CS156: The Calculus of quantifier-free formula (qff) G that is - PowerPoint PPT Presentation

Quantifier Elimination (QE) Algorithm for elimination of all quantifiers of formula F until CS156: The Calculus of quantifier-free formula (qff) G that is equivalent to F remains Note: Could be enough if F is equisatisfiable to G , that is F is Computation satisfiable iff G is satisfiable Zohar Manna A theory T admits quantifier elimination iff Winter 2010 there is an algorithm that given Σ-formula F returns a quantifier-free Σ-formula G that is T -equivalent to F . Chapter 7: Quantified Linear Arithmetic Page 1 of 40 Page 2 of 40 Example: ∃ x . 2 x = y About QE Algorithm For Σ Q -formula In developing a QE algorithm for theory T , we need only consider formulae of the form F : ∃ x . 2 x = y , ∃ x . F quantifier-free T Q -equivalent Σ Q -formula is for quantifier-free F . G : ⊤ Example: For Σ-formula For Σ Z -formula G 1 : ∃ x . ∀ y . ∃ z . F 1 [ x , y , z ] F : ∃ x . 2 x = y , � �� � there is no quantifier-free T Z -equivalent Σ Z -formula. F 2 [ x , y ] : ∃ x . ∀ y . F 2 [ x , y ] G 2 Let � T Z be T Z with divisibility predicates | . G 3 : ∃ x . ¬ ∃ y . ¬ F 2 [ x , y ] For � � �� � Σ Z -formula F 3 [ x ] F : ∃ x . 2 x = y , G 4 : ∃ x . ¬ F 3 [ x ] a quantifier-free � T Z -equivalent � Σ Z -formula is � �� � F 4 G : 2 | y . G 5 : F 4 G 5 is quantifier-free and T -equivalent to G 1 Page 3 of 40 Page 4 of 40

Augmented theory � Quantifier Elimination for T Z T Z Σ Z : { . . . , − 2 , − 1 , 0 , 1 , 2 , . . . , − 3 · , − 2 · , 2 · , 3 · , . . . , + , − , = , < } � Σ Z : Σ Z with countable number of unary divisibility predicates Lemma: for k ∈ Z + k | · Given quantifier-free Σ Z -formula F [ y ] s.t. free( F [ y ]) = { y } . Intended interpretations: S represents the set of integers k | x holds iff k divides x without any remainder S : { n ∈ Z : F [ n ] is T Z -valid } . Either S ∩ Z + or Z + \ S is finite. Example: Note: Z + is the set of positive integers. x > 1 ∧ y > 1 ∧ 2 | x + y is satisfiable (choose x = 2 , y = 2). Example: Σ Z -formula F [ y ] : ∃ x . 2 x = y ¬ (2 | x ) ∧ 4 | x S : even integers is not satisfiable. S ∩ Z + : positive even integers — infinite Z + \ S : positive odd integers — infinite Axioms of � T Z : axioms of T Z with additional countable set of Therefore, by the lemma, there is no quantifier-free T Z -formula axioms that is T Z -equivalent to F [ y ]. for k ∈ Z + ∀ x . k | x ↔ ∃ y . x = ky Thus, T Z does not admit QE. Page 5 of 40 Page 6 of 40 � T Z admits QE (Cooper’s method) Cooper’s Method: Step 1 Algorithm: Given � Σ Z -formula Put F [ x ] in Negation Normal Form (NNF) F 1 [ x ], so that ∃ x . F 1 [ x ] ◮ has negations only in literals (only ∧ , ∨ ) ∃ x . F [ x ] , ◮ is � T Z -equivalent to ∃ x . F [ x ] where F is quantifier-free, construct quantifier-free � Σ Z -formula Example: that is equivalent to ∃ x . F [ x ]. 1. Put F[x] into Negation Normal Form (NNF). ∃ x . ¬ ( x − 6 < z − x ∧ 4 | 5 x + 1 → 3 x < y ) 2. Normalize literals: s < t , k | t , or ¬ ( k | t ). is equivalent to 3. Put x in s < t on one side: hx < t or s < hx . 4. Replace hx with x ′ without a factor. ∃ x . x − 6 < z − x ∧ 4 | 5 x + 1 ∧ ¬ (3 x < y ) 5. Replace F [ x ′ ] by � F [ j ] for finitely many j . Note: ¬ ( A ∧ B → C ) ⇔ ( A ∧ B ∧ ¬ C ) Page 7 of 40 Page 8 of 40

