A framework for eager encoding

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A framework for eager encoding

• Daniel Kroening ETH, Switzerland
• Ofer Strichman Technion, Israel

(Executive summary) (submitted to Formal Aspects
of Computing)
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• A generic framework for reducing decidable logics
  to propositional logic (beyond NP).
• Instantiating the framework for a specific logic
  L, requires a deductive system for L that meets
  several criteria.
• Linear arithmetic, EUF, arrays etc all have it.

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• A proof rule
• A proof step (Rule, Antecedent, Proposition)

• Definition (Proof-step Constraint) let A1Ak be
  the Antecedents and p the Proposition of step.
  Then

Boolean encoding
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• A proof P (s1,, sn) is a set of Proof Steps,
• in which the Antecedence relation is acyclic
• The Proof Constraint c(P) induced by P is the
  conjunction of the constraints induced by its
  steps

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• Propositional skeleton
• Theorem 1 For every formula ? and any sound
  proof P, ? is satisfiable ) ?sk Æ c(P) is
  satisfiable.

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Complete proofs

• Definition (Complete proofs) A proof P is called
  complete with respect to ? if

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Sufficient condition for completeness 1

• Notation A assumption, B a proposition.
  denotes P proves B from A.
• Let ? be an unsatisfiable formula
• Theorem 2 A proof P is complete with respect to
  ? if for every full assignment ?

TL(?) Theory Literals corresponding to ?
Not constructive!
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• Projection of a variable x a set of proof steps
  that eliminate x and maintains satisfiability.
• Strong projection of a variable x a projection
  of x that maintains
• The projected consequences from each minimal
  unsatisfiable core of literals is unsatisfiable.

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Example strong projection
Consider the formula
U2
U1
Now strongly project x1

• Both sub-formulas are unsatisfiable and do not
  contain x1.

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• Let ?C be a conjunction of ?s literals.
• A proof construction procedure eliminate all
  variables in ?C through strong projection.
• Theorem 3 The constructed proof is complete
  for ?.

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• Goal for a given logic L,
• Find a strong projection procedure.
• Construct P
• Generate c(P)
• Check ?sk Æ c(P)

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Example Disjunctive Linear Arithmetic S02
e1 e2 e3
e4
?C x1 - x2 lt 0, x1 - x3 lt 0, -x1 2x3 x2 lt
0, -x3 lt -1
A proof P by (Strong) projection
e1 ? e3 ? e5
x1
e2 ? e3 ? e6
e4 ? e5 ? false
x3
4. Solve ? ?sk Æ c(P)
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What now ?

• It is left to show a strong projection method for
  each logic we are interested in integrating.
• Current eager procedures are far too wasteful.
  Need to find better ones.

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Optimizations

• Optimizations that were previously published in
  the eager encoding series can all be
  interpreted in this framework.
• Conjunction Matrices
• Simplifications and early detection
• Cross-theory learning

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Cross-theory learning

• ?C (T1)
• ?C (T2)
• From T1 we learn z1 z2 which we propagate to
  T2
• In T2 we get a contradiction on z1 gt 2, z21, z1
  z2
• This results in a conflict clause
• Which represents cross-theory learning

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Projection (by example)(Starting from a
conjunction of literals)

• Indeed,
• x1? var(x4 gt x4)
• ? (x2 gt x3) Æ (x4 gt x4) is equisatisfiable to ?

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? (x1 - x2 lt 0) ? (x1 - x3 lt 0) ? ((-x1 2x3
x2 lt 0) ? (-x3 lt -1))
?c (x1 - x2 lt 0) ? (x1 - x3 lt 0) ? (-x1 2x3
x2 lt 0) Æ (-x3 lt -1) ? (x1,x2,x3) Choose x1 ?
(x2,x3) Strong-project P (R, (2x3 0),
(x1 - x2 lt 0), (-x1 2x3 x2 lt 0), (R, (x2
x3 0), (x1 - x2 lt 0), (-x1 2x3 x2 lt
0) ?c (2x3 0) ? (x2 x3 0) ? (-x3 lt
-1)
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Example

• c(step) e(x5) Æ e(x 0) ! e(5 0)

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Example

• Prove validity of x ? 5 Ç x 0 by using atoms
  only

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Example (contd)
?sk Æ c(P) is unsatisfiable hence ? is valid
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Sufficient condition for completeness 2
For a partial assignment ? s.t. ? ² ?, ? is
minimal if 8v. ?nv 2 ?

• ? - an unsatisfiable formula.
• A - the set of minimal assignments that satisfy
  ?sk. A proof P is complete with respect to ? if
  8? 2 A,

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Sufficient condition for completeness 3

• ? - an unsatisfiable formula
• A - the set of minimal assignments that satisfy
  ?sk. A proof P is complete with respect to ? if
  8? 2 A, for some unsatisfiable core TLuc(?) µ
  TL(?)

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Proof-graph of P
A
A,B sets of propositions
P proves B using A
A
B
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