I have a (get-model) query for Z3 which returns this function:
(define-fun rules ((x!0 Tree)) Bool
(ite (= x!0 (node "mann" (cons (node "adam" nil) nil))) true
(ite (= x!0 (node "mensch" (cons (node "adam" nil) nil))) true
true)))
When using this code:
(declare-datatypes () ((Tree leaf (node (value String) (children TreeList)))
(TreeList nil (cons (car Tree) (cdr TreeList)))))
(declare-const list TreeList)
(declare-const fact1 Tree)
(declare-const fact2 Tree)
(assert (not (is-leaf fact1)))
(assert (not (is-leaf fact2)))
(assert (not (= list nil)))
(assert (= (value fact1) "mann"))
(assert (= (value fact2) "adam"))
(assert (= (children fact1) list))
(assert (= fact2 (car list)))
(declare-const list2 TreeList)
(declare-const fact3 Tree)
(declare-const fact4 Tree)
(assert (not (is-leaf fact3)))
(assert (not (is-leaf fact4)))
(assert (not (= list2 nil)))
(assert (= (value fact3) "mensch"))
(assert (= (value fact4) "adam"))
(assert (= (children fact3) list2))
(assert (= fact4 (car list2)))
(declare-fun rules (Tree) Bool)
(assert (= (rules fact1) true))
(assert (=> (rules fact1) (rules fact3)))
(check-sat)
(get-model)
The problem is I need the function "rules" to return false, whenever the argument is not one of the trees I have asserted rules for it to be true, but I can't find a way to edit the last "else" in the function. (get-model) seems to always use the most common answer of the function as it's answer if none of the rules work and since I only have rules for trees which make the answer true it uses true for the else as well, but I can't use the function that way.
It seems you're trying to model Prolog like "closed-world" assumption here. This is not how SMT solvers work: They will find a model that will satisfy all the requirements, and everything else is fair game. That is, if you want no other value to satisfy your rules, then you have to state that explicitly.
You might want to look at datalog modeling though, which seems closer to what you are trying to express: https://rise4fun.com/z3/tutorialcontent/fixedpoints
Related
I defined my own version of list concat based on Haskell as below:
(declare-datatypes ((MyList 1))
((par (T) ((cons (head T) (tail (MyList T))) (nil)))))
(declare-fun my-concat ( (MyList T1) (MyList T1) ) (MyList T1))
(assert (forall ((xs (MyList T1)) (ys (MyList T1)) (x T1))
(ite (= (as nil (MyList T1)) xs)
(= (my-concat xs ys) ys)
(= (my-concat (cons x xs) ys) (cons x (my-concat xs ys))))))
I am wondering why z3 is not able to reason about the following?
(assert (not (= (my-concat (cons 4 (as nil (MyList Int))) (as nil (MyList Int)))
(cons 4 (as nil (MyList Int))))))
(check-sat) ; runs forever
Loading your file
If I load your program to z3 as you've given it, it says:
(error "line 3 column 33: Parsing function declaration. Expecting sort list '(': unknown sort 'T1'")
(error "line 4 column 29: invalid sorted variables: unknown sort 'T1'")
(error "line 8 column 79: unknown function/constant my-concat")
That's because you cannot define "polymorphic" functions in SMTLib. At the user-level, only fully monomorphic functions are allowed. (Though internally SMTLib does provide polymorphic constants, there's no way for the user to actually create any polymorphic constants.)
So, I'm not sure how you got to load that file.
Monomorphisation
Looks like you only care about integer-lists anyhow, so let's just modify our program to work on lists of integers. This process is called monomorphisation, and is usually automatically done by some front-end tool before you even get to the SMT-solver, depending on what framework you're working in. That's just a fancy way of saying create instances of all your polymorphic constants at the mono-types they are used. (Since you mentioned Haskell, I'll just throw in that while monomorphisation is usually possible, it's not always feasible: There might be way too many variants to generate making it impractical. Also, if you have polymoprhic-recursion then monomorphisation doesn't work. But that's a digression for the time being.)
