Discussion #18 Resolution with Propositional Calculus; Prenex Normal Form

1. Discussion #18Resolution withPropositional Calculus;Prenex Normal Form

2. Topics • Motivation for resolution • Resolution • Why resolution works • Examples • Prenex normal form

3. Programming a Computer to do Proofs Too much work to program all the possibilities we have considered. We need a better way. • Better = more uniform  not so many cases (even though it may sometimes be longer). • Better = fewer rules of inference. • Better = a heuristic guide to lead us to the conclusion. • Better = easier to convert to an algorithm.

4. 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: • 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.

5. clauses: A and B will always be disjunctions of literals or just a literal or possibly missing. literals Resolution Rule This says: In two disjunctive clauses, if we have complementary literals, we can discard them and “OR” the remaining clauses. P  A P  B A  B

6. P A B (P  A)  (P  B)  A  B T T T T T F T T T T T F T F F F T T T F T T T F T T T T F F T F F F T F F T T T T T T T T F T F T T T T T T F F T F F T T T T F F F F F T T T F Resolution is Valid

7. Resolution Subsumes “the big 3” Inference Rules Modus ponens Modus tollens Hypothetical syllogism P  Q Q  R P  R P Q  P Q P P  Q Q 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!

8. Example #1 If P  Q, Q  R, P then R. • Negate the conclusion (R becomes another premise). • Convert to CNF: (P  Q)  (Q  R)  P  R • Write the premises as the first lines of the proof. • Do resolution. • P  Q premise • Q  R premise • P premise • R premise • P  R resolution 1,2 • R resolution 3,5 • resolution 4,6 } Sometimes called the support. empty = F

9. Example #2 If P  (Q  R), R  Q then P. • Negate conclusion: P  P • Convert to CNF: (P  Q)  (P  R)  R  Q  P • P  Q premise (not used  could discard) • P  R premise • R premise • Q premise (not used  could discard) • P premise • R resolution 2,5 • F 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.

10. Example #3 If P  Q, Q P, P Q then P  Q. • Negate conclusion: (P  Q)  (P  Q) • CNF: (P  Q)  (Q  P)  (P  Q)  (P  Q) • P  Q premise • Q  P premise • P  Q premise • P  Q premise • P resolution 1,2 (idemp. P  P  P) • Q resolution 3,5 • Q resolution 4,5 • F resolution 6,7

11. Example #4 If (P  Q)  (P R), P then Q  R. • Negate conclusion: (Q  R)  (Q  R) • CNF: ((P  Q)  (P  R))  P  (Q  R)  (P  Q  R)  P  Q  R • P  Q  R premise • P premise • Q premise • R premise • Q  R resolution 1,2 • R resolution 3,5 • resolution 4,6

12. 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. Remove  and  2. Move  in 3. Rectify (standardize all variables apart) 4. Move quantifiers to the front

13. Prenex Normal Form – Example y(xP(x)  xQ(x, y))  y(xP(x)  xQ(x, y)) implication  y(xP(x)  xQ(x, y)) xA  xA (de Morgan’s)  y(xP(x)  xQ(x, y)) de Morgan’s, double neg.  y(xP(x)  xQ(x, y)) xA  xA (de Morgan’s)  y(xP(x)  zQ(z, y)) rectification  yx(P(x)  zQ(z, y)) xAB  x(AB) (x not free in B)  yxz(P(x)  Q(z, y)) AzB  z(AB) (z not free in A)