Incremental input and assertion sets in Z3 - z3

I have a program that runs Z3 version 4.8.8 - 64 bit, with incremental input: the program starts Z3 once, executes many rounds of input-output to Z3, and then stops Z3. For performance reasons, running Z3 without incremental input is not an option.
Each round, the program inputs some (assert ...) statements to Z3, inputs (check-sat) to Z3, then gets the output of (check-sat) from Z3.
I have two rounds of input-output: the first round of inputs is as in z3.sat:
(declare-fun f () Int)
(declare-fun n () Int)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not false))
(check-sat)
which means: f is an even Int greater or equals to 2.
And the second round of inputs is as in z3.unsat:
(declare-fun f () Int)
(declare-fun n () Int)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not (exists ((alpha Int)) (= (* 2 alpha) f))))
(check-sat)
which means: if f is an even Int greater or equals to 2, then there exists an alpha where alpha=f/2.
I assume that running Z3 with incremental input is similar to concatenating the two rounds of input, z3.sat and z3.unsat, into one input, as in z3.combined:
(declare-fun f () Int)
(declare-fun n () Int)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not false))
(check-sat)
(declare-fun f () Int)
(declare-fun n () Int)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not (exists ((alpha Int)) (= (* 2 alpha) f))))
(check-sat)
Running:
z3 -smt2 z3.sat outputs sat
z3 -smt2 z3.unsat outputs unsat
z3 -smt2 z3.combined outputs errors, because the (assert ...) statements from the first round do not disappear:
sat
(error "line 8 column 21: invalid declaration, constant 'f' (with the given signature) already declared")
(error "line 9 column 21: invalid declaration, constant 'n' (with the given signature) already declared")
unknown
So it seems (push 1) and (pop 1) statements are needed for Z3 to forget previous assertion sets, so I added these statements at the start and end of z3.sat and z3.unsat, and re-concatenated z3.pushpop.sat and z3.pushpop.unsat to get z3.pushpop.combined.
z3.pushpop.sat:
(push 1)
(declare-fun f () Int)
(declare-fun n () Int)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not false))
(check-sat)
(pop 1)
z3.pushpop.unsat:
(push 1)
(declare-fun f () Int)
(declare-fun n () Int)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not (exists ((alpha Int)) (= (* 2 alpha) f))))
(check-sat)
(pop 1)
z3.pushpop.combined:
(push 1)
(declare-fun f () Int)
(declare-fun n () Int)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not false))
(check-sat)
(pop 1)
(push 1)
(declare-fun f () Int)
(declare-fun n () Int)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not (exists ((alpha Int)) (= (* 2 alpha) f))))
(check-sat)
(pop 1)
However, now running:
z3 -smt2 z3.pushpop.sat outputs sat
z3 -smt2 z3.pushpop.unsat outputs unknown
z3 -smt2 z3.pushpop.combined outputs:
sat
unknown
Why does z3 -smt2 z3.pushpop.unsat output unknown?

As Malte mentioned, the presence of pus/pop triggers "weaker" solvers in z3. (There are many technical reasons for this, but I agree from an end-user view-point, the change in behavior is unfortunate and can be rather confusing.)
But there are commands that let you do what you want without resorting to push and pop. Instead of it, simply insert:
(reset)
when you want to "start" a new session, and this will make sure it'll all work. That is, drop the push/pop and when you concatenate, insert a (reset) in between.
A slightly better approach
While the above will work, in general you only want to forget assertions, but not definitions. That is, you want to "remember" that you have an f and an n in the environment. If this is your use case, then put the following at the top of your script:
(set-option :global-declarations true)
and when you want to "switch" to a new problem, issue:
(reset-assertions)
This way, you won't have to "repeat" the declarations each time. That is, your entire interaction should look like:
(set-option :global-declarations true)
(declare-fun f () Int)
(declare-fun n () Int)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not false))
(check-sat)
(reset-assertions)
(assert (< 1 n))
(assert (<= 2 f))
(assert (= (+ (+ 1 f) 1) (+ n n)))
(assert (not (exists ((alpha Int)) (= (* 2 alpha) f))))
(check-sat)
which produces:
sat
unsat
Reference
All of this is documented in the official SMTLib document. See Section 3.9, pg. 44, for the descripton of global-declarations, and Section 4.2.2, pg. 59, for the description of (reset-assertions).