Cooper’s Method: Step 2 Cooper’s Method: Step 3 Replace (left to right) Collect terms containing x so that literals have the form s = t ⇔ s < t + 1 ∧ t < s + 1 hx < t , t < hx , k | hx + t , or ¬ ( k | hx + t ) , ¬ ( s = t ) ⇔ s < t ∨ t < s where t is a term (does not contain x ) and h , k ∈ Z + . The output ¬ ( s < t ) ⇔ t < s + 1 is the formula ∃ x . F 3 [ x ], which is � T Z -equivalent to ∃ x . F [ x ]. The output ∃ x . F 2 [ x ] contains only literals of form Example: k | t , or ¬ ( k | t ) , s < t , x + x + y < z + 3 z + 2 y − 4 x 5 | − 7 x + t where s , t are � T Z -terms and k ∈ Z + . ⇓ ⇓ Example: 6 x < 4 z + y 5 | 7 x − t ¬ ( x < y ) ∧ ¬ ( x = y + 3) ⇓ y < x + 1 ∧ ( x < y + 3 ∨ y + 3 < x ) Page 9 of 40 Page 10 of 40 Cooper’s Method: Step 4 I Cooper’s Method: Step 4 II Finally, construct Let 3 [ x ′ ] ∧ δ ′ | x ′ δ ′ = lcm { h : h is a coefficient of x in F 3 [ x ] } , ∃ x ′ . F ′′ � �� � F 4 [ x ′ ] where lcm is the least common multiple. Multiply atoms in F 3 [ x ] by constants so that δ ′ is the coefficient of x everywhere: ∃ x ′ . F 4 [ x ′ ] is equivalent to ∃ x . F [ x ] and each literal of F 4 [ x ′ ] has one of the forms: hx < t ⇔ δ ′ x < h ′ t where h ′ h = δ ′ (A) x ′ < t t < hx ⇔ h ′ t < δ ′ x where h ′ h = δ ′ (B) t < x ′ k | hx + t ⇔ h ′ k | δ ′ x + h ′ t where h ′ h = δ ′ (C) k | x ′ + t ¬ ( h ′ k | δ ′ x + h ′ t ) h ′ h = δ ′ ¬ ( k | hx + t ) ⇔ where (D) ¬ ( k | x ′ + t ) where t is a term that does not contain x , and k ∈ Z + . The result ∃ x . F ′ 3 [ x ], in which all occurrences of x in F ′ 3 [ x ] are in terms δ ′ x . 3 with a fresh variable x ′ to form Replace δ ′ x terms in F ′ F ′′ : F 3 { δ ′ x �→ x ′ } 3 Page 11 of 40 Page 12 of 40

