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The equivalence of the subsumption theorem and the refutation-completeness for unconstrained resolution

  • Shan-Hwei Nienhuys-Cheng
  • Ronald de Wolf
Knowledge Representation and Programming Languages
Part of the Lecture Notes in Computer Science book series (LNCS, volume 1023)

Abstract

The subsumption theorem is an important theorem concerning resolution. Essentially, it says that a set of clauses ∑ logically implies a clause C, iff C is a tautology, or a clause D which subsumes C can be derived from ∑ with resolution. It was originally proved in 1967 by Lee in [Lee67]. In Inductive Logic Programming, interest in this theorem is increasing since its independent rediscovery by Bain and Muggleton [BM92]. It provides a quite natural “bridge” between subsumption and logical implication. Unfortunately, a correct formulation and proof of the subsumption theorem are not available. It is not clear which forms of resolution are allowed. In fact, at least one of the current forms of this theorem is false. This causes a lot of confusion.

In this paper, we give a careful proof of the subsumption theorem for unconstrained resolution, and show that the well-known refutation-completeness of resolution is an immediate consequence of this theorem. On the other hand, we also show here that the subsumption theorem can be proved starting from the refutation-completeness. This establishes that these two results have equal strength.

Furthermore, we show that the subsumption theorem does not hold when only input resolution is used, not even in case ∑ contains only one clause. Since [Mug92, Ide93a] assume the contrary, some results (for instance results on nth roots and nth powers) in these articles should perhaps be reconsidered.

Subfield

Automated Reasoning (Inductive) Logic Programming 

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References

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Copyright information

© Springer-Verlag Berlin Heidelberg 1995

Authors and Affiliations

  • Shan-Hwei Nienhuys-Cheng
    • 1
  • Ronald de Wolf
    • 1
  1. 1.Department of Computer ScienceH4-19 Erasmus University of RotterdamDR RotterdamThe Netherlands

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