When using z3 to solve SMT2 files and applying a tactic, (e.g., "(apply qfbv)"), how do I set an option for that tactic? For example, QFBV has an option "cache-all" that by default is set to false. How can I set it to true using SMT2 files? Or is this not possible using the SMT2 language?
You can use the combinator using-params. The combinator ! is a shorthand for using-params.
Here is a small example using the tactic simplify (Try it online at: http://rise4fun.com/Z3/JaZ).
(declare-const x Int)
(declare-const y Int)
(assert (= (+ x 1) (+ y 3)))
(apply simplify)
(echo ">>>> Using arith-lhs := True, and eq2ineq := True")
(apply (using-params simplify :arith-lhs true :eq2ineq true))
;; ! is a shorthand for using-params
(apply (! simplify :arith-lhs true :eq2ineq true))
You can use using-params combinator. Typing (help-tactic) in Z3 SMT online gives me this fragment:
- (using-params <tactic> <attribute>*) executes the given tactic using the given attributes, where <attribute> ::= <keyword> <value>. ! is a
syntax sugar for using-params.
Here is a type-checked example (not sure that it makes sense):
(declare-const x (_ BitVec 16))
(declare-const y (_ BitVec 16))
(assert (= (bvor x y) (_ bv13 16)))
(assert (bvslt x y))
(apply (using-params qfbv :cache-all true))
Related
I am using Z3 to solve my horn clauses. In the body of Horn clauses uninterpreted predicates should be positive. However, I need negation of some of uninterpreted predicates.
I have seen some examples in which negation works fine. For instance Z3 would return sat for the following example:
(set-logic HORN)
(declare-fun inv (Int) Bool)
(assert (inv 0))
(assert (forall ((k Int)) (or (> k 10) (not (inv k)) (inv (+ k 1)))))
(check-sat)
But my example looks like the following for which Z3 returns unknown.
(set-logic HORN)
(declare-fun inv (Int ) Bool)
(declare-fun s ( Int ) Bool)
(assert (forall ((k Int) (pc Int))(=>(and (= pc 1)(= k 0)) (inv k ))))
(assert (forall ((k Int)(k_p Int)(pc Int)(pc_p Int))
(=>(and (inv k )(= pc 1)(= pc_p 2)(= k_p (+ k 1))(not (s pc ))(s pc_p ))
(inv k_p ))))
(check-sat)
I wonder if there is a way to rewrite my clauses to Horn clause fragment of Z3.
Your clauses are not in the Horn fragment because the predicate s is used with both polarities in the last assertion. So there are two occurrences of a predicate with positive polarity (both (s pc) and (inv k_p) are positive polarity).
A basic method to avoid polarity issues is to introduce an extra argument to s of type Bool. Consequently, you would also have to say what is the specification of s using Horn clauses so it all makes sense. The typical scenario is that s encodes the behavior of a recursive procedure and the extra Boolean argument to s would be the return value of the procedure s. Of course this encoding doesn't ensure that s is total or functional.
There is a second approach, which is to add an extra argument to "inv", where you let 's' be an array. Then the occurrences (not (s pc)) becomes (not (select s pc)), etc.
It all depends on the intent of your encoding for what makes sense.
The following code
(declare-fun f (Int) Real)
(declare-fun g (Int) Real)
(declare-const x Int)
(declare-const y Int)
(assert (= (* (f x) (g y)) 1.0))
(check-sat)
(get-model)
returns "unknown", even though there is an obvious solution. Eliminating arguments to f and g (effectively making them constants?) results in "sat" with expected assignments. I guess, my question is: what is special about arithmetic with uninterpreted functions?
BTW, replacing * with + also results in "sat", so the issue is not about uninterpreted functions, per se, but about how they are combined.
Additional thoughts
Making the domain (very) finite does not help, e.g.,
(declare-fun f (Bool) Real)
(declare-fun g (Bool) Real)
(declare-const x Bool)
(declare-const y Bool)
(assert (= (* (f x) (g y)) 1.0))
(check-sat)
(get-model)
returns "unknown". This is odd given that f:Bool->Real is essentially just two variables f(False) and f(True) (of course, the solver has to recognize this).
