1. Artificial Intelligence 8. The Resolution Method
Course V231
Department of Computing
Imperial College, London
Jeremy Gow

2. Soundness & Completeness
Want to prove theorem T from Axioms Ax
Chosen a set of inference rules R
A B means B is entailed by A
A B means B is derived from A using R
R should be sound: if A B then A B
Want R to be complete: if A B then A B

3. Choose Your Search Space
Soundness and completeness of R isn’t enough
We want a reasonable search space
R determines operators, so influences the space
Can I see where I’m going? (Heuristic measure)
Can I restrict options at each state? (Branching)
Three main approaches
Forwards chaining: no heuristic guidance
Backwards chaining: large branching (many things entail KB)
Proof by refutation: clear goal (false), forwards inference

4. Proof by Refutation Proof by contradiction, reductio ad absurdum
Negate the theorem & add to axioms (¬T,Ax)
Use rules of inference to derive the False
So sentences (¬T,Ax) can’t all be true (unsatisfiable)
But the axioms Ax are true
Hence the negated theorem ¬T must be false
Hence the theorem T must be true

5. The Resolution Method Proof by refutation with a single inference rule
No need to worry about choice of rule
Just how we apply the one rule
Resolution is complete for FOL
Refutation-complete [Robinson 1965]
If (¬T,Ax) unsatisfiable it will derive a contradiction
So if will prove any true theorem of FOL
Even so, it might take a long time (> universe)
Even fairly trivial theorems can take a long time
Can use search heuristics to speed it up (next lecture)
No guarantees if it’s not a theorem

6. Binary Resolution (Propositional)
Unit resolution rule (last lecture)
AB, ¬B A
Binary resolution rule
AB, ¬BC
AC
The literals B and ¬B are resolved

7. Binary Resolution (First-Order)
Binary resolution rule
AB, ¬CD
Subst(, AD)
if substitution  s.t. Subst(,B) = Subst(,C)
The literals B and ¬C are resolved
B and C have been made the same by 

8. Resolution in FOL (This Lecture)
What if KB contains non-disjuncts?
Preprocessing step: rewrite to CNF
How do we find substitution ?
The unification algorithm
But what if more than two disjuncts?
Extend binary resolution to full resolution

9. Conjunctive Normal Form
A conjunction of clauses
Each clause is a disjunction of literals
Prop. literal: proposition or negated proposition
FO literal: predicate or negated predicate
No quantifiers (all variables implicitly universal)
Example FO clause
likes(george, X)  ¬likes(tony, houseof(george))  ¬is_mad(maggie)
Any FO sentence can be rewritten as CNF

10. Converting FOL to CNF (see notes)
Eliminate implication/equivalence (rewrite)
Move ¬ inwards (rewrite)
Rename variables apart (substitution)
Move quantifiers outwards (rewrite)
Skolemise existential variables (substitution)
Distribute  over  (rewrite)
Flatten binary connectives (rewrite)
(Optional: Reintroduce implications)

11. Example CNF Conversion
Propositional example (B  (A  C))  (B  ¬A)
1. Remove implication:
¬(B  (A  C))  (B  ¬A)
2. Move ¬ inwards (De Morgan’s x 2):
(¬B  ¬(A  C))  (B  ¬A)
(¬B  (¬A  ¬C))  (B  ¬A)
(Skip 3 to 5 as no variables.)

12. Example CNF Conversion (Contd)
(¬B  (¬A  ¬C))  (B  ¬A)
6. Distribute  over :
(¬B  (B  ¬A))  ((¬A  ¬C)  (B  ¬A))
7. Flatten connectives
(¬B  B  ¬A)  (¬A  ¬C  B  ¬A)
Drop 1st clause (¬B  B), remove duplicate from 2nd:
¬A  ¬C  B

13. ¬A  ¬C  B becomes (A  C)  B
Kowalski Normal Form
Can reintroduce  to CNF, e.g.
¬A  ¬C  B becomes (A  C)  B
Kowalski form
(A1 … An)  (B1 … Bn)
Binary resolution…
AB, BC
AC
Resembles Horn clauses (basis for Prolog)

14. Skolemisation Replace V with a ‘something’ term
If no preceeding U use fresh Skolem constant
Otherwise fresh Skolem function
parameterised by all preceeding U
X Y (person(X)  has(X, Y)  heart(Y)) to
person(X)  has(X, f(X))  heart(f(X))
(The particular heart f(X) depends on person X)

15. Substitutions & Inference Rules
Propositional inference rules used in first-order logic
But in FOL we can make substitutions
Sometimes a substitution can allow a rule to be applied
cf. FO binary resolution
knows(john, X)  hates(john, X)
knows(john, mary)
Substitution + Modus Ponens: infer hates(john, mary)
Need to find substitution that makes literals equal
Known as a unifier

16. Unifying Predicates We unified these two predicates:
knows(john, X) and knows(john, mary)
By saying that X should be substituted by mary
Why? Because john = john and can {X\mary}
For knows(jack, mary) and knows(john, X)
john doesn’t match jack
So we cannot unify the two predicates
Hence we cannot use the rule of inference

