four flavors of entailment
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Four Flavors of Entailment ohle 1 , Roberto Sebastiani 2 , and Armin - PowerPoint PPT Presentation

Four Flavors of Entailment ohle 1 , Roberto Sebastiani 2 , and Armin Biere 1 Sibylle M 2 Department of Information Engineering 1 Institute for Formal Models and Verification LIT Secure and Correct Systems Lab and Computer Science The 23rd


  1. Four Flavors of Entailment ohle 1 , Roberto Sebastiani 2 , and Armin Biere 1 Sibylle M¨ 2 Department of Information Engineering 1 Institute for Formal Models and Verification LIT Secure and Correct Systems Lab and Computer Science The 23rd International Conference on Theory and Applications of Satisfiability Testing (SAT 2020) 3 – 10 July 2020

  2. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l ➤ o a l g l e p i r n c r e a o t a a l j e s model shrinking i C l o c m D t n i e i o C n n (Tibebu and Fey, DDECS’18) n g L t dual reasoning (M¨ ohle and Biere, ICTAI’18) logical entailment (Sebastiani, arXiv.org, 2020) 2

  3. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s model shrinking i C l o c m ➤ D t n i e i o C n n (Tibebu and Fey, DDECS’18) n g L t dual reasoning (M¨ ohle and Biere, ICTAI’18) logical entailment (Sebastiani, arXiv.org, 2020) 2

  4. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s model shrinking i C l o c m D t n i e i o C n n (Tibebu and Fey, DDECS’18) n g L t dual reasoning ➤ (M¨ ohle and Biere, ICTAI’18) logical entailment (Sebastiani, arXiv.org, 2020) 2

  5. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s model shrinking i C l o c m D t n i e i o C n n (Tibebu and Fey, DDECS’18) n g L t dual reasoning (M¨ ohle and Biere, ICTAI’18) logical entailment ➤ (Sebastiani, arXiv.org, 2020) 2

  6. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s model shrinking i C l o c m D t n i e i o C n n (Tibebu and Fey, DDECS’18) n g L t dual reasoning (M¨ ohle and Biere, ICTAI’18) logical entailment (Sebastiani, arXiv.org, 2020) Example F = ( x ∧ y ) ∨ ( x ∧ ¬ y ) ➤ F | x = y ∨ ¬ y � = 1 F | xy = F | x ¬ y = 1 ⇒ x | = = F 2

  7. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s model shrinking i C l o c m D t n i e i o C n n (Tibebu and Fey, DDECS’18) n g L t dual reasoning (M¨ ohle and Biere, ICTAI’18) logical entailment (Sebastiani, arXiv.org, 2020) Example F = ( x ∧ y ) ∨ ( x ∧ ¬ y ) F | x = y ∨ ¬ y � = 1 F | xy = F | x ¬ y = 1 ⇒ x | = = F But determining logical entailment is harder than it seems! ➤ 2

  8. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n i e i o C n n n g L t 2

  9. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n . . . pairwise disjoint models i e i o C n ➤ n n g L t add the negated models as blocking clauses variant of conflict analysis (Toda and Soh, ACM J. Exp. Algorithmics, 2016) chronological CDCL (Nadel and Ryvchin, SAT’18; M¨ ohle and Biere, SAT’19) 2

  10. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n . . . pairwise disjoint models i e i o C n n n g L t add the negated models as blocking clauses ➤ variant of conflict analysis (Toda and Soh, ACM J. Exp. Algorithmics, 2016) chronological CDCL (Nadel and Ryvchin, SAT’18; M¨ ohle and Biere, SAT’19) 2

  11. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n . . . pairwise disjoint models i e i o C n n n g L t add the negated models as blocking clauses variant of conflict analysis ➤ (Toda and Soh, ACM J. Exp. Algorithmics, 2016) chronological CDCL (Nadel and Ryvchin, SAT’18; M¨ ohle and Biere, SAT’19) 2

  12. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n . . . pairwise disjoint models i e i o C n n n g L t add the negated models as blocking clauses variant of conflict analysis (Toda and Soh, ACM J. Exp. Algorithmics, 2016) chronological CDCL ➤ (Nadel and Ryvchin, SAT’18; M¨ ohle and Biere, SAT’19) 2

  13. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n . . . pairwise disjoint models i e i o C n n n g L t 2