Cooper’s Method: Step 4 III Cooper’s Method: Step 4 IV Example: � T Z -formula Multiply when necessary: ∃ x . 30 x < 15 z + 90 ∧ 10 y − 10 < 30 x ∧ 24 | 30 x + 6 ∃ x . 3 x + 1 > y ∧ 2 x − 6 < z ∧ 4 | 5 x + 1 � �� � Replacing 30 x with fresh x ′ and adding divisibility conjunct: F [ x ] After step 3: ∃ x ′ . x ′ < 15 z + 90 ∧ 10 y − 10 < x ′ ∧ 24 | x ′ + 6 ∧ 30 | x ′ � �� � ∃ x . 2 x < z + 6 ∧ y − 1 < 3 x ∧ 4 | 5 x + 1 F 4 [ x ′ ] � �� � F 3 [ x ] ∃ x ′ . F 4 [ x ′ ] is equivalent to ∃ x . F [ x ]. Collecting coefficients of x (step 4): δ ′ = lcm(2 , 3 , 5) = 30 Multiply when necessary: ∃ x . 30 x < 15 z + 90 ∧ 10 y − 10 < 30 x ∧ 24 | 30 x + 6 Page 13 of 40 Page 14 of 40 Cooper’s Method: Step 5 Intuition of Step 5 I Construct left infinite projection F −∞ [ x ′ ] of F 4 [ x ′ ] by Property (Periodicity) (A) replacing literals x ′ < t by ⊤ if m | δ (B) replacing literals t < x ′ by ⊥ then m | n iff m | n + λδ for all λ ∈ Z That is, m |· cannot distinguish between m | n and m | n + λδ . Idea: very small numbers satisfy (A) literals but not (B) literals By the choice of δ (lcm of the k ’s) — no | literal in F 5 can Let � � k of (C) literals k | x ′ + t distinguish between n and n + λδ , for any λ ∈ Z . δ = lcm k of (D) literals ¬ ( k | x ′ + t ) δ δ � � � F 5 : F −∞ [ j ] ∨ F 4 [ t + j ] and B be the set of terms t appearing in (B) literals of F 4 [ x ′ ]. j =1 j =1 t ∈ B Construct � δ � δ � F 5 : F −∞ [ j ] ∨ F 4 [ t + j ] . j =1 j =1 t ∈ B F 5 is quantifier-free and � T Z -equivalent to ∃ x . F [ x ]. Page 15 of 40 Page 16 of 40

Intuition of Step 5 II Example of Step 5 I ◮ left disjunct � δ j =1 F −∞ [ j ] : Contains only | literals ∃ x . 3 x + 1 > y ∧ 2 x − 6 < z ∧ 4 | 5 x + 1 Asserts: no least n ∈ Z s.t. F 4 [ n ]. � �� � F [ x ] For if there exists n satisfying F −∞ , ⇓ then every n − λδ , for λ ∈ Z + , also satisfies F −∞ ∃ x ′ . x ′ < 15 z + 90 ∧ 10 y − 10 < x ′ ∧ 24 | x ′ + 6 ∧ 30 | x ′ ◮ right disjunct � δ � t ∈ B F 4 [ t + j ] : � �� � j =1 Asserts: There is least n ∈ Z s.t. F 4 [ n ]. F 4 [ x ′ ] For let t ∗ = { largest t | t < x ′ in (B) } . By step 5, If n ∈ Z is s.t. F 4 [ n ], then F −∞ [ x ′ ] : ⊤ ∧ ⊥ ∧ 24 | x ′ + 6 ∧ 30 | x ′ , ∃ j (1 ≤ j ≤ δ ) . t ∗ + j ≤ n ∧ F 4 [ t ∗ + j ] which simplifies to ⊥ . In other words, if there is a solution, then one must appear in δ interval to the right of t ∗ Page 17 of 40 Page 18 of 40 Example of Step 5 II Cooper’s Method: Example I Compute ∃ x . (3 x + 1 < 10 ∨ 7 x − 6 > 7) ∧ 2 | x δ = lcm { 24 , 30 } = 120 and B = { 10 y − 10 } . � �� � F [ x ] Then replacing x ′ by 10 y − 10 + j in F 4 [ x ′ ] produces Isolate x terms � � 120 � 10 y − 10 + j < 15 z + 90 ∧ 10 y − 10 < 10 y − 10 + j ∃ x . (3 x < 9 ∨ 13 < 7 x ) ∧ 2 | x , F 5 : ∧ 24 | 10 y − 10 + j + 6 ∧ 30 | 10 y − 10 + j j =1 so δ ′ = lcm { 3 , 7 } = 21 . which simplifies to After multiplying coefficients by proper constants, � � 120 � 10 y + j < 15 z + 100 ∧ ✘✘ 0 < j ✘ F 5 : . ∃ x . (21 x < 63 ∨ 39 < 21 x ) ∧ 42 | 21 x , ∧ 24 | 10 y + j − 4 ∧ 30 | 10 y − 10 + j j =1 we replace 21 x by x ′ : F 5 is quantifier-free and � T Z -equivalent to ∃ x . F [ x ]. ∃ x ′ . ( x ′ < 63 ∨ 39 < x ′ ) ∧ 42 | x ′ ∧ 21 | x ′ . � �� � F 4 [ x ′ ] Page 19 of 40 Page 20 of 40