If I monomorphise your program to Int's only, I get:
(declare-datatypes ((MyList 1))
((par (T) ((cons (head T) (tail (MyList T))) (nil)))))
(declare-fun my-concat ( (MyList Int) (MyList Int) ) (MyList Int))
(assert (forall ((xs (MyList Int)) (ys (MyList Int)) (x Int))
(ite (= (as nil (MyList Int)) xs)
(= (my-concat xs ys) ys)
(= (my-concat (cons x xs) ys) (cons x (my-concat xs ys))))))
(assert (not (= (my-concat (cons 4 (as nil (MyList Int))) (as nil (MyList Int)))
(cons 4 (as nil (MyList Int))))))
(check-sat)
(get-info :reason-unknown)
When I run z3 on this, I get:
unknown
(:reason-unknown "smt tactic failed to show goal to be sat/unsat (incomplete quantifiers)")
So, it doesn't loop at all like you mentioned; but your file didn't really load in the first place. So maybe you were working with some other contents in the file as well, but that's a digression for the current question.
But it still doesn't prove it!
Of course, you wanted unsat for this trivial formula! But z3 said it's too hard for it to deal with. The reason-unknown is "incomplete quantifiers." What does that mean?
In short, SMTLib is essentially logic of many-sorted first order formulas. Solvers are "complete" for the quantifier-free fragment of this logic. But adding quantifiers makes the logic semi-decidable. What that means is that if you give enough resources, and if smart heuristics are in play, the solver will eventually say sat for a satisifiable formula, but it may loop forever if given an unsat one. (It might get lucky and say unsat as well, but most likely it'll loop.)
There are very good reasons why above is the case, but keep in mind that this has nothing to do with z3: First-order logic with quantifiers is semi-decidable. (Here's a good place to start reading: https://en.wikipedia.org/wiki/Decidability_(logic)#Semidecidability)
What usually happens in practice, however, is that the solver will not even answer sat, and will simply give up and say unknown as z3 did above. Quantifiers are simply beyond the means of SMT-solvers in general. You can try using patterns (search stack-overflow for quantifiers and pattern triggers), but that usually is futile as patterns and triggers are tricky to work with and they can be quite brittle.
What course of action do you have:
Honestly, this is one of those cases where "give up" is good advice. An SMT solver is just not a good fit for this sort of a problem. Use a theorem-prover like Isabelle, ACL2, Coq, HOL, HOL-Light, Lean, ... where you can express quantifiers and recursive functions and reason with them. They are built for this sort of thing. Don't expect your SMT solver to handle these sorts of queries. It's just not the right match.
Still, is there anything I can do?
You can try SMTLib's recursive-function definition facilities. You'd write:
(declare-datatypes ((MyList 1))
((par (T) ((cons (head T) (tail (MyList T))) (nil)))))
(define-fun-rec my-concat ((xs (MyList Int)) (ys (MyList Int))) (MyList Int)
(ite (= (as nil (MyList Int)) xs)
ys
(cons (head xs) (my-concat (tail xs) ys))))
(assert (not (= (my-concat (cons 4 (as nil (MyList Int))) (as nil (MyList Int)))
(cons 4 (as nil (MyList Int))))))
(check-sat)
Note the define-fun-rec construct, which allows for recursive definitions.
And voila, we get:
unsat
But this does not mean z3 will be able to prove arbitrary theorems regarding this concat function. If you try anything that requires induction, it'll either give up (saying unknown) or loop-forever. Of course, as the capabilities improve some induction-like proofs might be possible in z3 or other SMT-solvers, but that's really beyond what they're designed for. So, use with caution.
