How to test inconsistency in the user theory plug-in?

1
votes

I am implementing a user theory plug-in. I want to test inconsistency in the user theory to prune some models.

More specifically, in my theory, x is a variable of the user sort of set and f is the function returning the size of the set. I have two assertions: x = set1 OR x = set2 and f x > 2. Suppose the size of set1 is 1 and the size of set2 is 3.

In the search, Z3 goes with x = set1 first. So I can add another assertion f x = 1, which will be inconsistent in the INT part. I want to test the inconsistency so that I can negate the current assignment, let Z3 backtrack and try the other option.

My question is that how I can do that.

I tried 3 approaches:

(1) Add the assertion f x = 1 directly using Z3_theory_assert_axiom(). Then the search terminates immediately returning UNSAT.

(2) I tried to use Z3_check_assumptions() with f x = 1 as the assumption to check with current context. But Z3_check_assumptions() doesn't allow such a compound formula. So, it can not be the solution.

(3) I firstly push the context, add the assertion f x = 1 with Z3_assert_cnstr(), test the consistency with Z3_check_and_get_model(), and then pop the context just pushed. In the test, if it's not consistent, I get current assignment with Z3_get_context_assignment(ctx) and assert the negated assignment to trigger the backtrack. What I observe is that Z3 does find the inconsistency but the current assignment only contains assertions about the size part like (= 1 (f x)). Or in other word, the assertions about the user theory like (= x set1) and not (= x set2) is missing. So even I negate the current assignment, after the backtracking, Z3 still tries the x = set1 instead of the other option x = set2.

Where I was wrong? Thanks!

1
z3

1 Answers

2
votes

You should use the first option only Z3_theory_assert_axiom() from user theory plugins in the callbacks. You should also only assert tautologies, that is axioms that are true independent of the current assignments. So instead of asserting 

f x = 1

you should assert

x = set1 => f x = 1

(assuming I understood) because this formula is true regardless. You could also assert:

f set1 = 1

(assuming, again I understood), which would also close branches where y = set1 & f y > 1 for some other variable holds. More generally, you want to assert the strongest theory axiom to prune out as many related branches. But the theory axioms have to be true and not just hold within the local branch of the search.