Consider these two formulae:
Exists y. Forall x. (y>x)
, which is unsat
.Forall x. Exists y. (y>x)
, which is sat
.Note that we cannot find “models” for the sat
formula, e.g., using Z3 it outputs Z3Exception: model is not available
for the following code:
phi = ForAll([x],Exists([y], lit))
s_def = Solver()
s_def.add(phi)
print(s_def.model())
Also, quantifier elimination does not output an elimination, but a satisfiability result
[[]]
(i.e., True
):
x, y = Reals('x, y')
t = Tactic("qe")
lit = (y>x)
ae = Goal()
ae.add(ForAll([x],Exists([y], lit)))
ae_qe = t(ae)
print(ae_qe)
I think this happens because the value of y
fully depends on x
(e.g., if x
is 5
then y
can be 6
). Thus, I have some questions:
True
or False
?y
that holds the constraint (y>x)
; e.g. f(x)=x+1
. In other words, does it make sense that the quantifier elimination of a Forall x. Exists y. Phi(x,y)
like the example would be Forall x. Phi(x,f(x))
?You get a model-not-available, because you didn't call check
. A model is only available after a call to check
. Try this:
from z3 import *
x, y = Ints('x y')
phi = ForAll([x], Exists([y], y > x))
s = Solver()
s.add(phi)
print(s.check())
print(s.model())
This prints:
sat
[]
Now, you're correct that you won't get a meaningful/helpful model when you have a universal-quantifier; because the model will depend on the choice of x
in each case. (You only get values for top-level existentials.) This is why the model is printed as the empty-list.
Side Note In certain cases, you can use the skolemization trick (https://en.wikipedia.org/wiki/Skolem_normal_form) to get rid of the nested existential and get a mapping function, but this doesn't always work with SMT solvers as they can only build "finite" skolem functions. (For your example, there isn't such a finite function; i.e., a function that can be written as a case-analysis on the inputs, or an if-then-else chain.) For your specific questions:
Yes; value of y
depends on x
and thus cannot be displayed as is. Sometimes skolemization will let you work around this, but SMT solvers can only build finite skolem-functions.
Model of a universally quantified formula is more or less meaningless. It's true for all values of the universally quantified variables. Only top-level existentials are meaningful in a model.
Quantifier elimination did work here; it got rid of the whole formula. Note that QE preserves satisfiability; so it has done its job.
This is the skolem function. As you noted, this skolem function is not finite (can't be written as a chain of finite if-then-elses on concrete values), and thus existing SMT solvers cannot find/print them for you. If you try, you'll see that the solver simply loops:
from z3 import *
x = Int('x')
skolem = Function('skolem', IntSort(), IntSort())
phi = ForAll([x], skolem(x) > x)
s = Solver()
s.add(phi)
print(s)
print(s.check())
print(s.model())
The above never terminates, unfortunately. This sort of problem is simply too complicated for the push-button approach of SMT solving.