Inability to handle non-linear arithmetic over real-valued uninterpreted functions is a very severe limitation because arrays are implemented as uninterpreted functions. So, for example,
(declare-const a (Array Int Real))
(assert (= (* (select a 1) (select a 1)) 1))
(check-sat)
(get-model)
returns "unknown". In other words, any non-linear algebraic expression on real array elements involving multiplication is unsolvable:'(
It's the non-liearity introduced by the multiplication that makes the problem hard to solve, not the uninterpreted functions. In fact, you can ask the solvers why by using the get-info command:
(set-logic QF_UFNIRA)
(declare-fun f (Int) Real)
(declare-fun g (Int) Real)
(declare-const x Int)
(declare-const y Int)
(assert (= (* (f x) (g y)) 1.0))
(check-sat)
(get-info :reason-unknown)
Here're some responses I got from various solvers:
Z3:
(:reason-unknown smt tactic failed to show goal to be sat/unsat (incomplete (theory arithmetic)))
CVC4:
(:reason-unknown incomplete)
MathSAT:
(error "sat, but with non-linear terms")
So, as the solvers themselves improve and start handling more nonliear arithmetic, you might get models eventually. But, in general, nonlinear problems will always be problematic as they are undecidable when integers are involved. (Since you can code Diophantine equations using non-linear terms.)
See also How does Z3 handle non-linear integer arithmetic? for a relevant discussion from 2012.
Removing uninterpreted functions
Even if you get rid of the uninterpreted functions, you'd still be in the unknown land, so long as you mix Int and Real types and non-linear terms:
(set-logic QF_NIRA)
(declare-const f Real)
(declare-const g Real)
(declare-const x Int)
(declare-const y Int)
(assert (= (+ (* f g) (to_real (+ x y))) 1.0))
(check-sat)
(get-info :reason-unknown)
Z3 says:
(:reason-unknown "smt tactic failed to show goal to be sat/unsat (incomplete (theory arithmetic))")
So, the issue would arise with mixed types and non-linear terms.
I'm trying to use Z3 to solve arithmetic equations using bitvector arithmetic. I was wondering if there is a way to handle also Real numbers. For example if I can specify a constant different from #x1 and use real number instead.
(set-option :pp.bv-literals false)
(declare-const x (_ BitVec 4))
(declare-const y (_ BitVec 4))
(assert (= (bvadd x y) #x1))
(check-sat)
(get-model)
Yes, both SMT-Lib (and Z3) fully support real numbers: http://smtlib.cs.uiowa.edu/theories-Reals.shtml
You can simply write your example as follows:
(declare-const x Real)
(declare-const y Real)
(assert (= (+ x y) 1))
(check-sat)
(get-model)
You can also mix/match Int/Real/Bitvector, so long as everything is properly typed. Here's an example showing how to use Ints and Reals together:
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(declare-const d Real)
(declare-const e Real)
(assert (> e (+ (to_real (+ a b)) 2.0)))
(assert (= d (+ (to_real c) 0.5)))
(assert (> a b))
(check-sat)
(get-model)
However, note that conversion from bit-vectors to integers is usually uninterpreted. See here for a discussion: Z3 int2bv operation
Given the following simplified quantifiers, with the Z3 options set according to those generated by Boogie (full details below), I get "unknown" as a result:
(declare-fun F (Int) Bool)
(declare-fun G (Int) Bool)
(assert (forall ((x Int)) (! (and
(F x) (G x))
:pattern ((F x))
)))
(assert (not (forall ((x Int)) (! (and
(G x) (F x))
:pattern ((F x))
))))
(check-sat)
My understanding for what (I think) Z3 would do with this problem, is skolemise the existential (not forall), which would yield ground instances of both F and G. Given these in the e-graph, we should be able to instantiate the other quantifier, and get unsat. I can see that Z3 probably has to case-split to do this, but I would expect this case-splitting to take place after removing the quantifier and populating the e-graph.
