Using Z3 4.4.1 or master, the following input gives unknown:
(declare-fun x () Int)
(declare-fun y () Int)
(declare-fun z () Int)
(assert (and (> x 0) (< x 10)))
(assert (and (> y 0) (< y 10)))
(assert (and (> z 0) (< z 10)))
(assert (= (* z z) (+ (* x x) (* y y))))
(check-sat-using (or-else qfnra-nlsat smt))
If I switch the order of the qfnra-nlsat and smt tactics, then Z3 will return sat, which is the correct solution. The way it works right now, I'm basically forced to use qfnra-nlsat at the very end of my strategy. However, this is not optimal for me.
Is the above behaviour a bug? It seems like qfnra-nlsat leaves some unwanted changes to the model, when I would expect the model to be unchanged if qfnra-nlsat fails.
Try instead
(check-sat-using (and-then qfnra-nlsat smt))
This is subtle:
The qfnra-nlsat tactic fails to establish whether the formula is satisfiable because the formula uses integers (and the model qfnra-nlsat finds a solution over the reals). It therefore acts as a no-op and produces the same subgoal it got. The first tactic in the or-else list that does not throw an exception becomes the tactic to decide the outcome. So qfnra-nlsat decides that the goal could not be reduced.
By using and-then instead you are basically reducing the goal using first qfnra-nlsat. If it succeeds, it produces either no sub-goals or a sub-goal with the formula "false". The smt tactic knows to handle these cases, where the sat and unsat cases are handled very efficiently.
There are other alternatives. In particular, if you want to use or-else as part of some larger set of alternatives, the approach is:
(check-sat-using (or-else (and-then qfnra-nlsat fail) (and-then smt fail)))
It causes a failure to be signaled if qfnra-nlsat creates non-true and non-false subgoals. Then upon failure, the smt tactic is applied.
Related
Given this formula,
(p & (x < 0)) | (~p & (x > 0)).
How could I get these 2 "parametric" models in Z3:
{p=true, x<0}
{p=false, x>0}
When I submit this SMTLIB program to Z3,
(declare-const p Bool)
(declare-const x Int)
(assert (or (and p (< x 0)) (and (not p) (> x 0))))
(check-sat)
(get-model)
(assert (or (not p) (not (= x -1))))
(check-sat)
(get-model)
(exit)
it gives me concrete models instead (e.g. {p=true, x=-1}, {p=true, x=-2}, ...).
You can't.
SMT solvers do not produce non-concrete models; it's just not how they work. What you want is essentially some form of "simplification" in steroids, and while you can use an SMT solver to help you in simplifying expressions, you'll have to build a tool on top that understands the kind of simplifications you'd like to see. Bottom line: What you'd consider "simple" as a person, and what an automated SMT-solver sees as "simple" are usually quite different from each other; and given lack of normal forms over arbitrary theories, you cannot expect them to do a good job.
If these sorts of simplifications is what you're after, you might want to look at symbolic math packages, such as sympy, Mathematica, etc.
I was working with z3 with the following example.
f=Function('f',IntSort(),IntSort())
n=Int('n')
c=Int('c')
s=Solver()
s.add(c>=0)
s.add(f(0)==0)
s.add(ForAll([n],Implies(n>=0, f(n+1)==f(n)+10/(n-c))))
The last equation is inconsistent (since n=c would make it indeterminate). But, Z3 cannot detect this kind of inconsistencies. Is there any way in which Z3 can be made to detect it, or any other tool that can detect it?
As far as I can tell, your assertion that the last equation is inconsistent does not match the documentation of the SMT-LIB standard. The page Theories: Reals says:
Since in SMT-LIB logic all function symbols are interpreted
as total functions, terms of the form (/ t 0) are meaningful in
every instance of Reals. However, the declaration imposes no
constraints on their value. This means in particular that
for every instance theory T and
for every value v (as defined in the :values attribute) and
closed term t of sort Real,
there is a model of T that satisfies (= v (/ t 0)).
Similarly, the page Theories: Ints says:
See note in the Reals theory declaration about terms of the form
(/ t 0).
The same observation applies here to terms of the form (div t 0) and
(mod t 0).
Therefore, it stands to reason to believe that no SMT-LIB compliant tool would ever print unsat for the given formula.
Z3 does not check for division by zero because, as Patrick Trentin mentioned, the semantics of division by zero according to SMT-LIB are that it returns an unknown value.
You can manually ask Z3 to check for division by zero, to ensure that you never depend division by zero. (This is important, for example, if you are modeling a language where division by zero has a different semantics from SMT-LIB.)
