(0) Obligation:
Clauses:
countstack(empty, 0).
countstack(S, X) :- ','(no(empty_stack(S)), ','(pop(S, nil), ','(popped(S, Pd), countstack(Pd, X)))).
countstack(S, s(X)) :- ','(no(empty_stack(S)), ','(pop(S, P), ','(no(empty_list(P)), ','(head(P, H), ','(tail(P, T), ','(popped(S, Pd), countstack(push(H, push(T, Pd)), X))))))).
pop(empty, X1).
pop(push(P, X2), P).
popped(empty, empty).
popped(push(X3, Pd), Pd).
head(nil, X4).
head(cons(H, X5), H).
tail(nil, nil).
tail(cons(X6, T), T).
empty_stack(empty).
empty_list(nil).
no(X) :- ','(X, ','(!, failure(a))).
no(X7).
failure(b).
Query: countstack(g,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
countstackA(push(nil, X1), X2) :- countstackA(X1, X2).
countstackA(push(cons(X1, X2), X3), s(X4)) :- countstackA(push(X1, push(X2, X3)), X4).
Clauses:
countstackcA(empty, 0).
countstackcA(push(nil, X1), X2) :- countstackcA(X1, X2).
countstackcA(push(cons(X1, X2), X3), s(X4)) :- countstackcA(push(X1, push(X2, X3)), X4).
Afs:
countstackA(x1, x2) = countstackA(x1)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
countstackA_in: (b,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
COUNTSTACKA_IN_GA(push(nil, X1), X2) → U1_GA(X1, X2, countstackA_in_ga(X1, X2))
COUNTSTACKA_IN_GA(push(nil, X1), X2) → COUNTSTACKA_IN_GA(X1, X2)
COUNTSTACKA_IN_GA(push(cons(X1, X2), X3), s(X4)) → U2_GA(X1, X2, X3, X4, countstackA_in_ga(push(X1, push(X2, X3)), X4))
COUNTSTACKA_IN_GA(push(cons(X1, X2), X3), s(X4)) → COUNTSTACKA_IN_GA(push(X1, push(X2, X3)), X4)
R is empty.
The argument filtering Pi contains the following mapping:
countstackA_in_ga(
x1,
x2) =
countstackA_in_ga(
x1)
push(
x1,
x2) =
push(
x1,
x2)
nil =
nil
cons(
x1,
x2) =
cons(
x1,
x2)
s(
x1) =
s(
x1)
COUNTSTACKA_IN_GA(
x1,
x2) =
COUNTSTACKA_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x1,
x3)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x1,
x2,
x3,
x5)
We have to consider all (P,R,Pi)-chains
Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
COUNTSTACKA_IN_GA(push(nil, X1), X2) → U1_GA(X1, X2, countstackA_in_ga(X1, X2))
COUNTSTACKA_IN_GA(push(nil, X1), X2) → COUNTSTACKA_IN_GA(X1, X2)
COUNTSTACKA_IN_GA(push(cons(X1, X2), X3), s(X4)) → U2_GA(X1, X2, X3, X4, countstackA_in_ga(push(X1, push(X2, X3)), X4))
COUNTSTACKA_IN_GA(push(cons(X1, X2), X3), s(X4)) → COUNTSTACKA_IN_GA(push(X1, push(X2, X3)), X4)
R is empty.
The argument filtering Pi contains the following mapping:
countstackA_in_ga(
x1,
x2) =
countstackA_in_ga(
x1)
push(
x1,
x2) =
push(
x1,
x2)
nil =
nil
cons(
x1,
x2) =
cons(
x1,
x2)
s(
x1) =
s(
x1)
COUNTSTACKA_IN_GA(
x1,
x2) =
COUNTSTACKA_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x1,
x3)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x1,
x2,
x3,
x5)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
COUNTSTACKA_IN_GA(push(cons(X1, X2), X3), s(X4)) → COUNTSTACKA_IN_GA(push(X1, push(X2, X3)), X4)
COUNTSTACKA_IN_GA(push(nil, X1), X2) → COUNTSTACKA_IN_GA(X1, X2)
R is empty.
The argument filtering Pi contains the following mapping:
push(
x1,
x2) =
push(
x1,
x2)
nil =
nil
cons(
x1,
x2) =
cons(
x1,
x2)
s(
x1) =
s(
x1)
COUNTSTACKA_IN_GA(
x1,
x2) =
COUNTSTACKA_IN_GA(
x1)
We have to consider all (P,R,Pi)-chains
(7) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COUNTSTACKA_IN_GA(push(cons(X1, X2), X3)) → COUNTSTACKA_IN_GA(push(X1, push(X2, X3)))
COUNTSTACKA_IN_GA(push(nil, X1)) → COUNTSTACKA_IN_GA(X1)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
COUNTSTACKA_IN_GA(push(cons(X1, X2), X3)) → COUNTSTACKA_IN_GA(push(X1, push(X2, X3)))
COUNTSTACKA_IN_GA(push(nil, X1)) → COUNTSTACKA_IN_GA(X1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(COUNTSTACKA_IN_GA(x1)) = 2·x1
POL(cons(x1, x2)) = 2 + 2·x1 + 2·x2
POL(nil) = 0
POL(push(x1, x2)) = 1 + 2·x1 + x2
(10) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(12) YES