CS156: The Calculus of E : { = , a , b , c , . . . , f , g , h , . . - PowerPoint PPT Presentation

The Theory of Equality T E CS156: The Calculus of Σ E : { = , a , b , c , . . . , f , g , h , . . . , p , q , r , . . . } Computation uninterpreted symbols: • constants a , b , c , . . . Zohar Manna • functions f , g , h , . . . Autumn 2008 • predicates p , q , r , . . . Example: x = y ∧ f ( x ) � = f ( y ) T E -unsatisfiable f ( x ) = f ( y ) ∧ x � = y T E -satisfiable f ( f ( f ( a ))) = a ∧ f ( f ( f ( f ( f ( a ))))) = a ∧ f ( a ) � = a T E -unsatisfiable Chapter 9: Quantifier-free Equality and Data x = g ( y , z ) → f ( x ) = f ( g ( y , z )) T E -valid Structures Page 1 of 48 Page 2 of 48 Axioms of T E Axiom schema 1. ∀ x . x = x (reflexivity) 5. for each positive integer n and n -ary predicate symbol p , 2. ∀ x , y . x = y → y = x (symmetry) � n � 3. ∀ x , y , z . x = y ∧ y = z → x = z (transitivity) � ∀ ¯ x , ¯ y . x i = y i → ( p (¯ x ) ↔ p (¯ y )) define = to be an equivalence relation. i =1 Axiom schema (predicate) 4. for each positive integer n and n -ary function symbol f , Thus, for unary p , the axiom is � n � � ∀ ¯ x , ¯ y . x i = y i → f (¯ x ) = f (¯ y ) ∀ x ′ , y ′ . x ′ = y ′ → ( p ( x ′ ) ↔ p ( y ′ )) i =1 (function) Therefore, a = b → ( p ( a ) ↔ p ( b )) For example, for unary f , the axiom is is T E -valid. ( x ′ → a , y ′ → b ). ∀ x ′ , y ′ . x ′ = y ′ → f ( x ′ ) = f ( y ′ ) Therefore, x = g ( y , z ) → f ( x ) = f ( g ( y , z )) is T E -valid. ( x ′ → x , y ′ → g ( y , z )). Page 3 of 48 Page 4 of 48

We discuss T E -formulae without predicates Equivalence and Congruence Relations: Basics For example, for Σ E -formula Binary relation R over set S • is an equivalence relation if F : p ( x ) ∧ q ( x , y ) ∧ q ( y , z ) → ¬ q ( x , z ) ◮ reflexive: ∀ s ∈ S . s R s ; introduce fresh constant • and fresh functions f p and f q , and ◮ symmetric: ∀ s 1 , s 2 ∈ S . s 1 R s 2 → s 2 R s 1 ; transform F to ◮ transitive: ∀ s 1 , s 2 , s 3 ∈ S . s 1 R s 2 ∧ s 2 R s 3 → s 1 R s 3 . G : f p ( x ) = • ∧ f q ( x , y ) = • ∧ f q ( y , z ) = • → f q ( x , z ) � = • . Example: Define the binary relation ≡ 2 over the set Z of integers m ≡ 2 n iff ( m mod 2) = ( n mod 2) That is, m , n ∈ Z are related iff they are both even or both odd. ≡ 2 is an equivalence relation • is a congruence relation if in addition n � ∀ s , t . s i R t i → f ( s ) R f ( t ) . i =1 Page 5 of 48 Page 6 of 48 Classes Quotient � equivalence � equivalence � � For relation R over set S , The quotient S / R of S by relation R is the congruence congruence � equivalence � equivalence � � the class of s ∈ S under R is partition of S into classes congruence congruence = { s ′ ∈ S : sRs ′ } . def [ s ] R S / R = { [ s ] R : s ∈ S } . Example: It satisfies total and disjoint conditions. The equivalence class of 3 under ≡ 2 over Z is Example: The quotient Z / ≡ 2 is a partition of Z . The set of [3] ≡ 2 = { n ∈ Z : n is odd } . equivalence classes Partitions {{ n ∈ Z : n is odd } , { n ∈ Z : n is even }} A partition P of S is a set of subsets of S that is � � � Note duality between relations and classes ◮ total S ′ = S S ′ ∈ P ◮ disjoint ∀ S 1 , S 2 ∈ P . S 1 � = S 2 → S 1 ∩ S 2 = ∅ Page 7 of 48 Page 8 of 48

