I have the following code in Z3py
import z3
x = z3.BitVec('x', 32)
a = z3.BitVec('a', 32)
solver=z3.Solver()
solver.set(unsat_core=True)
d=z3.Bool('d')
solver.assert_and_track(x!=a,d)
solver.add(x==a)
print solver.check(d)
print solver.check(z3.Not(d))
I would expect it to print unsat and sat. However it prints always unsat. Should the call solver.check(z3.Not(d)) effectively disable the assertion x!=a?
assert_and_track simply associates the answer-literal d with the formula x!=a in your example. Since the context contains x==a and x!=a at the same time, it is unsat. No assumptions will make it satisfiable.
But I do agree that this is very confusing. I think the issue goes back to the fact that assert_and_track is not really intended for what you are trying to do, but merely to make unsat-core-extraction possible. See the discussion here: https://stackoverflow.com/a/14565535/936310
To really achieve what you want, simply assert the implication yourself. That is, use:
solver.add(Implies(d, x!=a))
For instance, the following script:
from z3 import *
x = z3.BitVec('x', 32)
a = z3.BitVec('a', 32)
solver=z3.Solver()
solver.set(unsat_core=True)
d=z3.Bool('d')
solver.add(Implies(d, x!=a))
solver.add(x==a)
print solver.check(d)
print solver.unsat_core()
print solver.check(z3.Not(d))
print solver.model()
produces:
unsat
[d]
sat
[a = 0, d = False, x = 0]
which is what you are trying to achieve, I believe.
Related
Given the following simple example:
s = Solver()
Z = IntSort()
a = Const('a', Z)
s.add(a >= 0)
s.add(a < 10)
print(s.check(a > 5)) # sat
Up until the last line, a has an implied range of 0 <= a < 10 - for which a > 5 does not satisfy. However, .check() tells z3 that "this is true" - which is not what I'm after.
Is there a way to ask z3 to test if a > 5 given the existing set of constraints, and have z3 interpret this as "make sure this is true given everything else we know about a"?
EDIT: Okay I think I figured it out. Since z3 tries to find a model that satisfies all constraints, I should instead check for the inverse to see if it finds a solution - in which case, my check would not hold.
In this case, my "test" function would become:
def test(s, expr):
return s.check(Not(expr)) == unsat
thus...
s = Solver()
Z = IntSort()
a = Const('a', Z)
s.add(a >= 0)
s.add(a < 10)
print(s.check(a > 5)) # sat
print(s.check(Not(a > 5)) == unsat) # false, test failed
s.add(a > 8)
print(s.check(Not(a > 5)) == unsat) # true, test passed
Is that correct?
What you are doing is essentially correct, though not idiomatic. When you issue s.check(formula), you're telling z3 to show the satisfiability of all the other constraints you add'ed, along with formula; i.e., their conjunction. So, the way you set it up, it gives you the correct result.
Here's some more detail. Since what you want to prove looks like:
Implies(And(lower, upper), required)
the typical way of doing this would be to assert its negation, and then check if it is satisfiable. That is, you'd check if:
Not(Implies(And(lower, upper), required))
is satisfiable. Recall that Implies(a, b) is equivalent to Or(Not(a), b), so a little bit of Boolean logic transforms the above to:
And(And(lower, upper), Not(required))
What you are doing is to add the first conjunct above (i.e., And(lower, upper)) to the solver itself, and then use the second conjunct Not(required) as an argument to check to see if you get unsat. If you do, then you get a "proof" that required always holds. Otherwise you get a counterexample.
I should add, however, that this is not idiomatic usage of z3py and calls to check with a formula. The latter is usually used with a list of assumptions, and figuring out which subset of them was unsatisfiable with a call to get-unsat-assumptions. These applications typically come from model-checking problems. For details, see Section 4.2.5 of https://smtlib.cs.uiowa.edu/papers/smt-lib-reference-v2.6-r2021-05-12.pdf which discusses the semantics of check-sat-assuming, which is the SMTLib equivalent of calling check with extra assumptions.
