Running Z3 on the following sequence of propositions
(declare-const x Real)
(assert (= 1 (^ x (/ 1 2))))
(check-sat-using qfnra-nlsat)
(get-model)
(eval (= x (^ x (/ 1 2))))
produces
sat
(model
(define-fun x () Real
(- 1.0))
)
Z3(5, 25): ERROR: even root of negative number is not real
Note that the final line simply evaluates the equation from line 2 on the proposed solution for x, so Z3 seems to contradict itself. Is this a bug or am I missing something?
This example exposes some bugs in the facilities dealing with root objects. A fix has been checked into the master branch (Z3 now returns unknown for this tactic).
Related
I want to use Z3 to proof validity of statements like this:
∀ a b: ℤ, ~ b = 0 -> (a / b) ^ 2. = (a * a) / (b * b)
Or in SMT-LIB format:
(declare-fun b () Int)
(declare-fun a () Int)
(assert (=> (= b 0) false))
(assert (let ((a!1 (= (^ (/ (to_real a) (to_real b)) 2.0)
(/ (to_real (* a a)) (to_real (* b b))))))
(not a!1)))
(check-sat)
But I get timeout with the default tactic. I guess Z3 is wasting its time trying to instantiate numbers in order to find a solution. But I'm only interested in unsat output since the problem is generalized and a sat output doesn't mean anything. What combination of tactics I should use to find validity of simple algebraic statements like this?
When I run your script, I get unsat pretty quickly:
$ time z3 a.smt2
unsat
z3 a.smt2 0.17s user 0.01s system 97% cpu 0.191 total
Perhaps your z3 is too old? Here's the version I have:
$ z3 --version
Z3 version 4.12.0 - 64 bit
Try upgrading, I think the latest released version is 4.12.1.
If the issue persists and you have the latest z3, you should report this as a regression at https://github.com/Z3Prover/z3/issues
I tried QF_NRA in z3 and it gave me an abstract value about root-obj.
(set-logic QF_NRA)
(declare-const x Real)
(assert (= 2 (* x x)))
(check-sat)
sat
(get-model)
(
(define-fun x () Real
(root-obj (+ (^ x 2) (- 2)) 1))
)
I don’t quite understand its meaning.
In addition, the x seems defined in recursion but not by define-fun-rec.
Thanks.
Algebraic reals
Z3's Real theory supports what's known as algebraic reals. That is, it can express solutions in terms of the roots of polynomials with rational (equivalently, integer) valued coefficients. Note that such a polynomial can have complex roots. Z3 only supports those roots that are real, i.e., those with an imaginary part of 0. An algebraic real is essentially the real-root of a univariate polynomial with integer coefficients.
Dealing with root-obj's
In the example you posted, you're asking z3 to find a satisfying model for x*x == 2. And it's telling you that the solution is "a" zero-of-the polynomial (+ (^ x 2) (- 2)), or written in more familiar notation P(x) = x^2 -2. The index you get is 1 (the second argument to the root-obj), which says it's the "first" real-root of this polynomial. If you ask z3 to give you another solution, it'll give you the next one:
(set-logic QF_NRA)
(declare-const x Real)
(assert (= 2 (* x x)))
(assert (distinct x (root-obj (+ (^ x 2) (- 2)) 1)))
(check-sat)
(get-model)
This prints:
sat
(
(define-fun x () Real
(root-obj (+ (^ x 2) (- 2)) 2))
)
As you see, the "next" solution is the second root. What if we assert we want yet another solution?
(set-logic QF_NRA)
(declare-const x Real)
(assert (= 2 (* x x)))
(assert (distinct x (root-obj (+ (^ x 2) (- 2)) 1)))
(assert (distinct x (root-obj (+ (^ x 2) (- 2)) 2)))
(check-sat)
This prints:
unsat
as expected.
Note that algebraic reals do not include numbers such as pi, e, etc., i.e., they do not include transcendentals. Only those real numbers that can be expressed as the root of polynomials with integer coefficients. Leonardo's paper from 2012 explains this in great detail.
Getting approximations
z3 also allows you to get an approximation for such a root-obj solution, with as arbitrary a precision as you like. To do so, use the incantation:
(set-option :pp.decimal true)
(set-option :pp.decimal_precision 20)
where 20 in the second line is how many digits of precision you'd like, and you can change it as you see fit. If you add these two lines to your original script, z3 will respond:
sat
(
(define-fun x () Real
(- 1.4142135623730950?))
)
Note the ? at the end of the number. This is z3's way of telling you that the number it printed is an "approximation" to the value, i.e., it isn't the precise result.
A note on "recursion"
Your question suggests maybe x is defined recursively. This isn't the case. It just happens that you picked the variable name to be x and z3 always uses the letter x for the polynomial as well. If you picked y as the variable, you'd still get the exact same answer; the parameter to the polynomial has nothing to the with the variables in your program.
I'm new to Z3 and I'm trying to understand how it works, and what it can and cannot do. I know that Z3 has at least some support for exponentials through the power (^) operator (see Z3py returns unknown for equation using pow() function, How to represent logarithmic formula in z3py, and Use Z3 and SMT-LIB to define sqrt function with a real number). What I'm unclear on is how extensive this support is, and what kind of inferences z3 can make about exponentials.