Refinements Closures Two binary relations R 1 and R 2 over set S . Given binary relation R over S . The equivalence closure R E of R is the equivalence relation s.t. R 1 is a refinement of R 2 , R 1 ≺ R 2 , if ◮ R refines R E , i.e. R ≺ R E ; ∀ s 1 , s 2 ∈ S . s 1 R 1 s 2 → s 1 R 2 s 2 . ◮ for all other equivalence relations R ′ s.t. R ≺ R ′ , either R ′ = R E or R E ≺ R ′ R 1 refines R 2 . That is, R E is the “smallest” equivalence relation that “covers” R . Examples: Example: If S = { a , b , c , d } and R = { aRb , bRc , dRd } , then ◮ For S = { a , b } , • aR E b , bR E c , dR E d since R ⊆ R E ; R 1 : { aR 1 b } R 2 : { aR 2 b , bR 2 b } • aR E a , bR E b , cR E c by reflexivity; Then R 1 ≺ R 2 • bR E a , cR E b by symmetry; ◮ For set Z • aR E c by transitivity; R 1 : { xR 1 y : x mod 2 = y mod 2 } • cR E a by symmetry. R 2 : { xR 2 y : x mod 4 = y mod 4 } Then R 2 ≺ R 1 . Similarly, the congruence closure R C of R is the “smallest” congruence relation that “covers” R . Page 9 of 48 Page 10 of 48 T E -satisfiability and Congruence Classes I T E -satisfiability and Congruence Classes II Given Σ E -formula F Definition: For Σ E -formula F : s 1 = t 1 ∧ · · · ∧ s m = t m ∧ s m +1 � = t m +1 ∧ · · · ∧ s n � = t n F : s 1 = t 1 ∧ · · · ∧ s m = t m ∧ s m +1 � = t m +1 ∧ · · · ∧ s n � = t n with subterm set S F , F is T E -satisfiable iff there exists a the subterm set S F of F is the set that contains precisely congruence relation ∼ over S F such that the subterms of F . ◮ for each i ∈ { 1 , . . . , m } , s i ∼ t i ; Example: The subterm set of ◮ for each i ∈ { m + 1 , . . . , n } , s i �∼ t i . F : f ( a , b ) = a ∧ f ( f ( a , b ) , b ) � = a Such congruence relation ∼ defines T E -interpretation I : ( D I , α I ) of F . D I consists of | S F / ∼ | elements, one for each congruence is class of S F under ∼ . S F = { a , b , f ( a , b ) , f ( f ( a , b ) , b ) } . Note: we consider only quantifier-free conjunctive Σ E -formulae. Instead of writing I | = F for this T E -interpretation, we abbreviate Convert non-conjunctive formula F to DNF � i F i , where each ∼ | = F disjunct F i is a conjunction of =, � =. Check each disjunct F i . F is The goal of the algorithm is to construct the congruence relation T E -satisfiable iff at least one disjunct F i is T E -satisfiable. over S F , or to prove that no congruence relation exists. Page 11 of 48 Page 12 of 48

Congruence Closure Algorithm Congruence Closure Algorithm (Details) F : s 1 = t 1 ∧ · · · ∧ s m = t m ∧ s m +1 � = t m +1 ∧ · · · ∧ s n � = t n Initially, begin with the finest congruence relation ∼ 0 given by the � �� � � �� � partition generate congruence closure search for contradiction {{ s } : s ∈ S F } . Decide if F is T E -satisfiable. That is, let each term over S F be its own congruence class. Then, for each i ∈ { 1 , . . . , m } , impose s i = t i by merging the The algorithm performs the following steps: congruence classes 1. Construct the congruence closure ∼ of [ s i ] ∼ i − 1 and [ t i ] ∼ i − 1 { s 1 = t 1 , . . . , s m = t m } to form a new congruence relation ∼ i . over the subterm set S F . Then To accomplish this merging, ∼ | = s 1 = t 1 ∧ · · · ∧ s m = t m . ◮ form the union of [ s i ] ∼ i − 1 and [ t i ] ∼ i − 1 ◮ propagate any new congruences that arise within this union. 2. If for any i ∈ { m + 1 , . . . , n } , s i ∼ t i , return unsatisfiable. The new relation ∼ i is a congruence relation in which s i ∼ t i . 3. Otherwise, ∼| = F , so return satisfiable. How do we actually construct the congruence closure in Step 1? Page 13 of 48 Page 14 of 48 Congruence Closure Algorithm: Example 1 I Congruence Closure Algorithm: Example 1 II This partition represents the congruence closure of S F . Given Σ E -formula Is it the case that F : f ( a , b ) = a ∧ f ( f ( a , b ) , b ) � = a {{ a , f ( a , b ) , f ( f ( a , b ) , b ) } , { b }} | = F ? Construct initial partition by letting each member of the subterm No, as f ( f ( a , b ) , b ) ∼ a but F asserts that f ( f ( a , b ) , b ) � = a . set S F be its own class: Hence, F is T E -unsatisfiable. 1. {{ a } , { b } , { f ( a , b ) } , { f ( f ( a , b ) , b ) }} According to the first literal f ( a , b ) = a , merge { f ( a , b ) } { a } and to form partition 2. {{ a , f ( a , b ) } , { b } , { f ( f ( a , b ) , b ) }} According to the (function) congruence axiom, f ( a , b ) ∼ a , b ∼ b f ( f ( a , b ) , b ) ∼ f ( a , b ) , implies resulting in the new partition 3. {{ a , f ( a , b ) , f ( f ( a , b ) , b ) } , { b }} Page 15 of 48 Page 16 of 48