The Calculus of Computation: Decision Procedures with 2. - PowerPoint PPT Presentation

The Calculus of Computation: Decision Procedures with 2. First-Order Logic (FOL) Applications to Verification by Aaron Bradley Zohar Manna Springer 2007 2- 1 2- 2 First-Order Logic (FOL) quantifiers existential quantifier ∃ x . F [ x ] Also called Predicate Logic or Predicate Calculus “there exists an x such that F [ x ]” universal quantifier ∀ x . F [ x ] FOL Syntax “for all x , F [ x ]” variables x , y , z , · · · constants a , b , c , · · · FOL formula literal, application of logical connectives functions f , g , h , · · · ( ¬ , ∨ , ∧ , → , ↔ ) to formulae, terms variables, constants or or application of a quantifier to a formula n-ary function applied to n terms as arguments a , x , f ( a ) , g ( x , b ) , f ( g ( x , g ( b ))) predicates p , q , r , · · · atom ⊤ , ⊥ , or an n-ary predicate applied to n terms literal atom or its negation p ( f ( x ) , g ( x , f ( x ))) , ¬ p ( f ( x ) , g ( x , f ( x ))) Note: 0-ary functions: constant 0-ary predicates: P , Q , R , . . . 2- 3 2- 4

Example: FOL formula Translations of English Sentences into FOL ◮ The length of one side of a triangle is less than the sum of the ∀ x . p ( f ( x ) , x ) → ( ∃ y . p ( f ( g ( x , y )) , g ( x , y )) ) ∧ q ( x , f ( x )) lengths of the other two sides � �� � G � �� � ∀ x , y , z . triangle ( x , y , z ) → length ( x ) < length ( y )+ length ( z ) F The scope of ∀ x is F . ◮ Fermat’s Last Theorem. The scope of ∃ y is G . ∀ n . integer ( n ) ∧ n > 2 The formula reads: → ∀ x , y , z . “for all x, integer ( x ) ∧ integer ( y ) ∧ integer ( z ) if p ( f ( x ) , x ) ∧ x > 0 ∧ y > 0 ∧ z > 0 then there exists a y such that → x n + y n � = z n p ( f ( g ( x , y )) , g ( x , y )) and q ( x , f ( x ))” 2- 5 2- 6 FOL Semantics Example: F : p ( f ( x , y ) , z ) → p ( y , g ( z , x )) An interpretation I : ( D I , α I ) consists of: Interpretation I : ( D I , α I ) ◮ Domain D I D I = Z = {· · · , − 2 , − 1 , 0 , 1 , 2 , · · · } integers non-empty set of values or objects α I : { f �→ + , g �→ − , p �→ > } cardinality | D I | finite (eg, 52 cards), Therefore, we can write countably infinite (eg, integers), or F I : x + y > z → y > z − x uncountably infinite (eg, reals) ◮ Assignment α I (This is the way we’ll write it in the future!) ◮ each variable x assigned value x I ∈ D I Also ◮ each n-ary function f assigned α I : { x �→ 13 , y �→ 42 , z �→ 1 } f I : D n I → D I Thus F I : 13 + 42 > 1 → 42 > 1 − 13 In particular, each constant a (0-ary function) assigned value a I ∈ D I Compute the truth value of F under I ◮ each n-ary predicate p assigned 1 . | = x + y > z since 13 + 42 > 1 I p I : D n I → { true , false } 2 . I | = y > z − x since 42 > 1 − 13 3 . | = by 1, 2, and → I F In particular, each propositional variable P (0-ary predicate) assigned truth value (true, false) F is true under I 2- 7 2- 8

Semantics: Quantifiers Example For Q , the set of rational numbers, consider x variable. x -variant of interpretation I is an interpretation J : ( D J , α J ) such F I : ∀ x . ∃ y . 2 × y = x that Compute the value of F I ( F under I ): ◮ D I = D J Let ◮ α I [ y ] = α J [ y ] for all symbols y , except possibly x J 2 : J 1 ⊳ { y �→ v J 1 : I ⊳ { x �→ v } 2 } That is, I and J agree on everything except possibly the value of x x -variant of I y -variant of J 1 for v ∈ Q . Denote J : I ⊳ { x �→ v } the x -variant of I in which α J [ x ] = v for some v ∈ D I . Then Then ◮ I | = ∀ x . F iff for all v ∈ D I , I ⊳ { x �→ v } | = F since 2 × v ◮ I | = ∃ x . F iff there exists v ∈ D I s.t. I ⊳ { x �→ v } | = F 1 . J 2 | = 2 × y = x 2 = v 2 . | = ∃ y . 2 × y = x J 1 3 . I | = ∀ x . ∃ y . 2 × y = x since v ∈ Q is arbitrary 2- 9 2- 10 Satisfiability and Validity Second case 1 . I �| = ∀ x . p ( x ) assumption F is satisfiable iff there exists I s.t. I | = F 2 . | = ¬∃ x . ¬ p ( x ) assumption I F is valid iff for all I , I | = F 3 . I ⊳ { x �→ v } �| = p ( x ) 1 and ∀ , for some v ∈ D I 4 . I �| = ∃ x . ¬ p ( x ) 2 and ¬ F is valid iff ¬ F is unsatisfiable 5 . I ⊳ { x �→ v } �| = ¬ p ( x ) 4 and ∃ 6 . I ⊳ { x �→ v } | = p ( x ) 5 and ¬ Example: F : ( ∀ x . p ( x )) ↔ ( ¬∃ x . ¬ p ( x )) valid? Suppose not. Then there is I s.t. 3 and 6 are contradictory. 0 . I �| = ( ∀ x . p ( x )) ↔ ( ¬∃ x . ¬ p ( x )) Both cases end in contradictions for arbitrary I ⇒ F is valid. First case 1 . I | = ∀ x . p ( x ) assumption 2 . I �| = ¬∃ x . ¬ p ( x ) assumption 3 . I | = ∃ x . ¬ p ( x ) 2 and ¬ 4 . I ⊳ { x �→ v } | = ¬ p ( x ) 3 and ∃ , for some v ∈ D I 5 . I ⊳ { x �→ v } | = p ( x ) 1 and ∀ 4 and 5 are contradictory. 2- 11 2- 12

