Inductive Logic Programming: 10th International Conference, by David Page (auth.), James Cussens, Alan Frisch (eds.)

By David Page (auth.), James Cussens, Alan Frisch (eds.)

This e-book constitutes the refereed complaints of the tenth foreign convention on Inductive common sense Programming, ILP 2000, held in London, united kingdom in July 2000 as earlier of CL 2000. The 15 revised complete papers provided including an invited paper have been rigorously reviewed and chosen from 37 submissions. The papers handle all present matters in inductive common sense programming and inductive studying, from foundational facets to purposes in a variety of fields like facts mining, wisdom discovery, and ILP process layout.

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Extra info for Inductive Logic Programming: 10th International Conference, ILP 2000 London, UK, July 24–27, 2000 Proceedings

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If rhs(sx , cx ) =∅, then set [sx , cx ] to [sx , rhs(sx , cx )]. 4. Repeat until no more changes are made – For every term t appearing in sx , in increasing order of size, do • Let [sx , cx ] be the multi-clause obtained after removing from [sx , cx ] all those literals containing t. • If rhs(sx , cx ) =∅, then set [sx , cx ] to [sx , rhs(sx , cx )]. 5. Return [sx , cx ]. Fig. 3. The minimisation procedure counterexample’s antecedent: sx (line 2). This can be done by forward chaining using the hypothesis’ clauses, starting with the literals in antecedent(x).

Inductive Logic Programming (ILP) deals with learning first order logic programs. Very recently the expressiveness of the target language was extended to prenex conjunctive normal forms [20] by allowing existential quantifiers in the language. Description Logics (DL), on the other hand, are a different kind of knowledge representation language used for representing structural knowledge and J. Cussens and A. ): ILP 2000, LNAI 1866, pp. 40–59, 2000. c Springer-Verlag Berlin Heidelberg 2000 A Refinement Operator for Description Logics 41 concept hierarchies.

Ineq(st ) · θ · θt ⊆ ineq(sx ). Let t1 , t2 two distinct terms of st . We have to show that t1 · θ · θt and t2 · θ · θt are two different terms of sx and therefore their inequality appears in ineq(sx ). It is easy to see that they are terms of sx since st · θ · θt ⊆ sx . To see that they are also different terms, notice first that t1 · θ and t2 · θ are different terms of sx , since the clause ineq(st ), st → bt is captured by [sx , cx ]. It is sufficient to show that if t1 , t2 are any two distinct terms of sx , then t1 · θt and t2 · θt are also distinct terms.

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