Discussion #18
Resolution with
Propositional Calculus;
Prenex Normal Form
Discussion #18
1/13
Topics
•
•
•
•
•
Motivation for resolution
Resolution
Why resolution works
Examples
Prenex normal form
Discussion #18
2/13
Programming a Computer to do Proofs
Too much work to program all the
possibilities we have considered. We
need a better way.
1. Better = more uniform  not so many cases (even
though it may sometimes be longer).
2. Better = fewer rules of inference.
3. Better = a heuristic guide to lead us to the conclusion.
4. Better = easier to convert to an algorithm.
Discussion #18
3/13
Resolution
Resolution answers these “demands:”
Uniform: Only disjunctions of literals in every rule
Fewer Rules: Only one inference rule
Heuristic Guide: Reduce the number of literals with the goal
of reaching False
Algorithmic:
1.
2.
3.
4.
5.
Negate the conclusion and add it as a premise.
Convert the premises to CNF (conjunction of disjunction of
literals).
Write each premise (which is a disjunction of literals) as a line
of the proof.
Repeatedly apply resolution (the one inference rule) & simplify
as needed.
Success iff F is reached.
Discussion #18
4/13
Resolution Rule
literals
PA
P  B
AB
clauses: A and B will always be
disjunctions of literals or just a
literal or possibly missing.
This says: In two disjunctive clauses, if we
have complementary literals, we can discard
them and “OR” the remaining clauses.
Discussion #18
5/13
Resolution is Valid
P
T
T
T
T
F
F
F
F
A
T
T
F
F
T
T
F
F
Discussion #18
B
T
F
T
F
T
F
T
F
(P  A)
T
T
T
T
T
T
F
F

T
F
T
F
T
T
F
F
(P
F
F
F
F
T
T
T
T
 B)
T
F
T
F
T
T
T
T
 AB
T
T
T
T
T
T
T
F
T
T
T
T
T
T
T
F
6/13
Resolution Subsumes “the big 3”
Inference Rules
Modus ponens
Modus tollens
Hypothetical syllogism
P
PQ
Q
P
QP
Q
PQ
QR
PR
PF
P  Q
QF
P  F
Q  P
Q  F
P  Q
Q  R
P  R
We now have one rule to rule them all!
Discussion #18
7/13
Example #1
If P  Q, Q  R, P then R.
1. Negate the conclusion (R becomes another premise).
2. Convert to CNF: (P  Q)  (Q  R)  P  R
3. Write the premises as the first lines of the proof.
4. Do resolution.
1.
2.
3.
4.
5.
6.
7.
P  Q
Q  R
P
R
P  R
R
premise
premise
premise
premise
resolution 1,2
resolution 3,5
resolution 4,6
}
Sometimes called
the support.
empty = F
Discussion #18
8/13
Example #2
If P  (Q  R), R  Q then P.
1. Negate conclusion: P  P
2. Convert to CNF: (P  Q)  (P  R)  R  Q  P
1.
2.
3.
4.
5.
6.
7.
P  Q
P  R
R
Q
P
R
F
premise (not used  could discard)
premise
premise
premise (not used  could discard)
premise
resolution 2,5
resolution 3,6
Also, resolution 1,5 yields Q, which need not be added to the
derivation  already there.
Do we always need to use all the premises? If not, we can
discard them from the statement to be proved.
Discussion #18
9/13
Example #3
If P  Q, Q  P, P  Q then P  Q.
1. Negate conclusion: (P  Q)  (P  Q)
2. CNF: (P  Q)  (Q  P)  (P  Q)  (P  Q)
1.
2.
3.
4.
5.
6.
7.
8.
PQ
Q  P
P  Q
P  Q
P
Q
Q
F
Discussion #18
premise
premise
premise
premise
resolution 1,2 (idemp. P  P  P)
resolution 3,5
resolution 4,5
resolution 6,7
10/13
Example #4
If (P  Q)  (P  R), P then Q  R.
1. Negate conclusion: (Q  R)  (Q  R)
2. CNF: ((P  Q)  (P  R))  P  (Q  R)
 (P  Q  R)  P  Q  R
1.
2.
3.
4.
5.
6.
7.
Discussion #18
P  Q  R
P
Q
R
QR
R
premise
premise
premise
premise
resolution 1,2
resolution 3,5
resolution 4,6
11/13
Prenex Normal Form
• Prenex Normal Form  preparation to do
resolution in predicate calculus
– All quantifiers in front
– More formally: No quantifier in the scope of any
logical connector (, , , , )
• Algorithm to obtain prenex normal form:
1.
2.
3.
4.
Remove  and 
Move  in
Rectify (standardize all variables apart)
Move quantifiers to the front
Discussion #18
12/13
Prenex Normal Form – Example
y(xP(x)  xQ(x, y))
 y(xP(x)  xQ(x, y))
 y(xP(x)  xQ(x, y))
 y(xP(x)  xQ(x, y))
 y(xP(x)  xQ(x, y))
 y(xP(x)  zQ(z, y))
 yx(P(x)  zQ(z, y))
 yxz(P(x)  Q(z, y))
Discussion #18
implication
xA  xA (de Morgan’s)
de Morgan’s, double neg.
xA  xA (de Morgan’s)
rectification
xAB  x(AB) (x not free in B)
AzB  z(AB) (z not free in A)
13/13