Example: Prove Example: Show F : p ( a ) → ∃ x . p ( x ) is valid. F : ( ∀ x . p ( x , x )) → ( ∃ x . ∀ y . p ( x , y )) is invalid. Assume otherwise. Find interpretation I such that 1 . I �| = F assumption I | = ¬ [( ∀ x . p ( x , x )) → ( ∃ x . ∀ y . p ( x , y ))] 2 . | = p ( a ) 1 and → I 3 . I �| = ∃ x . p ( x ) 1 and → i.e. 4 . I ⊳ { x �→ α I [ a ] } �| = p ( x ) 3 and ∃ I | = ( ∀ x . p ( x , x )) ∧ ¬ ( ∃ x . ∀ y . p ( x , y )) Choose D I = { 0 , 1 } 2 and 4 are contradictory. Thus, F is valid. p I = { (0 , 0) , (1 , 1) } i.e. p I (0 , 0) and p I (1 , 1) are true p I (1 , 0) and p I (1 , 0) are false I falsifying interpretation ⇒ F is invalid. 2- 13 2- 14 Safe Substitution F σ Rename x by x ′ : replace x in ∀ x by x ′ and all free x in the scope of ∀ x by x ′ . Example: ∀ x ′ . G [ x ′ ] ∀ x . G [ x ] ⇔ scope of ∀ x � �� � F : ( ∀ x . p ( x , y ) ) → q ( f ( y ) , x ) Same for ∃ x տ free տ free bound by ∀ x ր free ր ∃ x ′ . G [ x ′ ] ∃ x . G [ x ] ⇔ where x ′ is a fresh variable free ( F ) = { x , y } substitution Proposition (Substitution of Equivalent Formulae) σ : { x �→ g ( x ) , y �→ f ( x ) , q ( f ( y ) , x ) �→ ∃ x . h ( x , y ) } σ : { F 1 �→ G 1 , · · · , F n �→ G n } F σ ? s.t. for each i , F i ⇔ G i 1. Rename F ′ : ∀ x ′ . p ( x ′ , y ) → q ( f ( y ) , x ) If F σ a safe substitution, then F ⇔ F σ ↑ ↑ where x ′ is a fresh variable 2. F ′ σ : ∀ x ′ . p ( x ′ , f ( x )) → ∃ x . h ( x , y ) 2- 15 2- 16

Formula Schema Substitution σ of H Formula σ : { F 1 �→ , . . . , F n �→ } ( ∀ x . p ( x )) ↔ ( ¬∃ x . ¬ p ( x )) mapping place holders F i of H to FOL formulae, (obeying the side conditions of H ) Formula Schema H 1 : ( ∀ x . F ) ↔ ( ¬∃ x . ¬ F ) Proposition (Formula Schema) ↑ place holder If H is valid formula schema and Formula Schema (with side condition) σ is a substitution obeying H ’s side conditions then H σ is also valid. H 2 : ( ∀ x . F ) ↔ F provided x / ∈ free ( F ) Example: Valid Formula Schema H : ( ∀ x . F ) ↔ F provided x / ∈ free ( F ) is valid H is valid iff valid for any FOL formula F i obeying the side conditions σ : { F �→ p ( y ) } obeys the side condition Therefore H σ : ∀ x . p ( y ) ↔ p ( y ) is valid Example: H 1 and H 2 are valid. 2- 17 2- 18 Proving Validity of Formula Schema Normal Forms 1. Negation Normal Forms (NNF) Example: Prove validity of Augment the equivalence with (left-to-right) H : ( ∀ x . F ) ↔ F provided x / ∈ free ( F ) Proof by contradiction. Consider the two directions of ↔ . ¬∀ x . F [ x ] ⇔ ∃ x . ¬ F [ x ] First case: ¬∃ x . F [ x ] ⇔ ∀ x . ¬ F [ x ] 1 . I | = ∀ x . F assumption Example 2 . �| = assumption I F 3 . I | = F 1, ∀ , since x �∈ free( F ) 4 . | = ⊥ 2, 3 I G : ∀ x . ( ∃ y . p ( x , y ) ∧ p ( x , z )) → ∃ w . p ( x , w ) . Second Case: 1. ∀ x . ( ∃ y . p ( x , y ) ∧ p ( x , z )) → ∃ w . p ( x , w ) 1 . I �| = ∀ x . F assumption 2. ∀ x . ¬ ( ∃ y . p ( x , y ) ∧ p ( x , z )) ∨ ∃ w . p ( x , w ) 2 . | = assumption I F F 1 → F 2 ⇔ ¬ F 1 ∨ F 2 3 . I | = ∃ x . ¬ F 1 and ¬ 3. ∀ x . ( ∀ y . ¬ ( p ( x , y ) ∧ p ( x , z ))) ∨ ∃ w . p ( x , w ) 4 . I | = ¬ F 3, ∃ , since x �∈ free( F ) ¬∃ x . F [ x ] ⇔ ∀ x . ¬ F [ x ] 5 . I | = ⊥ 2, 4 4. ∀ x . ( ∀ y . ¬ p ( x , y ) ∨ ¬ p ( x , z )) ∨ ∃ w . p ( x , w ) Hence, H is a valid formula schema. 2- 19 2- 20