You can read more about recursive definitions in Section 4.2.3 of http://smtlib.cs.uiowa.edu/papers/smt-lib-reference-v2.6-r2017-07-18.pdf
To sum up
Do not use an SMT-solver for reasoning with quantified or recursive definitions. They just don't have the necessary power to deal with such problems, and it's unlikely they'll ever get there. Use a proper theorem prover for these tasks. Note that most theorem-provers use SMT-solvers as underlying tactics, so you get the best of both worlds: You can do some manual work to guide the inductive proof, and have the prover use an SMT-solver to handle most of the goals automatically for you. Here's a good paper to read to get you started on the details: https://people.mpi-inf.mpg.de/~jblanche/jar-smt.pdf
alias makes valid points, but I don't completely agree with the final conclusion to " not use an SMT-solver for reasoning with quantified or recursive definitions". In the end, it depends on which kind of properties you need to reason about, what kind of answers you need (is unsat/unknown OK, or do you need unsat/sat and a model?), and how much work you're willing to invest :-)
For example, SMT-based program verifiers such as Dafny and Viper quickly verify the following assertions about lists:
assert [4] + [] == [4]; // holds
assert [4,1] + [1,4] == [4,1,1,4]; // holds
assert [4] + [1] == [1,4]; // fails
Both tools can be used online, but the websites are rather slow and not terribly reliable. You can find the Dafny example here and the Viper example here.
Here is the relevant SMT code that Viper generates:
(set-option :auto_config false) ; Usually a good idea
(set-option :smt.mbqi false)
;; The following definitions are an excerpt of Viper's sequence axiomatisation,
;; which is based on Dafny's sequence axiomatisation.
;; See also:
;; https://github.com/dafny-lang/dafny
;; http://viper.ethz.ch
(declare-sort Seq<Int>) ;; Monomorphised sort of integer sequences
(declare-const Seq_empty Seq<Int>)
(declare-fun Seq_length (Seq<Int>) Int)
(declare-fun Seq_singleton (Int) Seq<Int>)
(declare-fun Seq_index (Seq<Int> Int) Int)
(declare-fun Seq_append (Seq<Int> Seq<Int>) Seq<Int>)
(declare-fun Seq_equal (Seq<Int> Seq<Int>) Bool)
(assert (forall ((s Seq<Int>)) (!
(<= 0 (Seq_length s))
:pattern ((Seq_length s))
)))
(assert (= (Seq_length (as Seq_empty Seq<Int>)) 0))
(assert (forall ((s1 Seq<Int>) (s2 Seq<Int>)) (!
(implies
(and
(not (= s1 (as Seq_empty Seq<Int>)))
(not (= s2 (as Seq_empty Seq<Int>))))
(= (Seq_length (Seq_append s1 s2)) (+ (Seq_length s1) (Seq_length s2))))
:pattern ((Seq_length (Seq_append s1 s2)))
)))
(assert (forall ((s Seq<Int>)) (!
(= (Seq_append (as Seq_empty Seq<Int>) s) s)
:pattern ((Seq_append (as Seq_empty Seq<Int>) s))
)))
(assert (forall ((s Seq<Int>)) (!
(= (Seq_append s (as Seq_empty Seq<Int>)) s)
:pattern ((Seq_append s (as Seq_empty Seq<Int>)))
)))
(assert (forall ((s1 Seq<Int>) (s2 Seq<Int>) (i Int)) (!
(implies
(and
(not (= s1 (as Seq_empty Seq<Int>)))
(not (= s2 (as Seq_empty Seq<Int>))))
(ite
(< i (Seq_length s1))
(= (Seq_index (Seq_append s1 s2) i) (Seq_index s1 i))
(= (Seq_index (Seq_append s1 s2) i) (Seq_index s2 (- i (Seq_length s1))))))
:pattern ((Seq_index (Seq_append s1 s2) i))
:pattern ((Seq_index s1 i) (Seq_append s1 s2))
)))
(assert (forall ((s1 Seq<Int>) (s2 Seq<Int>)) (!
(=
(Seq_equal s1 s2)
(and
(= (Seq_length s1) (Seq_length s2))
(forall ((i Int)) (!
(implies
(and (<= 0 i) (< i (Seq_length s1)))
(= (Seq_index s1 i) (Seq_index s2 i)))
:pattern ((Seq_index s1 i))
:pattern ((Seq_index s2 i))
))))
:pattern ((Seq_equal s1 s2))
)))
(assert (forall ((s1 Seq<Int>) (s2 Seq<Int>)) (!
