Search in the semantic domain

1. Search in the semantic domain

2. Some definitions • atomic formula: smallest formula possible (no sub-formulas) • literal: atomic formula or negation of an atomic formula • clause: disjunction of literals • CNF: Conjunction of clauses literal (A Ç: B Ç C) Æ (D Ç B Ç E) Æ clause atomic

3. DPLL backtracking search algorithm • David-Puttnam-Logemann-Loveland • Algorithm: given a formula, return SAT or UNSAT • SAT: there some truth assignment that makes the formula true • UNSAT: formula is false on all truth assignments • Key idea • Pick a literal • Assign literal to true, simplify the formula, and recurse • Assign literal to false, simplify the formula, and recurse

4. In more detail • If formula is false, return UNSAT • else If formula is true, return SAT • else: • Pick a literal • Assign literal to true, simplify the formula, and recurse • If recursive call returns SAT, return SAT • Assign literal to false, simplify the formula, and recurse • If recursive call returns SAT, return SAT • If both recursive calls return UNSAT, return UNSAT

5. Example simplification A to true (A Ç: B Ç C) Æ (D Ç B Ç E) Æ (: A Ç D Ç: E) (A Ç: B Ç C) Æ (D Ç B Ç E) Æ (: A Ç D Ç: E) A to false (A Ç: B Ç C) Æ (D Ç B Ç E) Æ (: A Ç D Ç: E) (A Ç: B Ç C) Æ (D Ç B Ç E) Æ (: A Ç D Ç: E)

6. How do formulas become true or false? • Formula becomes true • when conjunction becomes empty • Formula becomes false • when clause becomes empty

7. Search tree (A Ç B) Æ (A Ç: B)

8. Search tree (A Ç B) Æ (A Ç: B)

9. Choice of literal matters C Æ (B Ç: C) Æ (A Ç: B) Æ : A

10. Choice of literal matters C Æ (B Ç: C) Æ (A Ç: B) Æ : A

11. Choice of literal matters C Æ (B Ç: C) Æ (A Ç: B) Æ : A

12. Some heuristics for picking literal • Pick literals that appear in unit clauses (called unit propagation) • Pick literals that always appear in the same polarity (A or : A) C Æ (B Ç: C) Æ (A Ç: B) Æ : A • Why? Because of the following optimization: • if pick A, don’t explore : A branch • if pick : A, don’t explore A branch (A Ç B) Æ (A Ç: B) Æ (C Ç B) Æ (: C Ç: B)

13. Some heuristics for picking literal • Pick literals for which the formula can be expressed as (R Ç A) Æ (Q Ç: A) Æ S • Can then merge both subtrees into just one subtree that checks (R Ç Q) Æ S • These are just a few simple heuristics • Many other heuristics have been developed • Decades of research on this

14. Extending backtracking search • Let’s assume we also have equality with uninterpreted function symbols, for example: ( f(f(a)) = a Ç: (f(a) = f(b)) ) Æ ( a = b Æ f(a) = f(f(b)) ) • Some observations • We can still simplify a formula based on a literal being T or F • But we can only simplify that literal • For instance, in the example above, once we’ve assumed a = b, how do we know that : (f(a) = f(b)) is false?

15. Keep an environment

16. Keep an environment • ( f(f(a)) = a Ç: (f(a) = f(b)) ) Æ • ( a = b Æ f(a) = f(f(b)) )

17. Keep an environment • ( f(f(a)) = a Ç: (f(a) = f(b)) ) Æ • ( a = b Æ f(a) = f(f(b)) )

18. Davis-Putnam paper • Semi-algorithm for first-order logic • Refutation based: negation formula, and show that formula is unsatisfiable • Uses successive SAT instances

19. Prenex normal form • Prenex normal form: all quantifiers on the outside • Some example conversions: • 9 x.P(x) Æ9 x. Q(x) • 8 x. P(x) Ç8 x. Q(x) • In general can convert any formula into prenex normal form

