Is there a way to simplify the following expression to "6 < var"?
According to Z3, these expressions are equivalent but simplification does not produce the latter.
I have tried the three parameters since they seem to be related to if-then-else but this also did not help.
(declare-fun var () Int)
(simplify
(exists ((bx Int))
(and
(exists ((byX Int))
(ite (> bx 5) (= byX 0) (&& (> bx 2) (= byX (+ byX 4)))))
(= bx (+ var 1))
(> var 6)
)
)
:push_ite_arith true
:pull_cheap_ite true
:ite_extra_rules true
)
(assert
(not
(iff
(exists ((bx Int))
(and
(exists ((by Int))
(ite (> bx 5) (= by 0) (&& (> bx 2) (= by (+ by 4)))))
(= bx (+ var 1))
(> var 6)
)
)
(< 6 var)
)
)
)
(check-sat)
Not in general, no.
Z3's simplifications and what you would consider "simple" are typically not the same, and it works more or less as a black-box. It won't produce output like what you would get from a symbolic math package or alike: The simplifications are more geared towards making the input "simpler" for further solving; not for "presenting it back to the user" purpose.
You can find many similar questions on stack-overflow, see: https://stackoverflow.com/search?q=%5Bz3%5D+simplify and in particular this answer from Leo: simplification in Z3
Related
(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.
I originally posted the question as shown below the dotted line, but since then I have an even simpler example:
(declare-fun f (Int) Int)
(assert (= (f 10) 1))
(check-sat)
(get-model)
produces an interpretation for f as expected. However change the constant to anything but 10, and Z3 just spins the arrowhead a couple of times but then prints nothing!
--------------------------------------- original question -----------------------------
I tried Z3 on the following input and the arrowhead turns a few times and stops but Z3 prints or says nothing. Why?
(declare-fun f (Int Int) Int)
(assert (>= (f 1 1) 1))
(assert (>= (f 1 2) 2))
(assert (>= (f 2 1) 2))
(assert (>= (f 2 2) 2))
(assert (= (f 1 1) 1))
(assert (= (f 2 2) 2))
(assert (or (= (f 1 2) 1) (= (f 1 2) 2)))
(assert (or (= (f 2 1) 1) (= (f 2 1) 2)))
(check-sat)
(get-model)
I feel like I'm missing something really obvious..
I am obtaining (using iZ3, Z3 unstable branch)
sat
(model
(define-fun f ((x!1 Int) (x!2 Int)) Int
(ite (and (= x!1 1) (= x!2 1)) 1
(ite (and (= x!1 2) (= x!2 2)) 2
(ite (and (= x!1 1) (= x!2 2)) 2
(ite (and (= x!1 2) (= x!2 1)) 2 2)))))
)
Run this example online here
I presume you're using Z3 on rise4fun? The version running there may be a little out of date. We have to manually update the binary there. If it doesn't reply, it's either because it times out, or because there was some other problem (e.g., segfault). It's quite possible that the version on rise4fun exhibits some bug that's already been fixed in other version of Z3 (e.g., unstable, iZ3, etc).
I have a pretty simple problem, I'm mentioning the relevant part here:
;; All variables are declared to be of type Real
(assert (and (<= 1.0 var1-r) (< var1-r 4.0)))
;;following defines var1-r
(assert (= var1-r (+ a b)))
;;following defines var1-e
(assert (=> (and (<= 1.0 var1-r) (< var1-r 2.0)) (= var1-e 8388608.0)))
(assert (=> (and (<= 2.0 var1-r) (< var1-r 4.0)) (= var1-e 4194304.0)))
;;following defines var1
(assert (= var1 (/ (foo (* var1-r var1-e)) var1-e)))
;;Similarly for var2-r, var2-e, var2
(assert (and (<= 1.0 var2-r) (< var2-r 4.0)))
(assert (= var2-r (+ b a)))
(assert (=> (and (<= 1.0 var2-r) (< var2-r 2.0)) (= var2-e 8388608.0)))
(assert (=> (and (<= 2.0 var2-r) (< var2-r 4.0)) (= var2-e 4194304.0)))
(assert (= var2 (/ (foo (* var2-r var2-e)) var2-e)))
Here, foo() is a simple interpreted function, eg., foo (x) = (to_real (to_int x))
Note that var1 and var2 are equal. Reason: var1-r and var2-r are equal (commutativity of Reals) and consequently var2-e and var1-e are equal, leading to var1 and var2 being equal. However, I am not able to prove unsatisfiability of (not (= var1 var2)) using z3. In fact, the same is true if var2-r is defined as (+ a b). [Note that var1 and var2 being equal is actually also independent of the definition of foo()].
Please look at here. I am obtaining
unsat
Some values of uninterpreted functions can be unconstrained during the search. For example, if in smt query only f(1) is called, then f(2), f(3) can be anything. Is there a way (some option may be) to know which values were not used during the solving and therefore can be anything?
For quantifier free problems, you can achieve that by using the option :model-partial to true.
Here is an example (also available here):
(set-option :model-partial true)
(declare-fun f (Int) Int)
(assert (> (f 0) 0))
(assert (< (f 1) 0))
(check-sat)
(get-model)
In this example, we get the output:
sat
(model
(define-fun f ((x!1 Int)) Int
(ite (= x!1 0) 1
(ite (= x!1 1) (- 1)
#unspecified)))
)
BTW, in the next release (Z3 4.3.2), this option is renamed to :model.partial. In the next release, the options are grouped in modules.
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)