Incremental mode forces Z3 to use different theory subsolvers, as explained by one of the developers in this SO answer. These "incremental mode" subsolvers are often less effective than the "regular" ones, or at least may behave differently. As far as I know, Z3 switches to incremental mode whenever an SMT program contains push-pop scopes or multiple check-sats.
You initially say that not using incremental mode is not an option, but at least your file z3.pushpop.combined looks easily splitable. Another option might be to reset Z3 (I think the SMT command (reset) exists for that purpose) in between, instead of having push-pop blocks. If what I claim above is correct, however, this wouldn't prevent Z3 from staying in non-incremental mode. You could consider asking the developers via a "question issue" on Z3's issue tracker.

Related

unknown when using a integer division in z3 smt2

I am trying to find a solution for the function penta(n) = (n * (3n -1)) / 2 and where penta (z) = penta (a) + penta(b) for all number positives. That works until the integer division (div) is part ofthe definition, but when it is added in the definition I either got a timeout or an unknown.
I would expect to get 8 , 7 , 4. Any idea on what I did wrongly?
(declare-const a Int)
(declare-const b Int)
(declare-const z Int)
(define-fun penta ((n Int)) Int (div (* (- (* 3 n ) 1) n) 2) )
(assert (= (penta z) (+ (penta a) (penta b)) ))
(assert (> a 1))
(assert (> b 1))
(assert (> z 1))
(check-sat)
(get-model)
I am using the version on the http://rise4fun.com/Z3 website and the version 4.1 (x64).
The main issue is that the problem uses integer multiplication between two non-numeric arguments. There are no decision procedures for general Diophantine problems so Z3 does a best effort, which does not favor model enumeration.
When you don't use integer division, Z3 will try a partial heuristic based on
converting the problem into finite domain bit-vectors to find models. It invokes
this heuristic by performing a syntactic check on the formulas. THe syntactic check fails when you use the operator (div .. 2).
You can encode (div x 2) so the heuristic picks up the problem
by introducing fresh variables and bounding them:
(declare-const penta_z Int)
(declare-const penta_a Int)
(declare-const penta_b Int)
(assert (or (= (* 2 penta_z) (penta z)) (= (+ 1 (* 2 penta_z)) (penta z))))
(assert (or (= (* 2 penta_a) (penta a)) (= (+ 1 (* 2 penta_a)) (penta a))))
(assert (or (= (* 2 penta_b) (penta b)) (= (+ 1 (* 2 penta_b)) (penta b))))
(assert (= penta_z (+ penta_a penta_b) ))
(assert (> a 1))
(assert (> b 1))
(assert (> z 1))
(assert (>= penta_z 0))
(assert (<= penta_z 100))
You can also directly encode your problem using bit-vectors although this starts getting error prone because you have to deal with how to handle overflows.

How to deal with recursive function in Z3?

(set-option :smt.mbqi true)
(declare-fun R(Int) Int)
(declare-const a Int)
(assert (= (R 0) 0))
(assert (forall ((n Int)) (=> (> n 0) (= (R n ) (+ (R (- n 1)) 1)))))
(assert (not (= a 5)))
(assert (not (= (R a) 5)))
(check-sat)
I have tried the above code in Z3,But Z3 unable to answer.Can you please guide me where i have made the mistake ?
As a general pattern don't expect MBQI to produce models
involving functions that
only have an infinite range of different values.
If you really must, then you can use the define-fun-rec construct to define
a recursive function. Z3 currently trusts that the definition
is well-formed (e.g., that the equation corresponding to the function
definition is satisfiable).
(set-option :smt.mbqi true)
(declare-fun F (Int) Int)
(define-fun-rec R ((n Int)) Int
(if (= n 0) 0
(if (> n 0) (+ (R (- n 1)) 1)
(F n))))
(declare-const a Int)
(assert (not (= a 5)))
(assert (not (= (R a) 5)))
(check-sat)
(get-model)
Z3 uses recursively defined functions passively during search: whenever
there is a candidate model for the ground portion of the constraints, it
checks that the function graph is adequately defined on the values of the candidate model. If it isn't, then the function definition is instantiated on the selected values until it is well defined on the values that are relevant
to the ground constraints.

