Semantic Minimization of 3-Valued Propositional Formulas

1. Semantic Minimizationof 3-Valued Propositional Formulas Thomas Reps Alexey Loginov University of Wisconsin Mooly Sagiv Tel-Aviv University

2. Semantic Minimization • pp = 1, right? • (A): Value of formula  under assignment A • In 3-valued logic, (A) may equal? pp([p 0]) = 1 pp([p?]) = ? pp([p 1]) = 1 • However, 1([p 0]) = 1 = pp([p 0]) 1([p?]) = 1  ? = pp([p?]) 1([p 1]) = 1 = pp([p 1])

3. Motivation • Dataflow analysis • Hardware verification • Symbolic trajectory evaluation • Shape analysis

4. Questions • What does “best” mean? • Can one find a best formula? • How?

5. Two-valued logic Three-valued logic {0,1} 0 1 {0} {1} Two- vs. Three-Valued Logic {0} {0,1} {1}  {0,1}

6. Two-valued logic Three-valued logic 1 1 {1} ½ 0 {0,1} 0 {0} Two- vs. Three-Valued Logic

7. Two-valued logic Three-valued logic 0 1 {0,1} {0} {1} Two- vs. Three-Valued Logic

8. Two-valued logic Three-valued logic 0 1 ½ 0 1 Two- vs. Three-Valued Logic 0 3½ 1 3½

9. Boolean Connectives [Kleene]

10. ½ Information order 0 1 Three-Valued Logic • 1: True • 0: False • 1/2: Unknown • A join semi-lattice: 0  1 = 1/2 0  ½ 1  ½

11. 1([p 0]) = 1 = pp([p 0]) 1([p½]) = 1 ½ = pp([p½]) 1([p 1]) = 1 = pp([p 1]) Semantic Minimization 2-valued logic: 1 is equivalent to pp 3-valued logic: 1 is better thanpp For a given , is there a best formula? Yes!

12. Minimal? x +x’ x x’ xy + x’z xy + x’y’ xy + x’z+ yz xy’+ x’z’+ yz No! Yes! No! Yes! Yes! No!

13. Rewrite Rules?    1    0

14. 2-Valued Propositional Meaning 0(a) = 0 1(a) = 1 xi(a) = a(xi) (a) = 1 – (a) 1  2(a) = min(1(a), 2(a)) 1  2(a) = max(1(a), 2(a))

15. 3-Valued Propositional Meaning ½(a) = ½ 0(a) = 0 1(a) = 1 xi(a) = a(xi) (a) = 1 – (a) 1  2(a) = min(1(a), 2(a)) 1  2(a) = max(1(a), 2(a))

16. 3-Valued Propositional Meaning ½(A) = ½ 0(A) = 0 1(A) = 1 xi(A) = A(xi) (A) = 1 – (A) 1  2(A) = min(1(A), 2(A)) 1  2(A) = max(1(A), 2(A))

17. [ p 0,q 0, r 1,s½ ] [ p½,q 0, r 1,s 0] [ p½,q 0, r 1,s 1] [ p 1, q 0, r 1,s½ ] [ p 0, q 0, r 1, s 0 ] [ p 0, q 0, r 1, s 1 ] [ p 1, q 0, r 1, s 0 ] [ p 1, q 0, r 1, s 1 ] Represented by A A = [ p½,q 0,r 1,s½ ]

18. ½ 0 1 The Right Definition of “Best”? Observation If for all A, (A)  (A),  is better than 

19. The Right Definition of “Best”? Observation If for all A, (A)  (A),  is better than  0(A) = 0  ½ =  ½ (A) 0 is better than ½ 1(A) = 1  ½ =  ½ (A) 1 is better than ½

20. 1 ½ ½ 0 1 0 Acceptance Device A   iff (A)  1 “Potentially accepts ”

21. 1 ½ ½ 0 1 0 Acceptance Device A   iff (A)  0 “Potentially rejects ”

22.    Acceptance Device 3-valued 2-valued  • Suppose that A represents a, and • a  2-valued assignments. We want: • Ifa  , then A   • Ifa  , then A  

23. ½ Acceptance Device 3-valued 2-valued • Suppose that A represents a, and • a  2-valued assignments. We want: • Ifa  ½, then A  0 • Ifa  ½, then A  0 Violated! 

