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; Copyright (C) 2016, Regents of the University of Texas
; Marijn Heule, Warren A. Hunt, Jr., and Matt Kaufmann
; License: A 3-clause BSD license. See the LICENSE file distributed with ACL2.
; See soundness.lisp. Here we prove a key lemma in support of that book,
; namely, that an addition (non-deletion) proof step preserves consistency: if
; the formula F is satisfiable and a proof clause C is verified again the
; formula, then the union F U {C} is satisfiable. Here is a sketch of the
; proof.
#||
Notation. Below, "unit-propagation" refers to any unit propagation,
even if it is partial (that is, with respect to a sequence of
clauses). We denote this operation by U(F,A), where F is a formula
and A is an assignment. If A is empty we may write U(F). We say that
U(F,A) = t if the indicated unit-propagation produces a contradiction,
that is, an assignment that falsifies some clause in F. For C a
clause, let N(C) be the assignment consisting of the negations of the
literals from C. We write "A |= P" to indicate that P, which is a
formula or a clause, is true under assignment A. We write A \ {L} to
denote (remove-literal L A).
Lemma (truth-monotone). For a clause or formula P and assignments A1
and A2 with A1 \subset A2, if A1 |= P then A2 |= P.
Proof sketch. First prove this by induction on \subset for P a
clause. Then prove this for formulas by induction on F. -|
Lemma (unit-propagation-identity). Suppose that F is a formula and
that assignment A |= F. Then U(F,A) = A (hence is not t).
Proof sketch. This is an easy induction on U. In the base case we
have a clause in F that is false under A, so it cannot be that A |=
F. The induction step is clear by truth-monotone. -|
Lemma (unit-propagation-t-monotone). For a formula F and assignments
A1 and A2 with A1 \subset A2, if U(F,A1) = t then U(F,A2) = t.
Proof sketch: By induction using a merge of the induction schemes for
(unit-propagation F indices A1) and (unit-propagation F indices A2). -|
Lemma (unit-propagation-monotonicity). For a formula F and
assignments A1 and A2 with A1 \subset A2, if U(F,A2) is not t, then
U(F,A1) is an assignment and U(F,A1) \subset U(F,A2).
Lemma (unit-propagation-correct). Suppose that F is a formula, C is a
clause, and assignment A |= F. Also suppose that U(F,N(C)) = t. Then
A |= C.
Proof sketch. Suppose (for a contradiction) otherwise; thus, the
assumptions hold but A is disjoint from C. Let B = A U N(C). Then B
is an assignment, and by truth-monotone, B |= F. Since U(F,N(C)) = t
then by unit-propagation-t-monotone, U(F,B) = t. We are done by the
lemma, unit-propagation-identity. -|
Finally, we outline the proof of the main result,
satisfiable-add-proof-clause. Assume that we have a formula F and a
clause C that is verified for addition to F. In our code we also have
a formula F' obtained by "shrinking" F. The following lemmas show
that this isn't a problem, at least in principle; but we will try to
reference either F' or F below, as appropriate, in order to guide us
in development of a mechanically-checkable proof.
Lemma (satisfiable-maybe-shrink-formula). F' is satisfiable if and
only if F is satisfiable. [We can surely also prove, if necessary,
that F' and F are true under the same assignments.]
Lemma (unit-propagation-maybe-shrink-formula). For all A and F,
U(F,A) = U(F',A). -|
There are two cases, according to the definition of verify-clause,
hence two main sub-lemmas. First, the easier one:
Case 1: U(F,N(C)) = t (Lemma satisfiable-add-proof-clause-rup).
Proof sketch. By unit-propagation-correct, any assignment A
satisfying F also satisfies C; thus A |= F' U {C} by a slight
variation of satisfiable-maybe-shrink-formula. -|
Case 2: U(F,N(C)) is not t and C is {L} U C' (Lemma
satisfiable-add-proof-clause-drat). We also assume that (*) F' U {C}
is not satisfiable, since otherwise we are done.
Proof sketch. Let A |= F. Thus A |= F'.
Claim 1 (Lemma satisfiable-add-proof-clause-drat-1): N(C) \subset A.
Proof sketch for Claim 1. Otherwise, taking P in N(C) \ A, we can
extend A by -P to get an assignment A' extending A; then A' |= F' by
truth-monotone, and A' |= C by choice of P, so A' |= F' U {C},
contradicting (*).
Let B = A \ {-L} U {L}. We will prove:
Claim 2 (Lemma satisfiable-add-proof-clause-drat-2): Let D \in F;
then B |= D.
Assume Claim 2 for the moment. It clearly implies that B |= F; thus
B|= F'. Also B |= C because L is in C. This contradicts (*), proving
Case 2, Lemma satisfiable-add-proof-clause-drat.
So it remains to prove Claim 2. First here are some useful lemmas.
Lemma (truth-remove-irrelevant-lit-1). For assignment A and clause C,
if A |= C and L is not in C then A \ {L} |= C.
Proof sketch: Easy induction. -|
Lemma (truth-remove-irrelevant-lit-2). For A and B as above and clause
C0 that does not contain -L or L, if A |= C0 then B |= C0.
Proof sketch: Easy induction. -|
Lemma (truth-remove-irrelevant-lit-3). For assignment A and clause C,
if A |= C \ {L} then A |= C.
Proof sketch: Easy induction. -|
Lemma (subsetp-preserves-assignment). If A is an assignment, B has
unique literals, and B \subset A, then B is an assignment. -|
Now to prove Claim 2, let D \in F. Then since A |= F, A |= D.
