I am trying to remove existential quantifiers in my theory using Skolemization. This means that I replace existential quantifiers with functions that are parameterized by the universally quantified variables in the scope of the existential quantifiers.
Here I found an explanation how to do this in Z3, but I am still having troubles doing it. Suppose the following two functions:
(define-fun f1 ((t Int)) Bool (= t 3))
(define-fun f2 () Bool (exists ((t Int)) (f1 t)))
I believe that f2 should be true, because there exists an integer t such that (f1 t) is true, namely t=3. I apply Skolemization by introducing a constant for the existentially quantified formula:
(define-const c Int)
Then the formula with the existential quantifier is rewritten to:
(define-fun f2 () Bool (f1 c))
This does not work, that is, the constant c does not have the value 3. I suspect it is because we have not given an interpretation to the constant c, because if we add (assert (= c 3)) it works fine, but this takes away the whole idea of the existential quantifier. Is there a way in which I give a less explicit interpretation to c so that this will work?
So, I think you have it about right actually, here's the script I used with automatic (via Z3's SNF tactic) and manual (via adding the constant c) skolemization, which gave the value 3 in the model for the skolem constant as expected (smt-lib script: http://rise4fun.com/Z3/YJy2 ):
(define-fun f1 ((t Int)) Bool (= t 3))
(define-fun f2 () Bool (exists ((t Int)) (f1 t)))
(declare-const c Int)
(define-fun f2a () Bool (f1 c))
(push)
(assert f2)
(check-sat) ; sat
(get-model) ; model gives t!0 = 3 (automatic skolemization in Z3)
(pop)
(push)
(assert f2a)
(check-sat) ; sat
(get-model) ; model gives c = 3 after manual skolemization
(pop)
Also, note that Z3 has a Skolem normal form (SNF) conversion tactic built in, and here's an example in z3py (link to script: http://rise4fun.com/Z3Py/ZY2D ):
s = Solver()
f1 = Function('f1', IntSort(), BoolSort())
t = Int('t')
f2 = Exists(t, f1(t))
f1p = ForAll(t, f1(t) == (t == 3)) # expanded define-fun macro (define-fun f1 ((t Int)) Bool (= t 3))
s.add(f1p)
s.add(f2)
print f1p
print f2
print s.check()
print s.model() # model has skolem constant = 3
g = Goal()
g.add(f1p)
g.add(f2)
t = Tactic('snf') # use tactic to convert to SNF
res = t(g)
print res.as_expr()
s = Solver()
s.add( res.as_expr() )
print s.check()
print s.model() # model has skolem constant = 3
Related
I list some assert about Quadratic function:
(declare-fun H () Int)
(assert (>= H 8000))
(assert (<= H 12000))
(minimize (- (^ H 2) H))
(check-sat)
but the answer is "unknown" and the reason for unknown is (incomplete (theory arithmetic)); I can't understand which is the lost one
In general z3 cannot deal with non-linear terms. (A term is non-linear if you multiply two variables together. In your case, that'd be (^ H 2).
This is especially true of the optimization engine: Nonlinear constraints over integers is most likely going to be beyond reach. But you're in luck: Your formula is rather simple so it can handle it fine. Rewrite it using multiplication:
(declare-fun H () Int)
(assert (>= H 8000))
(assert (<= H 12000))
(minimize (- (* H H) H))
(check-sat)
(get-model)
This prints:
sat
(model
(define-fun H () Int
8000)
)
I'm using Z3 theorem prover (using Z3Py: the Z3 API in Python) to create QBF (Quantified Boolean formula).
Is there any way in Z3 to directly convert your qbf formula into Prenex normal form ?
I don't think there's a tactic to convert to Prenex, but you can surely apply the quantifier-elimination tactic and further process your formulas. Note that the transformed formulas will not really look like the originals, as they are mechanically generated.
Here's an example:
from z3 import *
f = Function('f', IntSort(), IntSort(), IntSort())
x, y = Ints('x y')
p = ForAll(x, Or(x == 2, Exists(y, f (x, y) == 0)))
print Tactic('qe')(p)
Here qe is the quantifier elimination tactic. This produces:
[[Not(Exists(x!0,
Not(Or(x!0 == 2,
Exists(x!1,
And(f(x!0, x!1) <= 0,
f(x!0, x!1) >= 0))))))]]
For a nice tutorial on tactics, see here: http://ericpony.github.io/z3py-tutorial/strategies-examples.htm
You could use the skolemize tactic (snf) which will by definition be in prenex form. However it will also eliminate existential quantifiers which is not what you want. Here's an example.
(declare-fun a (Int) Bool)
(declare-fun b (Int) Bool)
(declare-fun c (Int) Bool)
(assert
(forall ((x Int))
(or
(exists ((y Int))
(a y)
)
(exists ((z Int))
(=>
(b z)
(c x)
)
)
)
)
)
(apply
(and-then
; mode (symbol) NNF translation mode: skolem (skolem normal form), quantifiers (skolem normal form + quantifiers in NNF), full (default: skolem)
(using-params snf :mode skolem)
)
:print_benchmark true
:print false
)
When Z3 is given the above it will responds with something like
(declare-fun c (Int) Bool)
(declare-fun b (Int) Bool)
(declare-fun a (Int) Bool)
(assert (forall ((x Int))
(or (exists ((y Int)) (a y)) (exists ((z Int)) (=> (b z) (c x))))))
(check-sat)
You can see the available tactics by running
echo "(help-tactic)" | ./z3 -in | less
from a bash shell.