Instead, the first quantifier doesn't get instantiated in the above problem. I've made a number of observations:
Swapping the order of the (F x) and (G x) terms in the first quantifier results in "unsat" without any quantifier instantiations (I suppose some simplification spots the similarity between the two quantified assertions?).
Swapping the order of the (G x) and (F x) terms in the second quantifier (as well as those in the first) results in "unsat" with a single quantifier instantiation (which is the behaviour I'd expect in general).
Changing the smt.case_split option affects the behaviour. Set to 3 (as chosen by Boogie) or 5, we get "unknown". Set to 0,1,2 or 4, I get "unsat".
It would be great to understand the scenarios above, and why (in the failing cases) these terms don't always make it to the e-graph after skolemisation. I'm not sure what the effects of changing the case_split option are in general. At the moment, I don't think Boogie allows that to be changed (and overrides any choice made on the command-line). But I have the feeling that the e-graph should get the information in all cases, ideally.
Here's the full file (removing most of the options set doesn't seem to make a difference to the failing cases, except for the smt.case_split one):
(set-option :print-success false)
(set-info :smt-lib-version 2.0)
(set-option :AUTO_CONFIG false)
;(set-option :MODEL.V2 true)
(set-option :smt.PHASE_SELECTION 0)
(set-option :smt.RESTART_STRATEGY 0)
(set-option :smt.RESTART_FACTOR |1.5|)
(set-option :smt.ARITH.RANDOM_INITIAL_VALUE true)
(set-option :smt.DELAY_UNITS true)
(set-option :NNF.SK_HACK true)
(set-option :smt.MBQI false)
(set-option :smt.QI.EAGER_THRESHOLD 100)
(set-option :smt.QI.COST |"(+ weight generation)"|)
(set-option :TYPE_CHECK true)
(set-option :smt.BV.REFLECT true)
(set-option :TIMEOUT 0)
(set-option :smt.QI.PROFILE true)
(set-option :smt.CASE_SPLIT 3)
; done setting options
(declare-fun F (Int) Bool)
(declare-fun G (Int) Bool)
(assert (forall ((x Int)) (! (and
(F x) (G x))
:pattern ((F x))
)))
(assert (not (forall ((x Int)) (! (and
(G x) (F x))
:pattern ((F x))
))))
(check-sat)
This is settled by the answer from https://stackoverflow.com/users/1096362/nikolaj-bjorner to this question:
Surprising behaviour when trying to prove a forall
The translation of the proof obligation into a disjunction, followed by corresponding relevancy of the clauses, explains why Z3 doesn't see both conjuncts as potential triggers.
Basically, I want to ask Z3 to give me an arbitrary integer whose value is greater than 10. So I write the following statements:
(declare-const x (Int))
(assert (forall ((i Int)) (> i 10)))
(check-sat)
(get-value(x))
How can I apply this quantifier to my model? I know you can write (assert (> x 10)) to achieve this. But I mean I want a quantifier in my model so every time I declare an integer constant whose value is guaranteed to be over 10. So I don't have to insert statement (assert (> x 10)) for every integer constant that I declared.
When you use (assert (forall ((i Int)) (> i 10))), i is a bounded variable and the quantified formula is equivalent to a truth value, which is false in this case.
I think you want to define a macro using quantifiers:
(declare-fun greaterThan10 (Int) Bool)
(assert (forall ((i Int)) (= (greaterThan10 i) (> i 10))))
And you can use them to avoid code repetition:
(declare-const x (Int))
(declare-const y (Int))
(assert (greaterThan10 x))
(assert (greaterThan10 y))
(check-sat)
It is essentially the way to define macros using uninterpreted functions when you're working with Z3 API. Note that you have to set (set-option :macro-finder true) in order that Z3 replaces universal quantifiers with bodies of those functions.
However, if you're working with the textual interface, the macro define-fun in SMT-LIB v2 is an easier way to do what you want:
(define-fun greaterThan10 ((i Int)) Bool
(> i 10))