In Z3-Py, I am performing quantifier elimination (QE) over the following formulae:
Exists y. Forall x. (x>=2) => ((y>1) /\ (y<=x))
Forall x. Exists y. (x>=2) => ((y>1) /\ (y<=x)),
where both x and y are Integers. I did QE in the following way:
x, y = Ints('x, y')
t = Tactic("qe")
negS0= (x >= 2)
s1 = (y > 1)
s2 = (y <= x)
#EA
ea = Goal()
ea.add(Exists([y],Implies(negS0, (ForAll([x], And(s1,s2))))))
ea_qe = t(ea)
print(ea_qe)
#AE
ae = Goal()
ae.add(ForAll([x],Implies(negS0, (Exists([y], And(s1,s2))))))
ae_qe = t(ae)
print(ae_qe)
Result QE for ae is as expected: [[]] (i.e., True). However, as for ea, QE outputs: [[Not(x, >= 2)]], which is a results that I do not know how to interpret since (1) it has not really performed QE (note the resulting formula still contains x and indeed does not contain y which is the outermost quantified variable) and (2) I do not understand the meaning of the comma in x, >=. I cannot get the model either:
phi = Exists([y],Implies(negS0, (ForAll([x], And(s1,s2)))))
s_def = Solver()
s_def.add(phi)
print(s_def.model())
This results in the error Z3Exception: model is not available.
I think the point is as follows: since (x>=2) is an implication, there are two ways to satisfy the formula; by making the antecedent False or by satisfying the consequent. In the second case, the model would be y=2. But in the first case, the result of QE would be True, thus we cannot get a single model (as it happens with a universal model):
phi = ForAll([x],Implies(negS0, (Exists([y], And(s1,s2)))))
s_def = Solver()
s_def.add(phi)
print(s_def.model())
In any case, I cannot 'philosophically' understand the meaning of a QE of x where x is part of the (quantifier-eliminated) answer.
Any help?
There are two separate issues here, I'll address them separately.
The mysterious comma This is a common gotcha. You declared:
x, y = Ints('x, y')
That is, you gave x the name "x," and y the name "y". Note the comma after the x in the name. This should be
x, y = Ints('x y')
I guess you can see the difference: The name you gave to the variable x is "x," when you do the first; i.e., comma is part of the name. Simply skip the comma on the right hand side, which isn't what you intended anyhow. And the results will start being more meaningful. To be fair, this is a common mistake, and I wish the z3 developers ignored the commas and other punctuation in the string you give; but that's just not the case. They simply break at whitespace.
Quantification
This is another common gotcha. When you write:
ea.add(Exists([y],Implies(negS0, (ForAll([x], And(s1,s2))))))
the x that exists in negS0 is not quantified over by your ForAll, since it's not in the scope. Perhaps you meant:
ea.add(Exists([y],ForAll([x], Implies(negS0, And(s1,s2)))))
It's hard to guess what you were trying to do, but I hope the above makes it clear that the x wasn't quantified. Also, remember that a top-level exist quantifier in a formula is more or less irrelevant. It's equivalent to a top-level declaration for all practical purposes.
Once you make this fix, I think things will become more clear. If not, please ask further clarifying questions. (As a separate question on Stack-overflow; as edits to existing questions only complicate the matters.)
Related
Consider these two formulae:
Exists y. Forall x. (y>x), which is unsat.
Forall x. Exists y. (y>x), which is sat.
Note that we cannot find “models” for the sat formula, e.g., using Z3 it outputs Z3Exception: model is not available for the following code:
phi = ForAll([x],Exists([y], lit))
s_def = Solver()
s_def.add(phi)
print(s_def.model())
Also, quantifier elimination does not output an elimination, but a satisfiability result
[[]] (i.e., True):
x, y = Reals('x, y')
t = Tactic("qe")
lit = (y>x)
ae = Goal()
ae.add(ForAll([x],Exists([y], lit)))
ae_qe = t(ae)
print(ae_qe)
I think this happens because the value of y fully depends on x (e.g., if x is 5 then y can be 6). Thus, I have some questions:
Am I right with this interpretation?
What would be the meaning of “a model of a universally quantified formula”?
Do we say a formula accepts quantifier elimination even if it never eliminates the quantifier but "only: evaluate to True or False?