Unfortunately I can't see one that states it does conversion to prenex.
Given the following code:
from z3 import *
a,b,c = BitVecs('a b c', 32)
f1 = Exists([a, b, c], And(a + b == c, a < b, c == 1337))
f2 = And(a + b == c, a < b, c == 1337)
prove(f1 == f2)
I would assume that z3 implicitly existential quantifies a, b and c, in this example. Why aren't the two formulas equal, what is the difference?
The way you formulated your query doesn't really check whether f1 equals f2. Your query is essentially asking the solver to find a, b, c such that the following fails to hold:
Exists([a, b, c], And(a + b == c, a < b, c == 1337))
=
And(a + b == c, a < b, c == 1337))
And indeed, you can instantiate the outer a, b, and c such that the right hand-side is false; but the left hand side is an existential which is true; thus failing the equivalence you asked for.
It might be easier to see this with a simpler example; with just one boolean variable. You're essentially asking:
x == (Exists [x], x)
You see that those xs are actually different, so we can rename the inner one to (say) y; we get:
x == (Exist [y]. y)
Now, the right-hand-side is clearly true since there is a y that makes (Exist [y]. y) true. So, you are essentially asking the prover to establish that no matter what x you pick, it is true. Which is definitely not the case when you pick x to be false.
Indeed, you can ask Z3 to give you the formula it's trying to prove, and this is what it returns for your original query:
(set-info :status unknown)
(declare-fun c () (_ BitVec 32))
(declare-fun b () (_ BitVec 32))
(declare-fun a () (_ BitVec 32))
(assert
(let (($x24 (exists ((a (_ BitVec 32))
(b (_ BitVec 32))
(c (_ BitVec 32)) )
(and (= (bvadd a b) c) (bvslt a b) (= c (_ bv1337 32))))))
(let (($x57 (= $x24 (and (= (bvadd a b) c) (bvslt a b) (= c (_ bv1337 32))))))
(and $x57))))
(check-sat)
Which is clearly satisfiable, by the above reasoning.
(See Z3: convert Z3py expression to SMT-LIB2 from Solver object for the code that converts a z3-python query to smt-lib.)
I have an enumeration type BoolT that contains Bool and Bot
(declare-datatypes () ((BoolT Bool Bot)))
and I want to define an equality function eq that given two BoolT returns Bot if one of the arguments is Bot, otherwise the equality between the two Bool arguments. However, I am not able to define the actual comparison between the boolean values. Until now I get the function
(define-fun eq ((x1 BoolT) (x2 BoolT)) BoolT
(ite (or (= x1 Bot) (= x2 Bot)) Bot Bool))
while I need something like
(define-fun eq ((x1 BoolT) (x2 BoolT)) BoolT
(ite (or (= x1 Bot) (= x2 Bot)) Bot
(or (and (= x1 true)(= x2 true)) (and (= x1 false)(= x2 false)))
)
or at least a correct definition of the following predicate
(define-fun is-True ((x1 BoolT)) Bool
(ite (= x1 true) true false)
)
Is there a way to model the eq function or the previous predicate on BoolT?
You will need tagged unions in Z3, similar to ML data-types. You can't just take the union of an existing type and included it in a datatype declaration.
So you will need to write something like:
(declare-datatypes () ((BoolT True False Bot)))
or
(declare-datatypes () ((BoolT (mkB (tf B)) Bot) (B True False)))
Then you can write:
(define-fun is-True1 ((x BoolT)) Bool
(= (mkB True) x))
or
(define-fun is-True2 ((x BoolT)) Bool
(and (is-mkB x) (= True (tf x))))
and assert
(declare-const a BoolT)
(assert (is-True2 a))
Declaring (B True False)
automatically declares predicates is-True and is-False
(declare-const b B)
(assert (is-True b))
I'm trying to influence the randomness of results for model values generated by Z3. As far as I understand, the options for this are very limited: in case of linear arithmetic, the simplex solver does not allow for random results that still satisfy the given constraints. However, there is an option smt.arith.random_initial_value ("use random initial values in the simplex-based procedure for linear arithmetic (default: false)") which I don't seem to get working:
from z3 import *
set_option('smt.arith.random_initial_value',True)
x = Int('x')
y = Int('y')
s = Solver()
s.add( x+y > 0)
s.check()
s.model()
This seems to always produce [y = 0, x = 1] as a result. Even model completion for variables unused in the given constraints seems to produce deterministic results all the time.
Any ideas or hints about how this option works?
Thanks for catching that! There was indeed a bug that caused the random seed not to be passed through to the arithmetic theory. This is now fixed in the unstable branch (fix here).
This example:
(set-option :smt.arith.random_initial_value true)
(declare-const x Int)
(declare-const y Int)
(assert (> (+ x y) 0))
(check-sat-using (using-params qflra :random_seed 1))
(get-model)
(check-sat-using (using-params qflra :random_seed 2))
(get-model)
(check-sat-using (using-params qflra :random_seed 3))
(get-model)
Now produces three different models:
sat
model
(define-fun y () Int
4294966763)
(define-fun x () Int
4294966337)
)
sat
(model
(define-fun y () Int
216)
(define-fun x () Int
4294966341)
)
sat
(model
(define-fun y () Int
196)
(define-fun x () Int
4294966344)
)
It looks like there may be another place where this option isn't passed through correctly (e.g., when using set-logic instead of calling the qflra tactic directly), we're still looking into that.