24. ½ Acceptance Device 3-valued 2-valued • Suppose that A represents a, and • a  2-valued assignments. We want: • Ifa  ½, then A  1 • Ifa  ½, then A  1  Violated!

25. The Right Definition of “Best”? Observation If for all A, (A)  (A),  is better than  Not all “better” formulas preserve potential acceptance of 2-valued assignments

26. What Does “Best” Mean? Supervaluational meaning (A) =  (a) a rep. by A

27. Truth-functional semantics Non-truth-functional semantics Minimization Semantic Minimization (A) = (A)

28. Example pp([p½]) =  pp(a) a{[p 0], [p 1]} = pp([p 0])  pp([p 1]) = 1  1 = 1 = 1([p½])

29. Example ½([p½]) =  ½(a) a{[p 0], [p 1]} = ½([p 0])  ½([p 1]) = ½ ½ = ½ =½([p½])

30. Truth-functional semantics Non-truth-functional semantics Minimization Semantic Minimization (A) = (A)  For all A, (A)  (A) “ is better than ”

31. Realization of aMonotonic Boolean Function[Blamey 1980] f  Formula[ f ] b a  a’b + 1b + ab + a1 + ab’  (a’b’)’

32. Realization of aMonotonic Boolean Function[Blamey 1980] f  Formula[ f ] b a  a’b + ab + a1 + ab’  (a’b’ + 1b)’

33. ([½, 1]) =  (a) a{[0,1], [1,1]} = ([0,0])  ([1,1]) = 1  1 = 1 Our Problem  Formula[] b a

34. Special Case: contains no occurrences of ½ or   contains no occurrences of ½ in corners b  a’b + 1b + ab + a1 + ab’  (a’b’)’ a  a’b + 1b + ab + a1 + ab’  (a’b’)’

35. Special Case: contains no occurrences of ½ or   contains no occurrences of ½ in corners b b a a

36. How Do We Obtain ? • Represent  with a pair • floor:    ½ = 0 • ceiling:    ½ = 1

37. How Do We Obtain (, )? 0  (a.0, a.0) 1  (a.1, a.1) ½  (a.0, a.1) xi (a.a(xi), a.a(xi)) ( f ,  f )  ( f ,  f ) ( f 1,  f1 )  ( f2,  f2 )  ( f 1   f2,  f1    f2 ) ( f 1,  f1 )  ( f2,  f2 )  ( f 1   f2,  f1    f2 ) BDD operations

38. Semantically Minimal Formula • General case  primes(  ) ( primes(   )) • When  contains no occurrences of ½ and  primes(  )

39. Example Original formula () xy’+ x’z’+ yz Minimal formula () x’y + x’z’+ yz + xy’+ xz + y’z’ A(A) (A) [x ½, y 0, z 0] 1 ½ [x 0, y 1, z ½] 1 ½ [x 1, y ½, z 1] 1 ½

40. Example Original formula ( = if x then y else z) xy + x’z Minimal formula () xy+ x’z+ yz A(A) (A) [x ½, y 1, z 1] 1 ½

41. Demo

42. Related Work • [Blamey 1980, 1986] • Realization of a monotonic Boolean function • [Godefroid & Bruns 2000] • Supervaluational (“thorough”) semantics for model checking partial Kripke structures • For propositional formulas Deciding “(A)  1?” is NP-complete

43. Our Questions • What does “best” mean? For all A, (A) = (A) • Can one find a best formula? Yes • How? Create (, ) Return  primes(  ) ( primes(   ))