(implies (Seq_equal s1 s2) (= s1 s2))
:pattern ((Seq_equal s1 s2))
)))
; ------------------------------------------------------------
; assert Seq(4) ++ Seq[Int]() == Seq(4)
(push)
(assert (not
(Seq_equal
(Seq_append (Seq_singleton 4) Seq_empty)
(Seq_singleton 4))))
(check-sat) ; unsat -- good!
(pop)
; assert Seq(4, 1) ++ Seq(1, 4) == Seq(4, 1, 1, 4)
(push)
(assert (not
(Seq_equal
(Seq_append
(Seq_append (Seq_singleton 4) (Seq_singleton 1))
(Seq_append (Seq_singleton 1) (Seq_singleton 4)))
(Seq_append
(Seq_append
(Seq_append (Seq_singleton 4) (Seq_singleton 1))
(Seq_singleton 1))
(Seq_singleton 4)))))
(check-sat) ; unsat -- good!
(pop)
; assert Seq(4) ++ Seq(1) == Seq(1, 4)
(push)
(assert (not (Seq_equal
(Seq_append (Seq_singleton 4) (Seq_singleton 1))
(Seq_append (Seq_singleton 1) (Seq_singleton 4)))))
(check-sat) ; unknown -- OK, since property doesn't hold
(pop)
(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.
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).
I have defined my own booleans, called boolean is SMT2, and the AND function boolean_and over them.
My conjecture is that AND is commutative:
(declare-sort boolean)
(declare-const sk_x boolean)
(declare-const sk_y boolean)
(declare-const boolean_false boolean)
(declare-const boolean_true boolean)
(declare-fun boolean_and (boolean boolean) boolean)
;; axiomatize booleans: false /= true and every bool is true or false
(assert (forall ((x boolean)) (or (= x boolean_false)
(= x boolean_true))))
(assert (not (= boolean_false boolean_true)))
;; definition of AND
(assert (forall ((a boolean)) (= (boolean_and boolean_false a) boolean_false)))
(assert (forall ((a boolean)) (= (boolean_and boolean_true a) a)))
;; try to prove that AND is commutative
(assert (not (= (boolean_and sk_x sk_y)
(boolean_and sk_y sk_x))))
(check-sat)
However, z3 reports unknown on this problem after a while, even though I
thought it should be able to use my case split assertions on the skolemised
variables sk_x and sk_y.
Curiously, if I remove my boolean axiomatization and replace it with
declare-datatypes, z3 will report unsat:
(declare-datatypes () ((boolean (boolean_true) (boolean_false))))
(declare-const sk_x boolean)
(declare-const sk_y boolean)
(declare-fun boolean_and (boolean boolean) boolean)
(assert (forall ((a boolean)) (= (boolean_and boolean_false a) boolean_false)))
(assert (forall ((a boolean)) (= (boolean_and boolean_true a) a)))
(assert (not (= (boolean_and sk_x sk_y)
(boolean_and sk_y sk_x))))
(check-sat)
What am I doing wrong? How can I get z3 to case split using my axiomatization?
You are not doing anything wrong. The official release (v4.3.1) may fail on problems containing cardinality constraints such as
(assert (forall ((x boolean)) (or (= x boolean_false)
(= x boolean_true))))
This constraint is asserting that the uninterpreted sort boolean has at most two elements.
I fixed this problem. The fix is already available in the unstable branch.
Here are some instructions on how to compile the unstable branch.
Tomorrow, the nightly builds will also contain this fix.
how can I make a datatype that contains a set of another objects. Basically, I am doing the following code:
(define-sort Set(T) (Array Int T))
(declare-datatypes () ((A f1 (cons (value Int) (b (Set B))))
(B f2 (cons (id Int) (a (Set A))))
))
But Z3 tells me unknown sort for A and B. If I remove "Set" it works just as the guide states.
I was trying to use List instead but it does not work. Anyone knows how to make it work?
You are addressing a question that comes up on a regular basis:
how can I mix data-types and arrays (as sets, multi-sets or
data-types in the range)?
As stated above Z3 does not support mixing data-types
and arrays in a single declaration.
A solution is to develop a custom solver for the
mixed datatype + array theory. Z3 contains programmatic
APIs for developing custom solvers.
