Copyright | (c) Levent Erkok |
---|---|
License | BSD3 |
Maintainer | erkokl@gmail.com |
Stability | experimental |
Safe Haskell | None |
Language | Haskell2010 |
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.
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 -> sAnd [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) .=> sNot (x .>= (0::SInteger))
Q.E.D.
This establishes that we indeed have an interpolant!