Reputation: 1347
Our problem here is to show that
using Kleene algebras with test.
In the case when the value of b is preserved by p, we have the commutativity condition bp = pb; and the equivalence between the two programs is reduced to the equation
In the case when the value of b is not preserved by p, we have the commutativity condition pc = cp; and the equivalence between the two programs is reduced to the equation
I am trying to prove the first equation using the following SMT-LIB code
(declare-sort S)
(declare-fun sum (S S) S)
(declare-fun mult (S S) S)
(declare-fun neg (S) S)
(assert (forall ((x S) (y S) (z S)) (= (mult x (sum y z)) (sum (mult x y) (mult y z))) ) )
(assert (forall ((x S) (y S) (z S)) (= (mult (sum y z) x) (sum (mult y x) (mult z x))) ) )
(assert (forall ((x S) (y S) (z S)) (= (mult x (mult y z)) (mult (mult x y) z)) ))
(check-sat)
(push)
(declare-fun b () S)
(declare-fun p () S)
(declare-fun q () S)
(declare-fun r () S)
(assert (= (mult b p) (mult p b)) )
(check-sat)
(pop)
but I am obtaining timeout
; it is to say Z3 is not able to hand the commutativity condition bp = pb. Please run this example online here.
Z3 is not able to prove these equations but Mathematica and Reduce are able. Z3 is not so powerful as a theorem prover. Do you agree?
Upvotes: 3
Views: 365
Reputation: 1347
The second equation is indirectly proved using Z3 with the following procedure
This indirect procedure is implemented with the following SMT-LIB code
(declare-sort S)
(declare-fun e () S)
(declare-fun O () S)
(declare-fun mult (S S) S)
(declare-fun sum (S S) S)
(declare-fun leq (S S) Bool)
(declare-fun negation (S) S)
(declare-fun test (S) Bool)
(assert (forall ((x S) (y S)) (= (sum x y) (sum y x ))))
(assert (forall ((x S) (y S) (z S)) (= (sum (sum x y) z) (sum x (sum y z)))))
(assert (forall ((x S)) (= (sum O x) x)))
(assert (forall ((x S)) (= (sum x x) x)))
(assert (forall ((x S) (y S) (z S)) (= (mult (mult x y) z) (mult x (mult y z)))))
(assert (forall ((x S)) (= (mult e x) x)))
(assert (forall ((x S)) (= (mult x e) x)))
(assert (forall ((x S) (y S) (z S)) (= (mult x (sum y z) ) (sum (mult x y) (mult x z)))))
(assert (forall ((x S) (y S) (z S)) (= (mult (sum x y) z ) (sum (mult x z) (mult y z)))))
(assert (forall ((x S)) (= (mult x O) O)))
(assert (forall ((x S)) (= (mult O x) O)))
(assert (forall ((x S) (y S)) (= (leq x y) (= (sum x y) y))))
(assert (forall ((x S) (y S)) (=> (and (test x) (test y) ) (= (mult x y) (mult y x))) ) )
(assert (forall ((x S)) (=> (test x) (= (sum x (negation x)) e) )))
(assert (forall ((x S)) (=> (test x) (= (mult x (negation x)) O) )))
(assert (forall ((x S)) (=> (test x) (test (negation x)) )) )
(assert (forall ((x S)) (=> (test x) (= (mult x x) x) )) )
(check-sat)
(push)
(declare-fun c () S)
(declare-fun b () S)
(declare-fun p () S)
(declare-fun q () S)
(declare-fun r () S)
(check-sat)
(assert (not (=> (and (test b) (test c))
(= (mult (sum (mult b c) (mult (negation b) (negation c)))
(sum (mult b (mult p q)) (mult (negation b) (mult p r) ) ))
(sum (mult b (mult c (mult p q)))
(mult (negation b) (mult (negation c) (mult p r) ) ) ))) ) )
(check-sat)
(pop)
(echo "Proved: part 1")
(push)
;;
(assert (=> (test c) (= (mult p c) (mult c p)) ))
(assert (=> (test c) (= (mult p (negation c)) (mult (negation c) p))))
(check-sat)
(assert (not (=> (test c)
(= (mult (sum (mult b c) (mult (negation b) (negation c)))
(mult p (sum (mult c q) (mult (negation c) r))))
(sum (mult b (mult c (mult p q)))
(mult (negation b) (mult (negation c) (mult p r) ) ) ))) ) )
(check-sat)
(pop)
(echo "Proved: part 2")
and the corresponding output is
sat
sat
unsat
Proved: part 1
sat
unsat
Proved: part 2
This output is obtained locally. When the code is executed online the output is
sat
sat
unsat
Proved: part 1
sat
timeout
Please run this example online here
Z3 is not able to prove the second equation directly but Mathematica and Reduce ane able.
Upvotes: 0
Reputation: 1347
The first equation is proved using Z3 with the following SMT-LIB code
(declare-sort S)
(declare-fun e () S)
(declare-fun O () S)
(declare-fun mult (S S) S)
(declare-fun sum (S S) S)
(declare-fun leq (S S) Bool)
(declare-fun negation (S) S)
(declare-fun test (S) Bool)
(assert (forall ((x S) (y S)) (= (sum x y) (sum y x ))))
(assert (forall ((x S) (y S) (z S)) (= (sum (sum x y) z) (sum x (sum y z)))))
(assert (forall ((x S)) (= (sum O x) x)))
(assert (forall ((x S)) (= (sum x x) x)))
(assert (forall ((x S) (y S) (z S)) (= (mult (mult x y) z) (mult x (mult y z)))))
(assert (forall ((x S)) (= (mult e x) x)))
(assert (forall ((x S)) (= (mult x e) x)))
(assert (forall ((x S) (y S) (z S)) (= (mult x (sum y z) ) (sum (mult x y) (mult x z)))))
(assert (forall ((x S) (y S) (z S)) (= (mult (sum x y) z ) (sum (mult x z) (mult y z)))))
(assert (forall ((x S)) (= (mult x O) O)))
(assert (forall ((x S)) (= (mult O x) O)))
(assert (forall ((x S) (y S)) (= (leq x y) (= (sum x y) y))))
(assert (forall ((x S) (y S)) (=> (and (test x) (test y) ) (= (mult x y) (mult y x))) ) )
(assert (forall ((x S)) (=> (test x) (= (sum x (negation x)) e) )))
(assert (forall ((x S)) (=> (test x) (= (mult x (negation x)) O) )))
(check-sat)
(push)
;; bpq + b`pr = p(bq + b`r)
(declare-fun b () S)
(declare-fun p () S)
(declare-fun q () S)
(declare-fun r () S)
(assert (=> (test b) (= (mult p b) (mult b p)) ))
(assert (=> (test b) (= (mult p (negation b)) (mult (negation b) p))))
(check-sat)
(assert (not (=> (test b) (= (sum (mult b (mult p q)) (mult (negation b) (mult p r) ))
(mult p (sum (mult b q) (mult (negation b) r))))) ) )
(check-sat)
(pop)
(echo "Proved: bpq + b`pr = p(bq + b`r)")
The output is
sat
sat
unsat
Proved: bpq + b`pr = p(bq + b`r)
Please run this proof online here
Upvotes: 3