For your example, this would look as follows:
(declare-fun f (Int) Int)
(declare-const c Int)
(assert (>= c 0))
(assert (= (f 0) 0))
; check for division by zero
(push)
(declare-const n Int)
(assert (>= n 0))
(assert (= (- n c) 0))
(check-sat) ; reports sat, meaning division by zero is possible
(get-model) ; an example model where division by zero would occur
(pop)
;; Supposing the check had passed (returned unsat) instead, we could
;; continue, safely knowing that division by zero could not happen in
;; the following.
(assert (forall ((n Int))
(=> (>= n 0)
(= (f (+ n 1))
(+ (f n) (/ 10 (- n c)))))))
Using smtlib I would like to make something like modulo using QF_UFNRA. This disables me from using mod, to_int, to_real an such things.
In the end I want to get the fractional part of z in the following code:
(set-logic QF_UFNRA)
(declare-fun z () Real)
(declare-fun z1 () Real)
(define-fun zval_1 ((x Real)) Real
x
)
(declare-fun zval (Real) Real)
(assert (= z 1.5));
(assert (=> (and (<= 0.0 z) (< z 1.0)) (= (zval z) (zval_1 z))))
(assert (=> (>= z 1.0) (= (zval z) (zval (- z 1.0)))))
(assert (= z1 (zval z)))
Of course, as I am asking this question here, implies, that it didn't work out.
Has anybody got an idea how to get the fractional part of z into z1 using logic QF_UFNRA?
This is a great question. Unfortunately, what you want to do is not possible in general if you restrict yourself to QF_UFNRA.
If you could encode such functionality, then you can decide arbitrary Diophantine equations. You would simply cast a given Diophantine equation over reals, compute the "fraction" of the real solution with this alleged method, and assert that the fraction is 0. Since reals are decidable, this would give you a decision procedure for Diophantine equations, accomplishing the impossible. (This is known as Hilbert's 10th problem.)
So, as innocent as the task looks, it is actually not doable. But that doesn't mean you cannot encode this with some extensions, and possibly have the solver successfully decide instances of it.
If you allow quantifiers and recursive functions
If you allow yourself quantifiers and recursive-functions, then you can write:
(set-logic UFNRA)
(define-fun-rec frac ((x Real)) Real (ite (< x 1) x (frac (- x 1))))
(declare-fun res () Real)
(assert (= (frac 1.5) res))
(check-sat)
(get-value (res))
To which z3 responds:
sat
((res (/ 1.0 2.0)))
Note that we used the UFNRA logic allowing quantification, which is required here implicitly due to the use of the define-fun-rec construct. (See the SMTLib manual for details.) This is essentially what you tried to encode in your question, but instead using the recursive-function-definition facilities instead of implicit encoding. There are several caveats in using recursive functions in SMTLib however: In particular, you can write functions that render your system inconsistent rather easily. See Section 4.2.3 of http://smtlib.cs.uiowa.edu/papers/smt-lib-reference-v2.5-draft.pdf for details.
If you can use QF_UFNIRA
If you move to QF_UFNIRA, i.e., allow mixing reals and integers, the encoding is easy:
(set-logic QF_UFNIRA)
(declare-fun z () Real)
(declare-fun zF () Real)
(declare-fun zI () Int)
(assert (= z (+ zF zI)))
(assert (<= 0 zF))
(assert (< zF 1))
(assert (= z 1.5))
(check-sat)
(get-value (zF zI))
z3 responds:
sat
((zF (/ 1.0 2.0))
(zI 1))
(You might have to be careful about the computation of zI when z < 0, but the idea is the same.)
Note that just because the encoding is easy doesn't mean z3 will always be able to answer the query successfully. Due to mixing of Real's and Integer's, the problem remains undecidable as discussed before. If you have other constraints on z, z3 might very well respond unknown to this encoding. In this particular case, it happens to be simple enough so z3 is able to find a model.
If you have sin and pi:
This is more of a thought experiment than a real alternative. If SMTLib allowed for sin and pi, then you can check whether sin (zI * pi) is 0, for a suitably constrained zI. Any satisfying model to this query would ensure that zI is integer. You can then use this value to extract the fractional part by subtracting zI from z.
But this is futile as SMTLib neither allows for sin nor pi. And for good reason: Decidability would be lost. Having said that, maybe some brave soul can design a logic that supported sin, pi, etc., and successfully answered your query correctly, while returning unknown when the problem becomes too hard for the solver. This is already the case for nonlinear arithmetic and the QF_UFNIRA fragment: The solver may give up in general, but the heuristics it employs might solve problems of practical interest.
