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Applied Computer Science II
Resolution in FOL
Luc De Raedt
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Clausal Form
Clauses are universally quantified
disjunctions of literals; all variables in a
clause are universally quantified
(x1 ,..., xn )(l1  ...  ln )
written as
l1  ...  ln
or
{l1 ,..., ln }
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Towards Resolution
• Examples:
{Nat(s(A)),Nat(A)} {Nat(s(A)),Nat(A)}
{Nat(A)}
{Nat(x)}
gives
gives
{Nat(s(A))}
{Nat(s(A))}
{Nat(s(s(x))),Nat(s(x))}
{Nat(s(A))}
gives
{Nat(s(s(A)))}
• We need to be able to work with variables !
• Unification of two expressions/literals
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Terms and instances
• Consider following atoms
P(x,f(y),B)
P(z,f(w),B) alphabetic variant
P(x,f(A),B) instance
P(g(z),f(A),B) instance
P(C,f(A),A) not an instance
• Ground expressions do not contain any variables
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A substitution s  {v1 / t1 ,..., vn / tn } substitutes
Substitution
terms ti for variables vi ( ti does NOT contain vi )
Applying a substitution s to an expression 
yields the expression  s which is 
with all occurrences of vi replaced by ti
P(x,f(y),B)
P(z,f(w),B)
P(x,f(A),B)
s ={x/z,y/w}
s ={y/A}
P(g(z),f(A),B)
s ={x/g(z),y/A}
P(C,f(A),A)
no substitution !
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Composing substitutions
• Composing substitutions s1 and s2 gives s1 s2
which is that substitution obtained by first
applying s2 to the terms in s1and adding
remaining term/vars pairs to s1
{z/g(x,y)}{x/A,y/B,w/C,z/D}=
{z/g(A,B),x/A,y/B,w/C}
• Apply to
P(x,y,z)
gives
P(A,B,g(A,B))
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s
Properties of substitutions
( s1 ) s2   ( s1s2 )
( s1s2 ) s3  s1 (s2 s3 ) associativity
s1s2  s2 s1 not commutative
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Unification
• Unifying a set of expressions {wi}
– Find substitution s such that wi s  wj s for all i, j
– Example
{P(x,f(y),B),P(x,f(B),B)}
s ={y/B,x/A} not the simplest unifier
s ={y/B} most general unifier (mgu)
• The most general unifier, the mgu, g of {wi} has
the property that if s is any unifier of {wi} then
there exists a substitution s’ such that
{wi}s={wi}gs’
• The common instance produced is unique up to
alphabetic variants (variable renaming)
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Disagreement set
The disagreement set of a set of expressions
{wi} is the set of subterms { ti } of {wi} at
the first position in {wi} for which the {wi}
disagree
{P(x,A,f(y)),P(w,B,z)} gives {x,w}
{P(x,A,f(y)),P(x,B,z)} gives {A,B}
{P(x,y,f(y)),P(x,B,z)} gives {y,B}
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Unification algorithm
Unify(Terms)
Initialize k  0;
Initialize Tk  Terms;
Initialize  k  {};
* If Tk is a singleton, then output  k . Otherwise, continue.
Let Dk be the disagreement set of Tk
If there exists a var vk and a term tk in D k such that vk
does not occur in tk , continue. Otherwise, exit with failure.
 k 1   k {vk / tk };
Tk 1  Tk {vk / tk };
k  k  1;
Goto *
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Predicate calculus Resolution
John Allan Robinson (1965)
Let C1 and C2 be two clauses with
literals l1  C1 and l2  C2 such that
C1 and C2 do not contain common variables,
and mgu (l1 , l2 )  
then C  [{ C1  {l1}} {C2  {l2 }}]
is a resolvent of C1 and C2
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Example
P(x) Q(f(x)) and R(g(x)) Q(f(A))
Standardizing the variables apart
P(x) Q(f(x)) and R(g(y)) Q(f(A))
Substitution  ={x/A}
Resolvent P(A) R(g(y))
P(x) Q(x,y) and P(A) R(B,z)
Standardizing the variables apart
Substitution  ={x/A}
Resolvent Q(A,y) R(B,z)
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A stronger version of resolution
Use more than one literal per clause
{P(u),P(v)} and {P(x),P(y)}
do not resolve to empty clause.
