistics. Benjamins, Amsterdam. Moortgat, M. 1988 Categorial Investigations. Logical and linguistic aspects of the Lambek cal- culus. ... of the searchspace inLambekTheorem Proving. Inter- nal report 88 ITI B 23, ITI-TNO, Delft (Dutch). Popowich, F. 1988, A Unification-Based Frame- work for Anaphora in Klein and van Benthem...
... gen-erally, for mathematically inclined readers interested in the formalization ofproofs, and the foundations of automatic theorem- proving. The book is self contained, and the level corresponds ... OF CONTENTS xix6.5 Craig’s Interpolation Theorem, 2876.5.1 Interpolants, 2876.5.2 Craig’s Interpolation Theorem Without Equality, 2886.5.3 Craig’s Interpolation Theorem With Equality, 292PROBLEMS, ... by swapping the accepting and the rejecting states ofM. Since for every input u, the computation C0, , Cnof M on input u halts in n ≤ p(|u|) steps, the modified machine M accepts Σ∗− A...
... 2.28)=ZAXZkdIP(Partial averaging)=ZAXdfIP(Lemma 2.28)Example 9.2 (Radon-Nikodym Theorem, continued) We show in Fig. 9.1 the values of the martingaleZk.We always haveZ0=1,sinceZ0= IEZ =ZZdIP ... equivalently,ZAXdfIP =ZAXZkdIP:Chapter 9Pricing in terms of Market Probabilities:The Radon-Nikodym Theorem. 9.1 Radon-Nikodym Theorem Theorem 1.27 (Radon-Nikodym) Let IP andfIPbe two ... self-financing value process (wealth process)consists ofX0, the non-random initial wealth, andXk+1=kSk+1+1+rkXk, kSk;k=0;::: ;n , 1:Then the following processes are martingales...
... continuous setting in resultssuch as (1.1) – that is to say, T – to the discrete, namely Z/NZ.Finally, we would like to remark that it is possible, indeed probable, thatRoth’s theoremin the ... Ruzsaand the author [9]. In the present context things look rather different howeverand, in the absence of anything which might be called a “grain”, we think theterminology of [9] no longer ... anythinglike this could be proved Theorem 1.4 would of course follow trivially.2. Preliminaries and an outline of the argumentAlthough the main results of this paper concern the primes in [N],...
... LP0grammars, since commutativity removes thebound on the number of alterations in L(a).Instead we exploit an assymmetry inherent in the Lifting operation.As noted in (Lambek, 1988), Lifting is a ... TreeCElim. Using hypothetical rea-soning and applying the Right Elimina-tion rule i < n times, we can obtain itimes the type a". All remaining a's canbe lifted to obtain itThus,forany71N+,U{(s, ... naming function)if G assigns types to the symbols in the stringsuch that these types can be combined to de-rive the distinguished type, normally written ass or t.Definition 1 A domain...
... lumped into a single quantity known as integral of the function, and in doing so any knowledge of that detailed structure is lost. Therefore, since the process of integration results in a loss ... field point r and the source point r’ respectively. Since the contribution to the force density from any pair of sources and field points will lie along the line joining the two points, 'r()()()()''n''2nn'rnrrrrrrrrr ... multiplying equation (N 1.2.2) by r and integrating over V produces the same result as multiplying equation (N 1.2.3) by r' and integrating over V'. Thus, doing this and adding equation...
... Jong for pointing out an error in the proof of Theorem 1.1 in the firstversion of the article, and the referee for forcing and helping us to restore thewhole strength of Theorem 1.1 in the corrected ... Deligne’s integrality theorem in unequal characteristic and rational points over finite fields By H´el`ene Esnault* INTEGRALITY 729of ⊕n0H∗(¯Vj, Q). Since integrality ... quotingLewis Carroll to express his “confusion”, is a P2blown up 7 times, crossingitself in 7 rational double curves, themselves crossing in 7 triple points Theorem 1.3 allows one to say (in...
... beexponentially small in M at the beginning of the argument. This parameterthen gets exponentiated again in any application of Proposition 5.1.We note that our proof in fact yields a version of Theorem 1.3 ... ob-tained in the case G = T by Helson [15] and G = Tdby Rudin [20]. See [19,Ch. 3] for a complete discussion of the theorem. When G is finite the idempotent theorem gives us no information, sinceM(G) ... within d(f, Z) + ε of an integer. Noting thatf∞ fA M, we see from Lemma 4.15 that ψSf(x) is not withind(f, Z) + ε/2 of an integer for any x ∈ x0+ Xε/40dM. Choosing, accordingto...
... a singular noun or a verbal infinitive, if possible, and looks it up again. Nouns remain in the singular, since the determiner of a noun provides the transla- tion routine with enough information ... for parenthesizing and grouping are stated in the following outline. OUTLINE OF THE PARENTHESIZING AND GROUPING ROUTINE The rules listed below are applied in sequence to an initially parenthesized ... procedure is described in greater detail in the following outline. OUTLINE OF THE PARSING ROUTINE Shelf 9 is input shelf, Shelf 6 is output shelf, Shelf 1 is for the partial parsings, Shelf 8 is...
... followingproposition. If a function of a real argument is continuous in someinterval and in this interval takes only integer values, then it takes onlyone value in the whole interval. Indeed, ... independent cycles in one of the ways indicated in Problem 190, then N contains all thepermutations splitting into independent cycles in this way.192. Prove that the group does not contain normal subgroupsexcept ... the pointof the plane having the coordinates So instead of the point corre-sponding to the complex number we shall say simply point233. Let the complex numbers be represented by the points...
... below).For information about publishing your research in Journal of Inequalities and Applications go tohttp://www.journalofinequalitiesandapplications.com/authors/instructions/For information ... sequence intheorem 1.1 obtained byHuang [12] but also removed the condition nP(|X1| > ηn) ≤ c(log n)ε0, 0 < ε0< 1 intheorem 1.1 of [12].Remark 1.4. If EX2< ∞, then X is in ... Foundation of China (11061012), the project supported by program to Sponsor Teams for Innovation in theConstruction of Talent Highlands in Guangxi Institutions of Higher Learning ([2011]47),...
... noted that in [20- 22], the asymptotically nonexpansive in the intermedi-ate sense mapping is required to be uniformly continuous. In this article, we assumethe continuity of T instead of un ... iform continuity. Chidume et al. [ 20] gave the w eakconvergence theorem for u niforml y continuous nonself mapping which is asymptoti-cally nonexpansive in the intermediate sense in uniformly ... nonselfnonexpansive mappings. Fixed Point Theory Appl. 2005,1–9 (2005)15. Shahzad, N: Approximating fixed points of non-self nonexpansive mappings in Banach spaces. Nonlinear Anal: Theory,Methods...
... point results: nondecreasing caseOur starting point is the following definition.Definition 2.1. Let ( X, ≤) be a partially ordered set and T: X ® X a mapping. Wesay that T is nondecreasing ... be seen using ε = d(x, y) in (1) when x <y.Proof of Theorem 2.3.IfTx0= x0, then the proof is finished.Suppose that x0<Tx0and T is a nondecreasing mapping, we obtain by induction ... d(z, Tz) = inf{d(x, Tx): x Î X} = 0 and, therefore, z is a fixed point of T.The uniqueness of the fixed point is proved as inTheorem 2.7.Remark 3.3. A parallel result in the nonincreasing case...