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Search in the semantic domain
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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
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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

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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

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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 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)
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How do formulas become T or F?

• Formula becomes true
• when conjunction becomes empty
• Formula becomes false
• when clause becomes empty

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Search tree
(A Ç B) Æ (A Ç B)
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Search tree
(A Ç B) Æ (A Ç B)
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Choice of literal matters
C Æ (B Ç C) Æ (A Ç B) Æ A
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Choice of literal matters
C Æ (B Ç C) Æ (A Ç B) Æ A
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Choice of literal matters
C Æ (B Ç C) Æ (A Ç B) Æ A
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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 literal is A, then pick A, dont explore A
  branch
• if literal is A, then pick A, dont explore A
  branch

(A Ç B) Æ (A Ç B) Æ (C Ç B) Æ ( C Ç B)
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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

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Extending backtracking search

• Lets 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 weve
  assumed a b, how do we know that (f(a)
  f(b)) is false?

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Keep an environment
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Keep an environment
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Keep an environment
( f(f(a)) a Ç (f(a) f(b)) ) Æ ( a b Æ
f(a) f(f(b)) )
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Keep an environment
( f(f(a)) a Ç (f(a) f(b)) ) Æ ( a b Æ
f(a) f(f(b)) )
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Davis-Putnam paper

• Semi-algorithm for first-order logic
• Refutation based negation formula, and show that
  formula is unsatisfiable
• Uses successive SAT instances

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Prenex normal form

• Prenex normal form all quantifiers on the
  outside
• Some example conversions
• 8 x. P(x) Æ 8 x. Q(x)
• 9 x.P(x) Ç 9 x. Q(x)
• 8 x. P(x) Ç 8 x. Q(x)
• In general can convert any formula into prenex
  normal form (might possibly strengthen)

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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))

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Herbrands 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
• HF is the minimal set that satisfies these
  constraints

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Quantifier free lines

• Instantiate body of a formula F with elements of
  HF
• Suppose F 8 x1, x2 R(x1, f(x1), x2)
• HF 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

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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 set of
  quantifier-free formulas is inconsistent iff some
  finite subset is inconsistent

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Example

• 8 x. P(x) Ç 9 x. P(x)

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Example

• 8 x. P(x) Ç 9 x. P(x)

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ATP using Lazy Proof Explication

• a b Æ ( (f(a) f(b)) Ç b c) Æ (f(a)
  f(c))

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ATP using Lazy Proof Explication

• a b Æ ( (f(a) f(b)) Ç b c) Æ (f(a)
  f(c))
• Assign proxies
• x1 Æ ( x2 Ç x3) Æ x4
• Use SAT solver if SAT solver says unsatisfiable,
  then original formula is unsatisfiable

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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)

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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 procedures
  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)

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Example continued

• Formula becomes
• x1 Æ ( x2 Ç x3) Æ x4 Æ ( x1 Ç 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

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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)

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Example continued

• New formula
• x1 Æ ( x2 Ç x3) Æ x4 Æ ( x1 Ç x2) Æ
• ( x1 Ç x3 Ç x4)
• SAT solver returns unsatisfiable, and so we know
  the original formula is unsatisfiable.

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Algorithm in more detail
function satisfy(Formula F) Monome while
(true) allocate proxy prop vars for atomic
formulas in F, and create mapping ? from
proxies to atomic formulas TruthAssignment
A SAT-solve(?-1(F)) if (A null) //
F is unsatisfiable return null
else Monome M ?(A) Formula
E check(M) if (E null) // M is
satisfiable, then so is F return M
else F F Æ E