sbv-7.13: SMT Based Verification: Symbolic Haskell theorem prover using SMT solving.

Copyright(c) Levent Erkok
LicenseBSD3
Maintainererkokl@gmail.com
Stabilityexperimental
Safe HaskellNone
LanguageHaskell2010

Documentation.SBV.Examples.Queries.Interpolants

Description

Demonstrates extraction of interpolants via queries.

N.B. As of Z3 version 4.8.0; Z3 no longer supports interpolants. You need to use the MathSAT backend for this example to work.

Synopsis

Documentation

example :: Symbolic String Source #

Compute the interpolant for the following sets of formulas:

{x - 3y >= -1, x + y >= 0}

AND

{z - 2x >= 3, 2z <= 1}

where the variables are integers. Note that these sets of formulas are themselves satisfiable, but not taken all together. The pair (x, y) = (0, 0) satisfies the first set. The pair (x, z) = (-2, 0) satisfies the second. However, there's no triple (x, y, z) that satisfies all these four formulas together. We can use SBV to check this fact:

>>> sat $ \x y z -> bAnd [x - 3*y .>= -1, x + y .>= 0, z - 2*x .>= 3, 2 * z .<= (1::SInteger)]
Unsatisfiable

An interpolant for these sets would only talk about the variable x that is common to both. We have:

>>> runSMTWith mathSAT example
"(<= 0 s0)"

Notice that we get a string back, not a term; so there's some back-translation we need to do. We know that s0 is x through our translation mechanism, so the interpolant is saying that x >= 0 is entailed by the first set of formulas, and is inconsistent with the second. Let's use SBV to indeed show that this is the case:

>>> prove $ \x y -> (x - 3*y .>= -1 &&& x + y .>= 0) ==> (x .>= (0::SInteger))
Q.E.D.

And:

>>> prove $ \x z -> (z - 2*x .>= 3 &&& 2 * z .<= 1) ==> bnot (x .>= (0::SInteger))
Q.E.D.

This establishes that we indeed have an interpolant!