Completeness of the SLD-Resolution

1. Completeness of the SLD-Resolution

2. 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

3. MGU LemmaLet 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.

4. MGU LemmaLet 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…‘nsuch that 1…n=‘12…n= ‘1…‘n .

5. Lifting Lemma LemmaLet 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.

6. Lifting Lemma ProofSuppose the first input clause for the refutation of P{G} is C1, the first mgu is 1and 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.

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

8. 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.

9. 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}.

10. Completeness - 1 TheoremLet 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}. ProofLet 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.

11. Completeness – 2: Every correct answer an instance of a computed answer LemmaLet 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. Proofintroduce 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 .

12. Completeness – 2: Every correct answer an instance of a computed answer TheoremLet 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 sketchLet 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 ofP{G}.For this reason we can specify 