Case 2.1: -L is not in D. Then A \ {-L} |= D by Lemma
truth-remove-irrelevant-lit-1, so by truth-monotone, B |= D and we are
done.
Now let D' = D \ {-L]; thus D = {-L} U D'.
Case 2.2: -L is in D and A |= D'. Then by
truth-remove-irrelevant-lit-2, B |= D'; then B |= D by
truth-remove-irrelevant-lit-3, and we are done.
Case 2.3: -L is in D and A does not satisfy D'.
Let RAT-A = A U N(D') and RAT-NC = U(F,N(C)) U N(D'). Technically:
RAT-A = rat-assignment(A,-L,D) and
RAT-NC = rat-assignment(U(F,N(C)),-L,D).
Lemma (clause-or-assignment-p-rat-assignment). Suppose that A is an
assignment and C is a clause. Then if rat-assignment(A,lit,C) is not t,
it is an assignment.
Lemma (rat-assignment-not-t). Suppose that A is an assignment, C is a
clause, and rat-assignment(A,lit,C) = t. Then A |= remove-lit(lit,C).
Claim 2.3.1. RAT-A is an assignment, by Lemmas
clause-or-assignment-p-rat-assignment.
CLaim 2.3.2. U(F,A) = A since A |= F, by unit-propagation-identity.
Claim 2.3.3. U(F,N(C)) \subset A.
Proof sketch. Note that since U(F,A) = A by Claim 2.3.2; so U(F,A) is
an assignment. Therefore since by Claim 1, N(C) \subset A, we have
U(F,N(C)) \subset U(F,A) by unit-propagation-monotonicity. We are
done by again using U(F,A) = A.
Lemma (rat-assignment-monotone). If A1 \ subset A2 and
rat-assignment(A2,-L,D) is not t, then rat-assignment(A1,-L,D) \subset
rat-assignment(A2,-L,D).
Claim 2.3.4. RAT-NC is an assignment and \subset RAT-A, by Claims
2.3.1 and 2.3.3 and rat-assignment-monotone.
Claim 2.3.5. Now U(F',RAT-NC) = t by the RAT check for D. Hence by
lemma unit-propagation-maybe-shrink-formula, U(F,RAT-NC) = t.
Claim 2.3.6. U(F,RAT-A) = t, by Claims 2.3.1, 2.3.4, and 2.3.5,
together with unit-propagation-t-monotone.
Claim 2.3.7. A |= N(RAT-A).
Proof sketch. This is clear from Claim 2.3.6 and our assumption that
A |= F, by unit-propagation-correct and an obvious lemma
(negate-clause-or-assignment-self-inverts) that implies N(N(RAT-A)) =
RAT-A by Claim 2.3.1.
Lemma (negate-rat-assignment-key). Let RAT-A = rat-assignment(A,x,D)
where A is an assignment, D is a clause, and x is a literal in D.
Suppose A |= N(RAT-A). Then A |= remove-literal(x,D).
Proof sketch: easy induction on rat-assignment(A,x,D), where for the
base case we need it to be false that A |= N(A). For that we prove by
induction that if B \subset A then evaluate-clause(N(B),A) = nil, and
then appeal to reflexivity of \subset.
Claim 2.3.8. A |= D', by Claim 2.3.7 and Lemma
negate-rat-assignment-key.
Claim 2.3.9. Hence B |= D' by truth-remove-irrelevant-lit-2.
So B |= D by truth-remove-irrelevant-lit-3. -|
||#
(in-package "LRAT")
(include-book "lrat-checker")
(local
(encapsulate
()
(local (include-book "satisfiable-add-proof-clause-rup"))
(local (include-book "satisfiable-add-proof-clause-drat"))
(set-enforce-redundancy t)
(defthm satisfiable-add-proof-clause-rup
(mv-let (ncls ndel new-formula)
(verify-clause formula entry ncls ndel)
(declare (ignore ncls ndel))
(implies (and (proof-entry-p entry)
(formula-p formula)
(satisfiable formula)
(equal (unit-propagation formula
(access add-step entry :rup-indices)
(negate-clause-or-assignment
(access add-step entry :clause)))
t))
(satisfiable (add-proof-clause
(access add-step entry :index)
(access add-step entry :clause)
new-formula))))
:rule-classes nil)
(defthm satisfiable-add-proof-clause-drat
(mv-let (ncls ndel new-formula)
(verify-clause formula entry ncls ndel)
(declare (ignore ndel))
(implies (and ncls
(proof-entry-p entry)
(not (proof-entry-deletion-p entry))
(formula-p formula)
(satisfiable formula)
(not (equal (unit-propagation formula
(access add-step entry
:rup-indices)
(negate-clause-or-assignment
(access add-step entry
:clause)))
t))
(consp (access add-step entry :clause)))
(satisfiable (add-proof-clause
(access add-step entry :index)
(access add-step entry :clause)
new-formula))))
:rule-classes nil)))
(defthm satisfiable-add-proof-clause
(mv-let (ncls ndel new-formula)
(verify-clause formula entry ncls ndel)
(declare (ignore ndel))
(implies (and ncls ; success
(proof-entry-p entry)
(not (proof-entry-deletion-p entry))
(formula-p formula)
(satisfiable formula))
(satisfiable (add-proof-clause
(access add-step entry :index)
(access add-step entry :clause)
new-formula))))
:hints (("Goal"
:use (satisfiable-add-proof-clause-rup
satisfiable-add-proof-clause-drat)
:in-theory (union-theories '(verify-clause)
(theory 'minimal-theory)))))
|
