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By James H. Martin

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X, SLD-resolution can be applied to test P'. The above argument is correct, but an important point is missing, namely that and in the above method, answer substitutions are lost. It is possible to try to argue that the above method can be refined to take care of this problem. However, a refinement that works in the general case will have to mimic our method. This is because, in order to return the correct answer substitution, it is necessary to keep track carefully of all uses of negations of clauses from the query.

Who is innocent? The above puzzle can be formalized as follows: shot(X, X). inhouse(X), ingarden(X). +- ingarden( dale), ingarden(peter). inhouse(j essica). suspect( dale). suspect(peter ). suspect(j essica). suspect( dave). ingarden(X) +- shot( dave, X). shot(dave, X) +- ingarden(X), suspect(X). +- +- ? - -'shot(dave, X). The Hornlog system returns the expected substitutions, X and the indefinite answer X = dale V X = peter. = dave, X = jessica, A Prolog solution to the above example has a very different character that does not reflect the negative content of the information.

Since (B 2 ,Bs) E E and Bs is marked, mark B2i 5. since (B 1 , B 2 ) E E and B2 is marked, mark Bl i 6. 1. 56 5 SOUNDNESS AND COMPLETENESS RESULTS I This pebbling is thus of length 6. 4 Let Gp = (V,E,L) be a ground H-graph for a set P of ground Horn clauses. 1 from {T} in Gp. They also give an algorithm, the traverse procedure, which identifies pebblings in linear time and which is shown to be sound and complete, in the sense that if it finds a pebbling in G p, P is (model-theoretically) unsatisfiable, and if P is unsatisfiable, the algorithm will find a pebbling.

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