Here's a simple example involving exponentials which z3 can analyze. We define an exponential function, and then ask it to verify that exp(0) == 1:
(define-fun exp ((x Real)) Real
(^ 2.718281828459045 x))
(declare-fun x1 () Real)
(declare-fun y1 () Real)
(assert (= y1 (exp x1)))
(assert (not (=> (= x1 0.0) (= y1 1.0))))
(check-sat)
(exit)
Z3 returns unsat, as expected. On the other hand, here's a simple example which Z3 can't analyze:
(define-fun exp ((x Real)) Real
(^ 2.718281828459045 x))
(declare-fun x1 () Real)
(declare-fun y1 () Real)
(assert (= y1 (exp x1)))
(assert (not (< y1 0.0)))
(check-sat)
(exit)
This should be satisfiable, since literally any value for x1 would give y1 > 0. However, Z3 returns unknown. Naively I might have expected that Z3 would be able to analyze this, given that it could analyze the first example.
I realize this question is a bit broad, but: can anyone give me any insight into how Z3 handles exponentials, and (more specifically) why it can solve the first example I gave but not the second?
It is hard to say in general, since non-linear solving is challenging, but the case you presented is actually not so mysterious. You wrote:
(assert (= y (exp x)))
(assert (not (=> (= x 0) (= y 1))))
Z3 is going to simplify the second assertion, yielding:
(assert (= y (exp x)))
(assert (= x 0))
(assert (not (= y 1)))
Then it will propagate the first equality, yielding:
(assert (= y (exp 0)))
(assert (not (= y 1)))
Now when exp is expanded, you have a case of constant^constant, which Z3 can handle (for integer exponents, etc).
For the second case, you are asking it a very very basic question about variable exponents, and Z3 immediately barfs. That's not too odd, since so many questions about variable exponents are either known uncomputable or unknown but hard.
Not Solvable in Z3. Why?
My Program:
;x = 2x + 2 (This on Underlaying DB is always increasing as X > Y in DB)
(declare-const x0 Real)
(declare-const xn Real)
(declare-const n Real)
(push)
(assert (= x0 42))
(assert (= xn (+ (* x0 (^ 2 n)) (* 2 (- (^ 2 n) 1)) ) )) ; recurrence relation
(assert (> xn 700))
(check-sat-using qfnra-nlsat)
(get-model); to find a satisfiable valuation
(pop); removes any assertion
-----------------------------
Z3 Answer:
-----------------------------
unknown
(model
(define-fun n () Real 0.0)
(define-fun xn () Real 42.0)
(define-fun x0 () Real 42.0)
)
I tried with integer also:
According to the values in database 'n' should come as 4, but it is coming as 0.
Please see into it and anyone please help me.
unknown means the model is "alleged," i.e., may or may not be correct. Since your problem contains non-linear terms (exponentiation), this is to be expected.
Indeed, the model suggests xn = 42, which contradicts the assumption xn > 700 in your program. The model given is bogus. But you cannot blame z3 for that, since it already told you unknown. It's beyond Z3's capabilities to solve your query.
In Non-linear arithmetic and uninterpreted functions, Leonardo de Moura states that the qfnra-nlsat tactic hasn't been fully integrated with the rest of Z3 yet. I thought that the situation has changed in two years, but apparently the integration is still not very complete.
In the example below, I use datatypes purely for "software engineering" purposes: to organize my data into records. Even though there are no uninterpreted functions, Z3 still fails to give me a solution:
(declare-datatypes () (
(Point (point (point-x Real) (point-y Real)))
(Line (line (line-a Real) (line-b Real) (line-c Real)))))
(define-fun point-line-subst ((p Point) (l Line)) Real
(+ (* (line-a l) (point-x p)) (* (line-b l) (point-y p)) (line-c l)))
(declare-const p Point)
(declare-const l Line)
(assert (> (point-y p) 20.0))
(assert (= 0.0 (point-line-subst p l)))
(check-sat-using qfnra-nlsat)
(get-model)
> unknown
(model
)
However, if I manually inline all the functions, Z3 finds a model instantly:
(declare-const x Real)
(declare-const y Real)
(declare-const a Real)
(declare-const b Real)
(declare-const c Real)
(assert (> y 20.0))
(assert (= 0.0 (+ (* a x) (* b y) c)))
(check-sat-using qfnra-nlsat)
(get-model)
> sat
(model
(define-fun y () Real
21.0)
(define-fun a () Real
0.0)
(define-fun x () Real
0.0)
(define-fun b () Real
0.0)
(define-fun c () Real
0.0)
)
My question is, is there a way to perform such an inlining automatically? I'm fine with either one of these workflows:
Launch Z3 with a tactic that says "Inline first, then apply qfnra-nlsat. I haven't found a way to do so, but maybe I wasn't looking well enough.
Launch Z3 using some version of simplify to do the inlining. Launch Z3 the second time on the result of the first invocation (the inlined version).
In other words, how to make qfnra-nlsat work with tuples?
Thank you!
That's correct, the NLSAT solver is still not integrated with the other theories. At the moment, we can only use it if we eliminate all datatypes (or elements of other theories) before running it. I believe there is no useful existing tactic inside of Z3 at the moment though, so this would have to be done beforehand. In general it's not hard to compose tactics, e.g., like this:
(check-sat-using (and-then simplify qfnra-nlsat))
but the simplifier is not strong enough to eliminate the datatype constants in this problem. (The respective implementation files are datatype_rewriter.cpp and datatype_simplifier_plugin.cpp.)