Is there a way to synthetise or construct a model/function that represents a y that holds the constraint (y>x); e.g. f(x)=x+1. In other words, does it make sense that the quantifier elimination of a Forall x. Exists y. Phi(x,y) like the example would be Forall x. Phi(x,f(x))?
You get a model-not-available, because you didn't call check. A model is only available after a call to check. Try this:
from z3 import *
x, y = Ints('x y')
phi = ForAll([x], Exists([y], y > x))
s = Solver()
s.add(phi)
print(s.check())
print(s.model())
This prints:
sat
[]
Now, you're correct that you won't get a meaningful/helpful model when you have a universal-quantifier; because the model will depend on the choice of x in each case. (You only get values for top-level existentials.) This is why the model is printed as the empty-list.
Side Note In certain cases, you can use the skolemization trick (https://en.wikipedia.org/wiki/Skolem_normal_form) to get rid of the nested existential and get a mapping function, but this doesn't always work with SMT solvers as they can only build "finite" skolem functions. (For your example, there isn't such a finite function; i.e., a function that can be written as a case-analysis on the inputs, or an if-then-else chain.)
For your specific questions:
Yes; value of y depends on x and thus cannot be displayed as is. Sometimes skolemization will let you work around this, but SMT solvers can only build finite skolem-functions.
Model of a universally quantified formula is more or less meaningless. It's true for all values of the universally quantified variables. Only top-level existentials are meaningful in a model.
Quantifier elimination did work here; it got rid of the whole formula. Note that QE preserves satisfiability; so it has done its job.
This is the skolem function. As you noted, this skolem function is not finite (can't be written as a chain of finite if-then-elses on concrete values), and thus existing SMT solvers cannot find/print them for you. If you try, you'll see that the solver simply loops:
from z3 import *
x = Int('x')
skolem = Function('skolem', IntSort(), IntSort())
phi = ForAll([x], skolem(x) > x)
s = Solver()
s.add(phi)
print(s)
print(s.check())
print(s.model())
The above never terminates, unfortunately. This sort of problem is simply too complicated for the push-button approach of SMT solving.
Is it possible to change the domain of a variable after it has been defined and used in statements?. Example
s = Solver()
x = Real('x')
s.add(x < 1)
Now I want to change the domain of x to Int or Bool.
thanks!
The short answer is no.
But why do you want to do this? SMTLib is based on a many-sorted first-order logic, and variables can only have one sort. So, even if you can change the domain, it would be meaningless. (Essentially a type-error.)
Having said that, there's nothing stopping you from saying:
x = Int ('x')
at the end of that script. But the new x would be totally independent of the old x; i.e., a different name with a different sort and you'd lose access to the first one. Clearly, this is neither useful nor advisable. To wit:
from z3 import *
s = Solver ()
x = Real ('x')
s.add (x < 1)
x = Bool ('x')
s.add (x)
print s.sexpr()
print s.check()
print s.model()
This prints:
(declare-fun x () Real)
(declare-fun x () Bool)
(assert (< x 1.0))
(assert x)
sat
[x = True, x = 0]
This is very confusing to read, till you realize those two xs are totally independent of each other. (And I'd say the s.sexpr() method is rather buggy since it doesn't print out valid smt2-lib, as what it prints would be rejected by a compliant SMT-solver, but that's a different issue.)
I suspect, perhaps, you're trying to ask for something else. If you describe what you are trying to do in detail, you might get a better answer!
The ctx-solver-simplify tactic only works for bool variables, so how would I deal with variables over finite domain (e.g., which tactics to use)? For example, if z can only take 3 values 0,1,2, then how to simplify Or(z==0,z==1,z==2) to true ?
Also, even for bool expressions, the tactic ctx-solver-simplify doesn't simplify completely. For example:
x,y,z = z3.Bools('x y z')
c1 = z3.And(x==True,y==True,z==True)
c2 = z3.And(x==True,y==True,z==False)
c3 = z3.And(x==True,y==False,z==True)
c4 = z3.And(x==True,y==True,z==False)
z3.Tactic('ctx-solver-simplify')(z3.Or([c1,c2,c3,c4]))
[[Or(And(x, z), And(x == True, y == True, z == False))]]
How do I get something like And(x, Or(z, y)) ?
Thanks !
Reducing Boolean (or other finite-domain) problems to a minimal form is a hard problem. The ctx-solver-simplify tactic is one of the more expensive simplifiers, but it doesn't go all the way to the provably smallest form.