It is still useful to develop this example
to illustrate the capabilities and limitations
of encoding theories with quantifiers and triggers.
Let me simplify your example by just using A.
As a work-around you can define an auxiliary sort.
The workaround is not ideal, though. It illustrates some
axiom 'hacking'. It relies on the operational semantics
of how quantifiers are instantiated during search.
(set-option :model true) ; We are going to display models.
(set-option :auto-config false)
(set-option :mbqi false) ; Model-based quantifier instantiation is too powerful here
(declare-sort SetA) ; Declare a custom fresh sort SetA
(declare-datatypes () ((A f1 (cons (value Int) (a SetA)))))
(define-sort Set (T) (Array T Bool))
Then define bijections between (Set A), SetA.
(declare-fun injSA ((Set A)) SetA)
(declare-fun projSA (SetA) (Set A))
(assert (forall ((x SetA)) (= (injSA (projSA x)) x)))
(assert (forall ((x (Set A))) (= (projSA (injSA x)) x)))
This is almost what the data-type declaration states.
To enforce well-foundedness you can associate an ordinal with members of A
and enforce that members of SetA are smaller in the well-founded ordering:
(declare-const v Int)
(declare-const s1 SetA)
(declare-const a1 A)
(declare-const sa1 (Set A))
(declare-const s2 SetA)
(declare-const a2 A)
(declare-const sa2 (Set A))
With the axioms so far, a1 can be a member of itself.
(push)
(assert (select sa1 a1))
(assert (= s1 (injSA sa1)))
(assert (= a1 (cons v s1)))
(check-sat)
(get-model)
(pop)
We now associate an ordinal number with the members of A.
(declare-fun ord (A) Int)
(assert (forall ((x SetA) (v Int) (a A))
(=> (select (projSA x) a)
(> (ord (cons v x)) (ord a)))))
(assert (forall ((x A)) (> (ord x) 0)))
By default quantifier instantiation in Z3 is pattern-based.
The first quantified assert above will not be instantiated on all
relevant instances. One can instead assert:
(assert (forall ((x1 SetA) (x2 (Set A)) (v Int) (a A))
(! (=> (and (= (projSA x1) x2) (select x2 a))
(> (ord (cons v x1)) (ord a)))
:pattern ((select x2 a) (cons v x1)))))
Axioms like these, that use two patterns (called a multi-pattern)
are quite expensive. They produce instantiations for every pair
of (select x2 a) and (cons v x1)
The membership constraint from before is now unsatisfiable.
(push)
(assert (select sa1 a1))
(assert (= s1 (injSA sa1)))
(assert (= a1 (cons v s1)))
(check-sat)
(pop)
but models are not necessarily well formed yet.
the default value of the set is 'true', which
would mean that the model implies there is a membership cycle
when there isn't one.
(push)
(assert (not (= (cons v s1) a1)))
(assert (= (projSA s1) sa1))
(assert (select sa1 a1))
(check-sat)
(get-model)
(pop)
We can approximate more faithful models by using
the following approach to enforce that sets that are
used in data-types are finite.
For example, whenever there is a membership check on a set x2,
we enforce that the 'default' value of the set is 'false'.
(assert (forall ((x2 (Set A)) (a A))
(! (not (default x2))
:pattern ((select x2 a)))))
Alternatively, whenever a set occurs in a data-type constructor
it is finite
(assert (forall ((v Int) (x1 SetA))
(! (not (default (projSA x1)))
:pattern ((cons v x1)))))
(push)
(assert (not (= (cons v s1) a1)))
(assert (= (projSA s1) sa1))
(assert (select sa1 a1))
(check-sat)
(get-model)
(pop)
Throughout the inclusion of additional axioms,
Z3 produces the answer 'unknown' and furthermore
the model that is produced indicates that the domain SetA
is finite (a singleton). So while we could patch the defaults
this model still does not satisfy the axioms. It satisfies
the axioms modulo instantiation only.
This is not supported in Z3. You can use arrays in datatype declarations, but they can't contain "references" to the datatypes you are declaring. For example, it is ok to use (Set Int).