17. Unification Want an algorithm which: Example:
Takes two FOL sentences as input
Outputs a substitution {X/mary, Y/Z, etc.}
Which assigns terms to variables in the sentences
So that the first sentence looks exactly like the second
Or fails if there is no way to unify the sentences
Example:
Unify(“knows(john, X)”, “knows(john,mary)”) = {X/mary}
Unify(“knows(john, X)”, “knows(jack,mary)”) = Fail

18. Functions Within the Unify Algorithm
isa_variable(x)
checks whether x is a variable
isa_list(x)
checks whether x is a list
head(x)
outputs the head of a list (first term)
e.g., head([a,b,c]) = a
tail(x)
outputs the elements other than the head in a list
e.g., tail([a,b,c]) = [b,c]

19. More Internal Functions (Compound Expressions)
isa_compound(x)
checks whether x is compound expression
(either a predicate, a function or a connective)
args(x)
finds the subparts of the compound expression x
arguments of a predicate, function or connective
Returns the list of arguments
op(x)
predicate name/function name/connective symbol

20. Two Parts of the Unification Algorithm
Algorithm is recursive (it calls itself)
Passes around a set of substitutions, called mu
Making sure that new substitutions are consistent with old ones
unify(x,y) = unify_internal(x,y,{})
x and y are either a variable, constant, list, or compound
unify_internal(x,y,mu)
x and y are sentences, mu is a set of substitutions
finds substitutions making x look exactly like y
unify_variable(var,x,mu)
var is a variable
finds a single substitution (which may be in mu already)

21. unify_internal unify_internal(x,y,mu) ---------------------- Cases
1.if (mu=failure) then return failure
2.if (x=y) then return mu.
3.if (isa_variable(x)) then return unify_variable(x,y,mu)
4.if (isa_variable(y)) then return unify_variable(y,x,mu)
5.if (isa_compound(x) and isa_compound(y)) then return
unify_internal(args(x),args(y),unify_internal(op(x),op(y),mu))
6.if (isa_list(x) and isa_list(y)) then return
unify_internal(tail(x),tail(y),unify_internal(head(x),head(y),mu))
7.return failure

22. unify_variable unify_variable(var,x,mu) ------------------------ Cases
1. if (a substitution var/val is in mu) then return unify_internal(val,x,mu)
2. if (a substitution x/val is in mu) then return unify_internal(var,val,mu)
3. if (var occurs anywhere in x) return failure
4. add var/x to mu and return

23. Notes on the Unification Algorithm
unify_internal will not match a constant to a constant, unless they are equal (case 2)
Case 5 in unify_internal checks that two compound operators are the same (e.g. same predicate name)
Case 6 in unify_internal causes the algorithm to recursivly unify the whole list
Cases 1 and 2 in unify_variable check that neither inputs have already been substituted

24. The Occurs Check Suppose we needed to substitute X with f(X,Y)
This would give us f(X,Y) instead of X
But there is still an X in there,
so the substitution isn’t complete:
we need f(f(X,Y),Y), then f(f(f(X,Y),Y),Y) and so on
We need to avoid this situation
Otherwise the algorithm won’t stop
Case 3 in unify_variable checks this
Known as the “occurs check”
Occurs check slows the algorithm down
Order (n2), where n is size of expressions being unified

25. An Example Unification
Suppose we want to unify these sentences:
1. p(X,tony)  q(george,X,Z)
2. p(f(tony),tony)  q(B,C,maggie)
By inspection, this is a good substitution:
{X/f(tony), B/george, C/f(tony), Z/maggie}
This makes both sentences become:
p(f(tony),tony)  q(george, f(tony), maggie)
Note that we want to substitute X for C
But we have substituted f(tony) for X already
See the notes for this as a worked example
Using the unification algorithm
Requires five iterations!

26. Binary Resolution (First-Order)
Binary resolution rule (using unification)
AB, ¬CD
Subst(, AD)
if Unify(B, C) = 
Unification algorithm finds Most General Unifier (MGU) 
Don’t substitute any more than need to

27. The Full Resolution Rule
If Unify(Pj, ¬Qk) =  (¬ makes them unifiable)
P1  …  Pm, Q1  …  Qn
Subst(, P1  … (no Pj) …  Pm  Q1  … (no Qk) ... Qn)
Pj and Qk are resolved
Arbitrary number of disjuncts
Relies on preprocessing into CNF

28. Using Full Resolution Sentences already in CNF Pick two clauses
Pick positive literal P from first
Pick negative literal N from second
Find MGU  of ¬P and N
Write both clauses as one big disjunction
With P and N missing
Apply  to the new clause

29. Resolution Proof Search
Keep on resolving clause pairs
Eventually result in zero-length clause
This indicates that some literal K was true at the same time as the literal ¬K
Only way to reduce sentence to be empty
Hence there was an inconsistency
Which proves the theorem
Topic of the next lecture

30. Coursework: War of Life
Two player version of Game of Life
Implement several strategies in Prolog
Run a tournament
On CATE this afternoon
Submit via CATE (more to follow…)
Deadline: 3 weeks today