To solve your problem in more idiomatic z3, one would instead write something like:
from z3 import *
s = Solver()
a = Int('a')
lower = a >= 0
upper = a < 10
required = a > 5
formula = Implies(And(lower, upper), required)
# assert the negation
s.add(Not(formula))
r = s.check()
if r == sat:
print("Not valid, counterexample:")
print(s.model())
elif r == unsat:
print("Valid!")
else:
print("Solver said:", r)
This prints:
Not valid, counterexample:
[a = 0]
And if you change to lower = a >= 8, it'll print:
Valid!
Bottom line, what you're doing is correct but not very idiomatic usage for check-sat-assuming. A simpler way is to simply assert the negation of the implication that your lower/upper bounds imply the required inequality and check if the result is unsat instead.
Suppose I have a set of Z3 expressions:
exprs = [A, B, C, D, E, F]
I want to check whether any of them are equivalent and, if so, determine which. The most obvious way is just an N×N comparison (assume exprs is composed of some arbitrarily-complicated boolean expressions instead of the simple numbers in the example):
from z3 import *
exprs = [IntVal(1), IntVal(2), IntVal(3), IntVal(4), IntVal(3)]
for i in range(len(exprs) - 1):
for j in range(i+1, len(exprs)):
s = Solver()
s.add(exprs[i] != exprs[j])
if unsat == s.check():
quit(f'{(i, j)} are equivalent')
Is this the most efficient method, or is there some way of quantifying over a set of arbitrary expressions? It would also be acceptable for this to be a two-step process where I first learn whether any of the expressions are equivalent, and then do a longer check to see which specific expressions are equivalent.
As with anything performance related, the answer is "it depends." Before delving into options, though, note that z3 supports Distinct, which can check whether any number of expressions are all different: https://z3prover.github.io/api/html/namespacez3py.html#a9eae89dd394c71948e36b5b01a7f3cd0
Though of course, you've a more complicated query here. I think the following two algorithms are your options:
Explicit pairwise checks
Depending on your constraints, the simplest thing to do might be to call the solver multiple times, as you alluded to. To start with, use Distinct and make a call to see if its negation is satisfiable. (i.e., check if some of these expressions can be made equal.) If the answer comes unsat, you know you can't make any equal. Otherwise, go with your loop as before till you hit the pair that can be made equal to each other.
Doing multiple checks together
You can also solve your problem using a modified algorithm, though with more complicated constraints, and hopefully faster.
To do so, create Nx(N-1)/2 booleans, one for each pair, which is equal to that pair not being equivalent. To illustrate, let's say you have the expressions A, B, and C. Create:
X0 = A != B
X1 = A != C
X2 = B != C
Now loop:
Ask if X0 || X1 || X2 is satisfiable.
If the solver comes back unsat, then all of A, B, and C are equivalent. You're done.
If the solver comes back sat, then at least one of the disjuncts X0, X1 or X2 is true. Use the model the solver gives you to determine which ones are false, and continue with those until you get unsat.
Here's a simple concrete example. Let's say the expressions are {1, 1, 2}:
Ask if 1 != 1 || 1 != 2 || 1 != 2 is sat.
It'll be sat. In the model, you'll have at least one of these disjuncts true, and it won't be the first one! In this case the last two. Drop them from your list, leaving you with 1 != 1.
Ask again if 1 != 1 is satisfiable. The answer will be unsat and you're done.
In the worst case you'll make Nx(N-1)/2 calls to the solver, if it happens that none of them can be made equivalent with you eliminating one at a time. This is where the first call to Not (Distinct(A, B, C, ...)) is important; i.e., you will start knowing that some pair is equivalent; hopefully iterating faster.
Summary
My initial hunch is that the second algorithm above will be more performant; though it really depends on what your expressions really look like. I suggest some experimentation to find out what works the best in your particular case.
