Transcript Folie 1 - Universität Koblenz · Landau

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Completeness of the
SLD-Resolution
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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
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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.
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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 .
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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.
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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.
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Success Set and Least Herbrand Model
Theorem
The success set of a definite program P is equal
to its least Herbrand model.
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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.
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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}.
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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 SLDrefutation, we can also find a restricted SLD-refutation.
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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 .
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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}.
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this reason we can specify 
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