Why is Z3 not able to solve this instance without a seemingly trivial modification?

The original problem is:
(declare-const a Real)
(declare-const b Bool)
(declare-const c Int)
(assert (distinct a 0.))
(assert (= b (distinct (* a a) 0.)))
(assert (= c (ite b 1 0)))
(assert (not (distinct c 0)))
(check-sat)
The result is unknown.
But the last two constraints, taken together, are equivalent to (assert (= b false)), and after performing this rewrite by hand
(declare-const a Real)
(declare-const b Bool)
(declare-const c Int)
(assert (distinct a 0.))
(assert (= b (distinct (* a a) 0.)))
(assert (= b false))
;(assert (= c (ite b 1 0)))
;(assert (not (distinct c 0)))
(check-sat)
Z3 is now able to solve this instance (it is unsat).
Why can Z3 solve the second instance but not the first one, even though the first instance can be simplified to the second?
edit:
While locating the problem I found something very strange.
Z3 solves the following instance and returns "unsat":
(declare-fun a() Real)
(declare-fun b() Bool)
(declare-fun c() Int)
(assert (distinct a 0.0))
(assert (= b (distinct (* a a) 0.0)))
(assert (= b false))
;(assert (= c 0))
(check-sat)
But if I uncomment (assert (= c 0)), the solver returns "unknown", even though c=0 has nothing to do with the above assertions.
The problem here is that expressions like (* a a) are non-linear and Z3's default solver for non-linear problems gives up because it thinks it's too hard. Z3 does have another solver, but that one has very limited theory combination, i.e., you won't be able to use it for mixed Boolean, bit-vector, array, etc, problems, but only for arithmetic problems. It's easy to test by replacing the (check-sat) command with (check-sat-using qfnra-nlsat).

slow invariant inference with Horn clauses in Z3

I've played with the following example in Z3/Horn (unstable branch)
(set-logic HORN)
(declare-fun inv (Int) Bool)
(assert (inv 0))
(assert (forall ((I Int)) (=> (and (<= I 1000) (inv I)) (inv (+ I 1)))))
(assert (forall ((I Int)) (=> (inv I) (<= I 10000))))
(check-sat)
(get-model)
It takes 8.5s to infer the invariant x≤1001. This is unexpectedly long...
Time increases to 19 seconds if I replace 1000 by 1500 and to 34 seconds if I replace 1000 by 2000. This seems to indicate quadratic behaviour with respect to the loop bound.
I find it curious that it takes so much time to verify an assertion that is clearly inductive...
To close the loop on this question.
First of all, it is a very good example to illustrate some points.
The PDR engine in Z3 uses a monolithic strategy for generating
intermediary assertions. Intuitively, it is based on under-approximating
strongest post-conditions. It does not attempt to search within the spectrum
of interpolant strengths.
The example converges much faster (instantly)
if applying a magic set transformation (e.g, inverting the transition system):
(set-logic HORN)
(declare-fun inv (Int) Bool)
(assert (forall ((I Int)) (=> (not (<= I 10000)) (inv I))))
(assert (forall ((I Int)) (=> (and (<= I 1000) (inv (+ I 1))) (inv I))))
(assert (forall ((I Int)) (=> (inv I) (not (= I 0)))))
(check-sat)
(get-model)

Calculating Absolute Value in Z3

I have 3 variables a, b and c. I need to calculate c = absolute(b-a).
I encode this statement in Z3 as
(assert (>= c 0))
(assert (or (= c (- a b) (= c (- b a))))
I was thinking, is there a more efficient way of writing it in Z3?
Does Z3 have internal support for calculating absolute value?
Also, I hope there won't be any performance penalty for writing code like this, rather than using some other way.
Your encoding is correct. However, users usually encode the absolute value function using
(define-fun absolute ((x Int)) Int
(ite (>= x 0) x (- x)))
Then, they can write constraints such as:
(assert (= c (absolute (- a b))))
Here is the complete example (also available online at rise4fun):
(define-fun absolute ((x Int)) Int
(ite (>= x 0) x (- x)))
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(assert (= a 3))
(assert (= b 4))
(assert (= c (absolute (- a b))))
(check-sat)
(get-model)

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