A Python solution
Here's the algorithm coded:
from z3 import *
exprs = [IntVal(i) for i in [1, 2, 3, 4, 3, 2, 10, 10, 1]]
s = Solver()
bools = []
for i in range(len(exprs) - 1):
for j in range(i+1, len(exprs)):
b = Bool(f'eq_{i}_{j}')
bools.append(b)
s.add(b == (exprs[i] != exprs[j]))
# First check if they're all distinct
s.push()
s.add(Not(Distinct(*exprs)))
if(s.check()== unsat):
quit("They're all distinct")
s.pop()
while True:
# Be defensive, bools should not ever become empty here.
if not bools:
quit("This shouldn't have happened! Something is wrong.")
if s.check(Or(*bools)) == unsat:
print("Equivalent expressions:")
for b in bools:
print(f' {b}')
quit('Done')
else:
# Use the model to keep bools that are false:
m = s.model()
bools = [b for b in bools if not(m.evaluate(b, model_completion=True))]
This prints:
Equivalent expressions:
eq_0_8
eq_1_5
eq_2_4
eq_6_7
Done
which looks correct to me! Note that this should work correctly even if you have 3 (or more) items that are equivalent; of course you'll see the output one-pair at a time. So, some post-processing might be needed to clean that up, depending on the needs of the upstream algorithm.
Note that I only tested this for a few test values; there might be corner case gotchas. Please do a more thorough test and report if there're any bugs!
I want to find the range of valid values that a variable can have, given some constraints. Eg,
x = Int('x')
s = Solver()
s.add(x >= 1)
s.add(x < 5+2)
Is there some way that I can get z3 to print 1..6 for this variable?
I tried using the following, but range() applies only to declarations.
print("x.range():", x.range()) # this does not work
Note: 1. This question seems to ask the same, but I did not understand its answers, and I am looking for python answer.
in reply to #Malte: I am not looking for all the answers, I just want to simplify multiple constraints in to a valid range. If constraints on both sides of the variable cannot be merged, then at least only on one side as is mentioned in above mentioned question.
This question comes up occasionally, and the answer isn't very trivial, unfortunately. It really depends on what your constraints are and exactly what you are trying to do. See:
Is it possible to get a legit range info when using a SMT constraint with Z3
And
(Sub)optimal way to get a legit range info when using a SMT constraint with Z3
Essentially, the problem is too difficult (and I'd say not even well defined) if you have multiple variables. If you have exactly one variable, you can use the optimizer to some extent, assuming the variable is indeed bounded. In case you have multiple variables, one idea might be to fix all but one to satisfying constants, and compute the range of that last variable based on the constant assignment to the others. But again, it depends on what you're really trying to achieve.
Please take a look at the above two answers and see if it helps you. If not, please show us what you tried: Stack-overflow works the best when you post some code and see how it can be improved/fixed.
As a SAT/SMT solver, Z3 "only" needs to find a single model (satisfying assignment) to show that a formula is satisfiable. Finding all models is therefore not directly supported.
The question comes up regularly, though, and the solution is to repeatedly find and then block (assume in negated form) models until no further model can be found. For example, for your snippet of code:
x = Int('x')
s = Solver()
s.add(x >= 1)
s.add(x < 5+2)
result = s.check()
while result == sat:
m = s.model()
print("Model: ", m)
v_x = m.eval(x, model_completion=True)
s.add(x != v_x)
result = s.check()
print(result, "--> no further models")
Executing the script yields the solution you asked for, albeit in a less concise form:
Model: [x = 1]
Model: [x = 2]
Model: [x = 3]
Model: [x = 4]
Model: [x = 5]
Model: [x = 6]
unsat --> no further models
In general,
you would have iterate over all variables (here: just x)
model completion is necessary for variables whose value doesn't affect satisfiability; since any value will do, they won't be explicit in the model
Related questions whose answers provide additional details:
(Z3Py) checking all solutions for equation
Why Z3Py does not provide all possible solutions
Getting all solutions of a boolean expression in Z3Py never ends
I have tried to implement Floor and Ceiling Function as defined in the following link
https://math.stackexchange.com/questions/3619044/floor-or-ceiling-function-encoding-in-first-order-logic/3619320#3619320
But Z3 query returning counterexample.
Floor Function
_X=Real('_X')
_Y=Int('_Y')
_W=Int('_W')
_n=Int('_n')
_Floor=Function('_Floor',RealSort(),IntSort())
..