  14. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n . . . pairwise disjoint models i e i o C n n n g L t . . . projection ➤ F ( X , Y ) where X ∩ Y = ∅ relevant variables X irrelevant variables Y ∃ Y [ F ( X , Y ) ] project F ( X , Y ) onto X 2

  15. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n . . . pairwise disjoint models i e i o C n n n g L t . . . projection 2

  16. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n . . . pairwise disjoint models i e i o C n n n g L t . . . projection 2

  17. Motivation We need. . . c h l r o o g n i d c o . . . short (partial) models u a l o a l g l e p i r n c r e a o t a a l j e s i C l o c m D t n . . . pairwise disjoint models i e i o C n n n g L t . . . projection We get. . . . . . Disjoint Sum-of-Products (DSOP) ➤ DSOP 2

  18. Main Idea (Partial) Assignment I Check assignment Formula F SAT solver DSOP M Next assignment 3

  19. Our Contribution F | I = 1 F | I ≈ 1 ➤ F | I ≡ 1 ∀ X ∃ Y [ F | I ] = 1 (Partial) Assignment I Check assignment Formula F SAT solver DSOP M Next assignment 3

  20. Logical Entailment Test under Projection ➤ Given formula over variables in X ∪ Y F trail over variables in X ∪ Y I 4

  21. Logical Entailment Test under Projection Given formula over variables in X ∪ Y F trail over variables in X ∪ Y I Quantified entailment condition ➤ In ϕ = ∀ X ∀ Y [ F | I ] the unassigned variables in X ∪ Y are quantified ϕ = 1: all possible total extensions of I satisfy F 4

  22. Logical Entailment Test under Projection Given formula over variables in X ∪ Y F trail over variables in X ∪ Y I Quantified entailment condition In ϕ = ∀ X ∀ Y [ F | I ] the unassigned variables in X ∪ Y are quantified ϕ = 1: all possible total extensions of I satisfy F Entailment under projection onto the set of variables X ➤ Does for each J X exist one J Y such that F | I ′ = 1 where I ′ = I ∪ J X ∪ J Y ? 4

  23. Logical Entailment Test under Projection Given formula over variables in X ∪ Y F trail over variables in X ∪ Y I Quantified entailment condition In ϕ = ∀ X ∀ Y [ F | I ] the unassigned variables in X ∪ Y are quantified ϕ = 1: all possible total extensions of I satisfy F Entailment under projection onto the set of variables X Does for each J X exist one J Y such that F | I ′ = 1 where I ′ = I ∪ J X ∪ J Y ? QBF ( ϕ ) = 1 where ϕ = ∀ X ∃ Y [ F | I ] = 1? ➤ 4

  24. Four Flavors of Logical Entailment under Projection ➤ 1) F | I = 1 ( syntactic check ) F = ( x 1 ∨ y ∨ x 2 ) X = { x 1 , x 2 } Y = { y } I = x 1 : F | I = 1 = ⇒ I | = F 5

  25. Four Flavors of Logical Entailment under Projection 1) F | I = 1 ( syntactic check ) F = x 1 y ∨ yx 2 X = { x 1 , x 2 } Y = { y } 2) F | I ≈ 1 ( incomplete check in P ) I = x 1 x 2 : F | I = y ∨ y � = 1 but is valid ➤ 0 ∈ BCP ( ¬ F , I ) ⇒ x 1 x 2 | I = x 1 x 2 y : = = F 5

  26. Four Flavors of Logical Entailment under Projection F = x 1 ( x 2 y ∨ x 2 y ∨ x 2 y ∨ x 2 y ) X = { x 1 , x 2 } Y = { y } 1) F | I = 1 ( syntactic check ) I = x 1 : I ( F ) = x 2 y ∨ x 2 y ∨ x 2 y ∨ x 2 y � = 1 but is valid 2) F | I ≈ 1 ( incomplete check in P ) P = CNF( F ) 3) F | I ≡ 1 ( semantic check in coNP ) ➤ N = CNF( ¬ F ): P | I and N | I are non-constant and contain no units N | I = ( x 2 ∨ y )( x 2 ∨ y )( x 2 ∨ y )( x 2 ∨ y ): SAT ( N ∧ I ) = 0 ⇒ I | = = F 5

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