Problems from other domains (e.g., enumerations like z \in {0, 1, 2}) would have to be converted to Booleans first to use this tactic, but perhaps other tactics would be better suited, and perhaps a bit-vector encoding would help too.
I want to create an expression that selects one of a given set of expressions. Given an array of expressions
Expr[] availableExprs = ...;
with statically known length, I want Z3 to select any one of these (like a switch statement). In case the problem is SAT I need a way to find out which of these was selected in the model (its index in the array).
What is the fastest way to encode this?
I considered these approaches so far:
Have an integer restricted to [0, arrayLength) and use ITE to select one of those expressions. The model allows me to extract this integer. Unfortunately, this introduces the integer theory to the model (which previously did not use integers at all).
Have one boolean for each possible choice. Use ITE to select an expression. Assert that exactly one of those booleans is true. This strategy does not need any special theory (I think) but the encoding might be too verbose.
Store the expressions into an array expression and read from that array using an integer. This saves the ITE chain but introduces the array theory.
Clearly, all of these work, but they all seem to have drawbacks. What is the best strategy?
If all you want is to encode is that an element v is in a finite set {e1, ..., en} (with sort U), you can do this using smtlib2 as follows:
(declare-fun v () U)
(declare-fun p1 () Bool)
...
(assert (= p1 (= v e1)))
(assert (= p2 (= v e2)))
...
(assert (= pn (= v en)))
(assert (or p1 ... pn))
The variable v will be equal to one of the elements in "the array" of {e1 ... en}. And pi must be true if the selection variable v is equal to ei. This is basically a restatement of Nikolaj's suggestion, but recast for arbitrary sorts.
Note that multiple pi may be set to true as there is no guarantee ei != ej. If you need to ensure no two elements are both selected, you will need to figure out what semantics you want. If the {e1... en} are already entailed to be distinct, you don't need to add anything. If the "array" elements must be distinct but are not yet entailed to be distinct, you can assert
(assert (distinct e1 ... en))
(This will probably be internally expanded to something quadratic in n.)
You can instead say that no 2 p variables can be true as once. Note that this is a weaker statement. To see this, suppose v = e1 = 1 and e2 = e3 = 0. Then p1 = true and p2 = p3 = false. The obvious encoding for these constraints is quadratic:
(assert (or (not pi) (not pj))) ; for all i < j
If you need a better encoding for this, try taking a look at how to encode "p1+ ... + pn <= 1" in Translating Pseudo-Boolean Constraints into SAT section 5.3. I would only try this if the obvious encoding fails though.
I think you want to make sure that each expression is quantifier free and uses only functions and predicates already present in the formula. If this is not the case then introduce a fresh propositional variable p_i for each index and assert ctx.MkIff(p_i, availableExprs[i]) to the solver.
When Z3 produces a model, use model.Eval(p_i) and check if the result is the expression "True".
I have 2 formulas F1 and F2. These two formulas share most variables, except some 'temporary' (or I call them 'free') variables having different names, that are there for some reasons.
Now I want to prove F1 == F2, but prove() method of Z3 always takes into account all the variables. How can I tell prove() to ignore those 'free' variables, and focuses only on a list of variables I really care about?
I mean with all the same input to the list of my variables, if at the output time, F1 and F2 have the same value of all these variables (regardless the values of 'free' variables), then I consider them 'equivalence'
I believe this problem has been solved in other researches before, but I dont know where to look for the information.
Thanks so much.
We can use existential quantifiers to capture 'temporary'/'free' variables.
For example, in the following example, the formulas F and G are not equivalent.
x, y, z, w = Ints('x y z w')
F = And(x >= y, y >= z)
G = And(x > z - 1, w < z)
prove(F == G)
The script will produce the counterexample [z = 0, y = -1, x = 0, w = -1].
If we consider y and w as 'temporary' variables, we may try to prove:
prove(Exists([y], F) == Exists([w], G))
Now, Z3 will return proved. Z3 is essentially showing that for all x and z, there is a y that makes F true if and only if there is a w that makes G true.
Here is the full example.
Remark: when we add quantifiers, we are making the problem much harder for Z3. It may return unknown for problems containing quantifiers.
Apparently, I cannot comment, so I have to add another answer. The process of "disregarding" certain variables is typically called "projection" or "forgetting". I am not familiar with it in contexts going beyond propositional logic, but if direct existential quantification is possible (which Leo described), it is conceptually the simplest way to do it.