Completeness of the SLD-Resolution

Soundness of SLD Resolution Completeness of SLD Resolution Structure - 1 Preliminaries Definite Programs Semantics Soundness of SLD Resolution Completeness of SLD Resolution Normal Programs Add Negative Information and Finite Failure Soundness of SLDNF Resolution Completeness of SLDNF Resolution

MGU Lemma Let P be a definite Program and G a definite goal MGU Lemma Let P be a definite Program and G a definite goal. Suppose that P{G} has an unrestricted SLD-refutation (1, 2, … are not neccessarily mgu's). Then P{G} has an SLD-refutation of the same length with the mgu's ‘1… ‘n and there exists a substitution , such that 1… n = ‘1… ‘n. Proof (basis) Induction over the length of the unrestricted refutation. n=1: G0=G, G1=□ with input clause C1 and unifier 1. Suppose ‘1 is an mgu of the selected atom in G and the head of the input clause C1. Then 1= ‘1 for some . Furthermore, P{G} has a refutation G0=G, G1=□ with input clause C1 and MGU ‘1.

Proof (Inductive step) MGU Lemma Let P be a definite Program and G a definite goal. Suppose that P{G} has an unrestricted SLD-refutation (1, 2, … are not neccessarily mgu's). Then P{G} has an SLD-refutation of the same length with the mgu's ‘1… ‘n and there exists a substitution , such that 1… n = ‘1… ‘n. Proof (Inductive step) Hypothesis: result holds for n-1. Suppose P{G} has an unrestricted SLD-refutation G0,G1,…Gn =□ of length n with the input clauses C1… Cn and Unifiers ‘1,2,3…,n such that G1=G‘1. By the inductive hypothesis: P{G‘1} has a refutation G‘1,…,G‘n =□ with mgu's ‘2…‘n such that 2…n= ‘2…‘n  for some . Then P{G} has a refutation G0=G, G1,…,Gn =□ with mgu's ‘1…‘n such that 1…n=‘12…n= ‘1…‘n .

Lifting Lemma Lemma Let P be a definite program, G a definite goal and  a substitution. Suppose P{G} has an SLD-refutation with substitutions 1…n, such that the variables of the input clause are distinct from the variables in  and G. Then there exists an SLD-refutation of P{G} of the same length and mgu's ‘1…‘n and a substitution  such that 1…n=‘1…‘n.

Lifting Lemma Proof Suppose the first input clause for the refutation of P{G} is C1, the first mgu is 1 and G1 is the goal which results from the first step Now 1 is a unifier for the head of C1 and the atom in G which corresponds to the selected atom in G. The result of resolving G and C1 using 1 is exactly G1. We obtain an unrestricted refutation of P{G} – which looks almost like the given refutation of P{G}. Now apply the mgu lemma.

Success Set and Least Herbrand Model Theorem The success set of a definite program P is equal to its least Herbrand model.

Proof : corollary (slide 24, previous slide set). : Show that the least Herbrand model of P is contained in the success set of P. Show per induction over n, that ATPn implies, that P{A} has a refutation and hence A is in the success set. Basis: ATP1 means that A is a ground instance of a unit clause of P. Thus there exists a refutation.

Proof (continued) : Show that the least Herbrand model of P is contained in the success set of P. Show per induction over n, that ATPn implies, that P{A} has a refutation and hence A is in the success set. Inductive step: Hypothesis holds for n-1. Let ATPn. By the definition of TP, there exists a ground instance of a clause BB1,…,Bk such that A=B and {B1,…,Bk}  TP(n-1) for some . By the induction hypothesis P{Bi} has a refutation for i=1,…,k. Because each Bi is ground these refutations can be combined into a refutation of P{(B1 …Bk)}. Thus P{A} has an unrestricted refutation and we can apply the mgu lemma to obtain a refutation of P{A}.

Completeness - 1 Theorem Let P be a definite program and G be a definite goal. Suppose P{G} is unsatisfiable. Then there exists an SLD-refutation of P{G}. Proof Let G= A1,…,Ak. G is false wrt MP. Hence some ground instance Gist false wrt MP. Thus {A1,…,Ak} MP. Therefore there exists a refutation for each.Ai. Because Ai are ground, they can be combined to an SLD-refutation of G. By the lifting lemma, since there is an unrestricted SLD- refutation, we can also find a restricted SLD-refutation. Theorem 8.4

Completeness – 2: Every correct answer an instance of a computed answer Lemma Let P be a definite program and A an atom. Suppose (A) is a logical consequence of P, then there exists an SLD-refutation of P{A} with the identity substitution as the computed answer. Proof introduce new constants for all variables xi in A(={… xi/ai …}). Because of the premise P{A} has a refutation. Because A is ground the corresponding calculated answer is . Since the ai do not appear in P or A, we can keep the xi and obtain the calculated answer .

Completeness – 2: Every correct answer an instance of a computed answer Theorem Let P be a definite Program and G a definite goal. For every correct answer  for P{G} there exists a computed answer  for P{G} and a substitution  such that  and  have the same effect on all variables in G. Proof sketch Let G=A1,…,Ak. Since  is correct,((A1… Ak)) is a logical consequence. Hence there exists a SLD-refutation for P{ Ai} with  as calculated answer. Thus there exists a SLD-refutation for P{(A1… Ak)} with  as calculated answer. Let the composed mgu's be 1…n. By the lifting lemma we have: 1…n=‘1…‘n with mgu's ‘1…‘n for the refutation of P{G}. For this reason we can specify 