Deriving a proof of false #
linarith uses an untrusted oracle to produce a certificate of unsatisfiability.
It needs to do some proof reconstruction work to turn this into a proof term.
This file implements the reconstruction.
Main declarations #
The public facing declaration in this file is proveFalseByLinarith.
Typesafe conversion of n : ℕ to Q($α).
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Auxiliary functions for assembling proofs #
A typesafe version of mulExpr.
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mulExpr n e creates an Expr representing n*e.
When elaborated, the coefficient will be a native numeral of the same type as e.
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A type-safe analogue of addExprs.
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addExprs L creates an Expr representing the sum of the elements of L, associated left.
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If our goal is to add together two inequalities t1 R1 0 and t2 R2 0,
addIneq R1 R2 produces the strength of the inequality in the sum R,
along with the name of a lemma to apply in order to conclude t1 + t2 R 0.
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mkLTZeroProof coeffs pfs takes a list of proofs of the form táµ¢ Ráµ¢ 0,
paired with coefficients cáµ¢.
It produces a proof that ∑cᵢ * tᵢ R 0, where R is as strong as possible.
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mkNegOneLtZeroProof tp returns a proof of -1 < 0,
where the numerals are natively of type tp.
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addNegEqProofsIdx l inspects a list l of pairs (h, i) where h proves
táµ¢ Ráµ¢ 0 and i records the original index of the hypothesis. For each
equality proof t = 0 in the list, it appends a proof of -t = 0 with the
same index i. All other entries are preserved.
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proveEqZeroUsing tac e tries to use tac to construct a proof of e = 0.
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The main method #
proveFalseByLinarith is the main workhorse of linarith.
Given a list l of proofs of táµ¢ Ráµ¢ 0,
it tries to derive a contradiction from l and use this to produce a proof of False.
oracle : CertificateOracle is used to search for a certificate of unsatisfiability.
The returned certificate is a map m from hypothesis indices to natural number coefficients.
If our set of hypotheses has the form {táµ¢ Ráµ¢ 0},
then the elimination process should have guaranteed that
1.\ ∑ (m i)*tᵢ = 0,
with at least one i such that m i > 0 and Ráµ¢ is <.
We have also that
2.\ ∑ (m i)*tᵢ < 0,
since for each i, (m i)*tᵢ ≤ 0 and at least one is strictly negative.
So we conclude a contradiction 0 < 0.
It remains to produce proofs of (1) and (2). (1) is verified by calling the provided discharger
tactic, which is typically ring. We prove (2) by folding over the set of hypotheses.
transparency : TransparencyMode controls the transparency level with which atoms are identified.