The Calculus of Computation: Decision Procedures with Applications to Verification. Part I: FOUNDATIONS. by Aaron Bradley Zohar Manna

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1 The Calculus of Computation: Decision Procedures with Applications to Verification Part I: FOUNDATIONS by Aaron Bradley Zohar Manna 1. Propositional Logic(PL) Springer Propositional Logic(PL) PL Syntax Atom truth symbols ( true ) and ( false ) propositional variables P,Q,R,P 1,Q 1,R 1, Literal atom α or its negation α Formula literal or application of a logical connective to formulae F,F 1,F 2 F not (negation) F 1 F 2 and (conjunction) F 1 F 2 or (disjunction) F 1 F 2 implies (implication) F 1 F 2 if and only if (iff) formula F : (P Q) ( Q) atoms: P,Q, literal: Q subformulas: P Q, Q abbreviation F : P Q Q

2 PL Semantics (meaning) Sentence F + Interpretation I = Interpretation Evaluation of F under I: F F Truth value (true, false) I : {P true,q false, } where 0 corresponds to value false 1 true F : P Q P Q I : {P true,q false} P Q Q P Q P Q F F evaluates to true under I 1 = true 0 = false F 1 F 2 F 1 F 2 F 1 F 2 F 1 F 2 F 1 F Inductive Definition of PL s Semantics if F evaluates to true under I false Base Case: I = I = I = P iff I[P] = true I = P iff I[P] = false Inductive Case: I = F iff 1 F 2 iff 1 and 2 1 F 2 iff 1 or 2 1 F 2 iff, if 1 then 2 1 F 2 iff, 1 and 2, or 1 and 2 Note: 1 F 2 iff 1 and F : P Q P Q I : {P true, Q false} 1. I = P since I[P] = true 2. I = Q since I[Q] = false 3. I = Q by 2 and 4. I = P Q by 2 and 5. I = P Q by 1 and 6. by 4 and Why? Thus, F is true under I. 1-8

3 Satisfiability and Validity F satisfiable iff there exists an interpretation I such that. F valid iff for all interpretations I,. Method 1: Truth Tables F is valid iff F is unsatisfiable Example F : P Q P Q P Q P Q Q P Q F Example F : P Q P Q P Q P Q P Q F satisfying I falsifying I Thus F is satisfiable, but invalid. Thus F is valid Method 2: Semantic Argument Proof rules I = F G I = G and G I = G G I = G G G G I = F G I = G տ or G I = G G I = G G G I = F G Example 1: Prove F : P Q P Q is valid. Let s assume that F is not valid and that I is a falsifying interpretation. 1. I = P Q P Q assumption 2. I = P Q 1 and 3. I = P Q 1 and 4. I = P 2 and 5. I = P 3 and 6. I = 4 and 5 are contradictory Thus F is valid. I =

4 Example 2: Prove F : (P Q) (Q R) (P R) Let s assume that F is not valid. 1. assumption 2. I = (P Q) (Q R) 1 and 3. I = P R 1 and 4. I = P 3 and 5. I = R 3 and 6. I = P Q 2 and of 7. I = Q R 2 and of is valid. Two cases from 6 and Two cases from 7 and 8a. I = P 6 and 9a. I = 4 and 8a are contradictory 8b. I = Q 6 and 9ba. I = Q 7 and 10ba. I = 8b and 9ba are contradictory 9bb. I = R 7 and 10bb. I = 5 and 9bb are contradictory Our assumption is incorrect in all cases F is valid Example 3: Is F : P Q P Q Let s assume that F is not valid. valid? Equivalence F 1 and F 2 are equivalent (F 1 F 2 ) iff for all interpretations I, 1 F 2 Two options 1. I = P Q P Q assumption 2. I = P Q 1 and 3. I = P Q 1 and 4a. I = P 2 and 5a. I = Q 3 and We cannot derive a contradiction. F is not valid. 4b. I = Q 2 and 5b. I = P 3 and To prove F 1 F 2 show F 1 F 2 is valid. F 1 implies F 2 (F 1 F 2 ) iff for all interpretations I, 1 F 2 F 1 F 2 and F 1 F 2 are not formulae! Falsifying interpretation: I 1 : {P true, Q false} I 2 : {Q true, P false} We have to derive a contradiction in both cases for F to be valid

5 Normal Forms 1. Negation Normal Form (NNF) Negations appear only in literals. (only,, ) To transform F to equivalent F in NNF use recursively the following template equivalences (left-to-right): F 1 F 1 } (F 1 F 2 ) F 1 F 2 De Morgan s Law (F 1 F 2 ) F 1 F 2 F 1 F 2 F 1 F 2 F 1 F 2 (F 1 F 2 ) (F 2 F 1 ) Convert F : (P (P Q)) to NNF F : ( P (P Q)) to F : P (P Q) De Morgan s Law F : P P Q F is equivalent to F (F F) and is in NNF Disjunctive Normal Form (DNF) Disjunction of conjunctions of literals l i,j for literals l i,j i j To convert F into equivalent F in DNF, transform F into NNF and then use the following template equivalences (left-to-right): } (F 1 F 2 ) F 3 (F 1 F 3 ) (F 2 F 3 ) dist F 1 (F 2 F 3 ) (F 1 F 2 ) (F 1 F 3 ) Convert F : (Q 1 Q 2 ) ( R 1 R 2 ) into DNF F : (Q 1 Q 2 ) (R 1 R 2 ) in NNF F : (Q 1 (R 1 R 2 )) (Q 2 (R 1 R 2 )) dist F : (Q 1 R 1 ) (Q 1 R 2 ) (Q 2 R 1 ) (Q 2 R 2 ) dist F is equivalent to F (F F) and is in DNF Conjunctive Normal Form (CNF) Conjunction of disjunctions of literals l i,j for literals l i,j i j To convert F into equivalent F in CNF, transform F into NNF and then use the following template equivalences (left-to-right): (F 1 F 2 ) F 3 (F 1 F 3 ) (F 2 F 3 ) F 1 (F 2 F 3 ) (F 1 F 2 ) (F 1 F 3 ) Davis-Putnam-Logemann-Loveland (DPLL) Algorithm Decides the satisfiability of PL formulae in CNF In book, efficient conversion of F to F where F is in CNF and F and F are equisatisfiable (F is satisfiable iff F is satisfiable) Decision Procedure DPLL: Given F in CNF let rec dpll F = let F = bcp F in if F = then true else if F = then false else let P = choose vars(f ) in (dpll F {P }) (dpll F {P }) Don t choose only-positive or only-negative variables for splitting

6 Boolean Constraint Propagation (BCP) Based on unit resolution l throughout C[ l] clause C[ ] where l = P or l = P F : ( P Q R) ( Q R) ( Q R) (P Q R) Branching on Q By unit resolution F {Q } : (R) ( R) (P R) R ( R) On the other branch F {Q } : ( P R) F {Q, R, P } = true F is satisfiable with satisfying interpretation I : {P false, Q false, R true} F (R) ( R) (P R) ( P R) R Q ( R) Q P R P I : {P false, Q false, R true} F {Q } = false