However, ground instances
{P(A)} and {P(A)} resolve to empty clause
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Factors
Let C1 be a clause such that there exists
a substitution  that is a mgu of a set of literals
in C1. Then C1 is a factor of C1
Each clause is a factor of itself.
Also, {P(f(y)),R(f(y),y)}is a factor of {P(x),P(f(y)),R(x,y)}
with   {x / f ( y )}
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Resolution
Let C1 and C2 be two clauses and let
C1' and C2 ' be factors of C1 and C2 that do not common variables,
literals l1  C1' and l2  C2' such that
mgu (l1 , l2 )  
then C  [{ C '1  {l1}} {C2 ' {l2 }}]
is a resolvent of C1 and C2
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Resolution
• Properties
– Resolution is sound
• If there is a resolution derivation of a clause C from a database
KB of clauses, then KB |=c
– Incomplete
Given P(A)
Infer {P(A),P(B)}
– But fortunately it is refutation complete
• If KB is unsatisfiable then KB |- 
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Refutation
To decide whether a formula KB |= w do
• Convert KB to clausal form KB’
• Convert w to clausal form w’
• Combine w’ and KB’ to give 
• Iteratively apply resolution to  and add the
results back to  until either no more
resolvents can be added, or until the empty
clause is produced.
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Answer extraction
Suppose we wish to prove whether KB |=
(w)f(w)
We are probably interested in knowing the w
for which f(w) holds.
Add Ans(w) literal to each clause coming
from the negation of the theorem to be
proven; stop resolution process when there
is a clause containing only Ans literal
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{1  \  0}
{0  \  1}
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Resolution Search Strategies
• Ordering strategies
– In what order to perform resolution ?
– Breadth-first, depth-first, iterative deepening ?
– Unit-preference strategy :
• Prefer those resolution steps in which at least one clause is a
unit clause (containing a single literal)
• Refinement strategies
– Input : at least one of the clauses being resolved is a
member of the original set of clauses
• (complete for horn clauses, incomplete in general)
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• Refinement strategies (ctd.)
– Linear : at least one of the parents is a member of the
original set of clauses or is an ancestor of the other
parent
• Refutation complete
– Ancestor : c2 is a descendant of c1 iff c2 is a resolvent
of c1 (and another clause) or if c2 is a resolvent of a
descendant of c1 (and another clause); c1 is an
ancestor of c2
– Set of support : the set of clauses coming from the
negation of the theorem (to be proven) and their
descendants
– Set of support strategy : require that at least one of the
clauses in each resolution step belongs to the set of
support
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Herbrand base and
interpretations
Let KB be a set of clauses.
Let O( KB) denote the set of constants in KB, if there
are no constant symbols, then O( KB )  { A}
Let F ( KB)denote the set of function symbols.
Let P( KB )denote the set of predicate symbols.
Now, the Herbrand Universe H ( KB) is the set of
all legal ground terms that can be formed using O( KB)
and F ( KB).
A Herbrand interpretation for a set of clauses KB
is an interpretation that
1) has H ( KB)as its domain
2) assigns true or false to all ground atoms
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The Herbrand base for a set of clauses KB
is the set of ground clauses obtained by replacing
the variables in KB by all terms in the Herbrand Universe
Theorem (Herbrand's theorem)
If a finite set of clauses KB is unsatisfiable
then the Herbrand base for KB is unsatisfiable.
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Conclusions
• Logic
– A sound basis for reasoning, modeling and
computer science
– Syntax and Semantics
– Propositional and First Order
– Soundness and Completeness
– Resolution
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