Restriction to Rationals
As a theoretical aside, it turns out that if you restrict yourself to rationals only (instead of actual reals) then you can indeed write a first-order formula to recognize integers. The encoding is not for the faint of heart, however: http://math.mit.edu/~poonen/papers/ae.pdf. Furthermore, since the encoding involves quantifiers, it's probably quite unlikely that SMT solvers will do well with a formulation based on this idea.
[Incidentally, I should extend thanks to my work colleagues; this question made for a great lunch-time conversation!]
I have a test.smt2 file:
(set-logic QF_IDL)
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(assert (or (< a 2) (< b 2 )) )
(check-sat)
(get-model)
(exit)
Is there anyway to tell Z3 to only output a=1 (or b=1)? Because when a is 1, b's value does not matter any more.
I executed z3 smt.relevancy=2 -smt2 test.smt2
(following How do I get Z3 to return minimal model?, although smt.relevancy seems has default value 2), but it still outputs:
sat
(model
(define-fun b () Int
2)
(define-fun a () Int
1)
)
Thank you!
The example given in the answer to the question referred to is slightly out of date. A Solver() will pick a suitable tactic to solve the problem, and it appears that it picks a different one now. We can still get that behavior by using a SimpleSolver() (at a possibly significant performance loss). Here's an updated example:
from z3 import *
x, y = Bools('x y')
s = SimpleSolver()
s.set(auto_config=False,relevancy=2)
s.add(Or(x, y))
print s.check()
print s.model()
Note that the (check-sat) command will not execute the same tactic as the SimpleSolver(); to get the same behavior when solving SMT2 files, we need to use the smt tactic, i.e., use (check-sat-using smt). In many cases it will be beneficial to additionally run the simplifier on the problem first which we can achieve by constructing a custom tactic, e.g., (check-sat-using (then simplify smt))
I am trying to prove an inductive fact in Z3, an SMT solver by Microsoft. I know that Z3 does not provide this functionality in general, as explained in the Z3 guide (section 8: Datatypes), but it looks like this is possible when we constrain the domain over which we want to prove the fact. Consider the following example:
(declare-fun p (Int) Bool)
(assert (p 0))
(assert (forall ((x Int))
(=>
(and (> x 0) (<= x 20))
(= (p (- x 1)) (p x) ))))
(assert (not (p 20)))
(check-sat)
The solver responds correctly with unsat, which means that (p 20) is valid. The problem is that when we relax this constraint any further (we replace 20 in the previous example by any integer greater than 20), the solver responds with unknown.
I find this strange because it does not take Z3 long to solve the original problem, but when we increase the upper limit by one it becomes suddenly impossible. I have tried to add a pattern to the quantifier as follows:
(declare-fun p (Int) Bool)
(assert (p 0))
(assert (forall ((x Int))
(! (=>
(and (> x 0) (<= x 40))
(= (p (- x 1)) (p x) )) :pattern ((<= x 40)))))
(assert (not (p 40)))
(check-sat)
Which seems to work better, but now the upper limit is 40. Does this mean that I can better not use Z3 to prove such facts, or am I formulating my problem incorrectly?
Z3 uses many heuristics to control quantifier instantiation. One one them is based on the "instantiation depth". Z3 tags every expression with a "depth" attribute. All user supplied assertions are tagged with depth 0. When a quantifier is instantiated, the depth of the new expressions is bumped. Z3 will not instantiate quantifiers using expressions tagged with a depth greater than a pre-defined threshold. In your problem, the threshold is reached: (p 40) is depth 0, (p 39) is depth 1, (p 38) is depth 2, etc.
To increase the threshold, you should use the option:
(set-option :qi-eager-threshold 100)
Here is the example with this option: http://rise4fun.com/Z3/ZdxO.
Of course, using this setting, Z3 will timeout, for example, for (p 110).
In the future, Z3 will have better support for "bounded quantification". In most cases, the best approach for handling this kind of quantifier is to expand it.
With the programmatic API, we can easily "instantiate" expressions before we send them to Z3.
Here is an example in Python (http://rise4fun.com/Z3Py/44lE):
p = Function('p', IntSort(), BoolSort())
s = Solver()
s.add(p(0))
s.add([ p(x+1) == p(x) for x in range(40)])
s.add(Not(p(40)))
print s.check()
Finally, in Z3, patterns containing arithmetic symbols are not very effective. The problem is that Z3 preprocess the formula before solving. Then, most patterns containing arithmetic symbols will never match. For more information on how to use patterns/triggers effectively, see this article. The author also provides a slide deck.