_s.add(_X>=0)
_s.add(_Y>=0)
_s.add(Implies(_Floor(_X)==_Y,And(Or(_Y==_X,_Y<_X),ForAll(_W,Implies(And(_W>=0,_W<_X),And(_W ==_Y,_W<_Y))))))
_s.add(Implies(And(Or(_Y==_X,_Y<_X),ForAll(_W,Implies(And(_W>=0,_W<_X),And(_W==_Y,_W<_Y))),_Floor(_X)==_Y))
_s.add(Not(_Floor(0.5)==0))
Expected Result - Unsat
Actual Result - Sat
Ceiling Function
_X=Real('_X')
_Y=Int('_Y')
_W=Int('_W')
_Ceiling=Function('_Ceiling',RealSort(),IntSort())
..
..
_s.add(_X>=0)
_s.add(_Y>=0)
_s.add(Implies(_Ceiling(_X)==_Y,And(Or(_Y==_X,_Y<_X),ForAll(_W,Implies(And(_W>=0,_W<_X),And(_W ==_Y,_Y<_W))))))
_s.add(Implies(And(Or(_Y==_X,_Y<_X),ForAll(_W,Implies(And(_W>=0,_W<_X),And(_W==_Y,_Y<_W)))),_Ceiling(_X)==_Y))
_s.add(Not(_Ceilng(0.5)==1))
Expected Result - Unsat
Actual Result - Sat
[Your encoding doesn't load to z3, it gives a syntax error even after eliminating the '..', as your call to Implies needs an extra argument. But I'll ignore all that.]
The short answer is, you can't really do this sort of thing in an SMT-Solver. If you could, then you can solve arbitrary Diophantine equations. Simply cast it in terms of Reals, solve it (there is a decision procedure for Reals), and then add the extra constraint that the result is an integer by saying Floor(solution) = solution. So, by this argument, you can see that modeling such functions will be beyond the capabilities of an SMT solver.
See this answer for details: Get fractional part of real in QF_UFNRA
Having said that, this does not mean you cannot code this up in Z3. It just means that it will be more or less useless. Here's how I would go about it:
from z3 import *
s = Solver()
Floor = Function('Floor',RealSort(),IntSort())
r = Real('R')
f = Int('f')
s.add(ForAll([r, f], Implies(And(f <= r, r < f+1), Floor(r) == f)))
Now, if I do this:
s.add(Not(Floor(0.5) == 0))
print(s.check())
you'll get unsat, which is correct. If you do this instead:
s.add(Not(Floor(0.5) == 1))
print(s.check())
you'll see that z3 simply loops forever. To make this usefull, you'd want the following to work as well:
test = Real('test')
s.add(test == 2.4)
result = Int('result')
s.add(Floor(test) == result)
print(s.check())
but again, you'll see that z3 simply loops forever.
So, bottom line: Yes, you can model such constructs, and z3 will correctly answer the simplest of queries. But with anything interesting, it'll simply loop forever. (Essentially whenever you'd expect sat and most of the unsat scenarios unless they can be constant-folded away, I'd expect z3 to simply loop.) And there's a very good reason for that, as I mentioned: Such theories are just not decidable and fall well out of the range of what an SMT solver can do.
If you are interested in modeling such functions, your best bet is to use a more traditional theorem prover, like Isabelle, Coq, ACL2, HOL, HOL-Light, amongst others. They are much more suited for working on these sorts of problems. And also, give a read to Get fractional part of real in QF_UFNRA as it goes into some of the other details of how you can go about modeling such functions using non-linear real arithmetic.
Using Z3Py, I tried to build a program which Z3 would decide means that the sort Human is empty.
from z3 import *
from z3_helper import Z3Helper
Human = DeclareSort("Human")
is_mortal = Function("is_mortal", Human, BoolSort())
h = Const('h', Human)
s = Solver()
s.add([
ForAll([h], And(is_mortal(h), Not(is_mortal(h))))
])
print s.check()
s.model()
But instead of returning a model where Human is empty, it returns unsat. Why is this?
If I remove the "all men are mortal" axiom, it returns an empty set as the model.
Is the problem that the existence of const h means that the existence of at least one Human is required?
SMT-LIB and Z3 take the view that simply typed first-order logic assumes that all sorts are non-empty. See also http://smtlib.cs.uiowa.edu/papers/smt-lib-reference-v2.6-draft-3.pdf, section 5.1 onwards.