20. Getting rid of existentials • Replace existential with a function symbol that takes as parameters the enclosing universally quantified variables • Transform: 8 x1. 9 x2. 8 x3. 9 x4 R(x1, x2,x3,x4) • Into 8 x1. 8 x3. R(x1, f2(x1),x3,f4(x1, x3))

21. Herbrand’s universe of a formula • Given a formula F, we call HF the Herbrand universe of the formula • All constants in F belong to HF (if F does not have constants, then HF includes a fresh constant a) • For any function symbol of arity n occurring in F, and for any t1, …, tn belonging to HF, f(t1, …, tn) also belongs to HF • H_F is the minimal set that satisfies these constraints

22. Quantifier free lines • Instantiate body of a formula F with elements of HF • Suppose F = 8 x1, x2 R(x1, f(x1), x2) • H_F = { a, f(a), f(f(a)), … } • Quantifier free lines: • R(a, f(a), a) • R(a, f(a), f(a)) • R(f(a), f(f(a)), a) • … • Each line is implied by original formula • As a result, if the conjunction of some quantifier free lines is inconsistent, so is the original formula

23. Quantifier free lines • Each quantifier free line is implied by original formula • As a result, if the conjunction of some quantifier free lines is inconsistent, so is the original formula • If the conjunction of the first n quantifier free lines is consistent, for any n, then the original formula is consistent • Follows from the fact that an infinite sets of quantifier-free formulas is inconsistent iff some finite subset is inconsistent

24. Example • 8 x. : P(x) Ç9 x. P(x)

25. Example • 8 x. : P(x) Ç9 x. P(x)

26. ATP using Lazy Proof Explication • a = b Æ (: (f(a) = f(b)) Ç b = c) Æ: (f(a) = f(c))

27. ATP using Lazy Proof Explication • a = b Æ (: (f(a) = f(b)) Ç b = c) Æ: (f(a) = f(c)) • Assign proxies: • x1Æ (: x2Ç x3) Æ: x4 • Use SAT solver: if SAT solver says unsatisfiable, then original formula is unsatisfiable

28. ATP using Lazy Proof Explication • In this case, say SAT solver comes back with x1 set to true, and x2, x3, and x4 set to false • In the propositional world, this is a valid truth assignment • But when considering the underlying meaning of the proxies, we notice that x1 being true and x2 being false is an inconsistency • If the backtracking search is not aware of this, it will continue considering truth assignments with this same inconsistency (for example x1 = x3 = true, x2 = x4 = false)

29. Key idea • Have decision procedures return an explicating proof as to why the inconsistency occurred. • The new formula becomes: F Æ proof • The proof reflects the decision procedure’s knowledge back into the propositional world, and can then be used in the prop world to prune the search • In the example, the proof is: a = b ) f(a) = f(b)

30. Example continued • Formula becomes: x1Æ (: x2Ç x3) Æ: x4Æ (: x1Ç x2) • Note that SAT solver cannot find the original satisfying assignment (x1 set to true, and x2, x3, and x4 set to false) • Nor can it come back with any assignment that has x1 set to true and x2 set to false

31. Example continued • So SAT solver comes back with: x1, x2, x3 set to true, and x4 set to false • This assignment is also inconsistent when considering the underlying meaning of proxies • Explicating proof: (a = b Æ b = c) ) f(a) = f(c)

32. Example continued • New formula: x1Æ (: x2Ç x3) Æ: x4Æ (: x1Ç x2) Æ (: x1Ç: x3Ç x4) • SAT solver returns unsatisfiable, and so we know the original formula is unsatisfiable.

33. Algorithm in more detail functionsatisfy(FormulaF): Monome { while (true) “allocate proxy prop vars for atomic formulas in F, and create mapping  from proxies to atomic formulas” TruthAssignmentA := SAT-solve(-1(F)); if (A = null) { // F is unsatisfiable returnnull } else MonomeM := (A); Formula E := check(M); if (E = null) { // M is satisfiable, and so is F returnM; } else { F := FÆE; } } }