(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).
Queries:
countstack(g,a).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:
countstack1(empty, 0).
countstack1(push(nil, T10), T5) :- countstack1(T10, T5).
countstack1(push(cons(T24, T25), T26), s(T13)) :- countstack1(push(T24, push(T25, T26)), T13).
Queries:
countstack1(g,a).
(3) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
countstack1_in: (b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
countstack1_in_ga(empty, 0) → countstack1_out_ga(empty, 0)
countstack1_in_ga(push(nil, T10), T5) → U1_ga(T10, T5, countstack1_in_ga(T10, T5))
countstack1_in_ga(push(cons(T24, T25), T26), s(T13)) → U2_ga(T24, T25, T26, T13, countstack1_in_ga(push(T24, push(T25, T26)), T13))
U2_ga(T24, T25, T26, T13, countstack1_out_ga(push(T24, push(T25, T26)), T13)) → countstack1_out_ga(push(cons(T24, T25), T26), s(T13))
U1_ga(T10, T5, countstack1_out_ga(T10, T5)) → countstack1_out_ga(push(nil, T10), T5)
The argument filtering Pi contains the following mapping:
countstack1_in_ga(
x1,
x2) =
countstack1_in_ga(
x1)
empty =
empty
countstack1_out_ga(
x1,
x2) =
countstack1_out_ga(
x2)
push(
x1,
x2) =
push(
x1,
x2)
nil =
nil
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
cons(
x1,
x2) =
cons(
x1,
x2)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
s(
x1) =
s(
x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
countstack1_in_ga(empty, 0) → countstack1_out_ga(empty, 0)
countstack1_in_ga(push(nil, T10), T5) → U1_ga(T10, T5, countstack1_in_ga(T10, T5))
countstack1_in_ga(push(cons(T24, T25), T26), s(T13)) → U2_ga(T24, T25, T26, T13, countstack1_in_ga(push(T24, push(T25, T26)), T13))
U2_ga(T24, T25, T26, T13, countstack1_out_ga(push(T24, push(T25, T26)), T13)) → countstack1_out_ga(push(cons(T24, T25), T26), s(T13))
U1_ga(T10, T5, countstack1_out_ga(T10, T5)) → countstack1_out_ga(push(nil, T10), T5)
The argument filtering Pi contains the following mapping:
countstack1_in_ga(
x1,
x2) =
countstack1_in_ga(
x1)
empty =
empty
countstack1_out_ga(
x1,
x2) =
countstack1_out_ga(
x2)
push(
x1,
x2) =
push(
x1,
x2)
nil =
nil
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
cons(
x1,
x2) =
cons(
x1,
x2)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
s(
x1) =
s(
x1)
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
COUNTSTACK1_IN_GA(push(nil, T10), T5) → U1_GA(T10, T5, countstack1_in_ga(T10, T5))
COUNTSTACK1_IN_GA(push(nil, T10), T5) → COUNTSTACK1_IN_GA(T10, T5)
COUNTSTACK1_IN_GA(push(cons(T24, T25), T26), s(T13)) → U2_GA(T24, T25, T26, T13, countstack1_in_ga(push(T24, push(T25, T26)), T13))
COUNTSTACK1_IN_GA(push(cons(T24, T25), T26), s(T13)) → COUNTSTACK1_IN_GA(push(T24, push(T25, T26)), T13)
The TRS R consists of the following rules:
countstack1_in_ga(empty, 0) → countstack1_out_ga(empty, 0)
countstack1_in_ga(push(nil, T10), T5) → U1_ga(T10, T5, countstack1_in_ga(T10, T5))
countstack1_in_ga(push(cons(T24, T25), T26), s(T13)) → U2_ga(T24, T25, T26, T13, countstack1_in_ga(push(T24, push(T25, T26)), T13))
U2_ga(T24, T25, T26, T13, countstack1_out_ga(push(T24, push(T25, T26)), T13)) → countstack1_out_ga(push(cons(T24, T25), T26), s(T13))
U1_ga(T10, T5, countstack1_out_ga(T10, T5)) → countstack1_out_ga(push(nil, T10), T5)
The argument filtering Pi contains the following mapping:
countstack1_in_ga(
x1,
x2) =
countstack1_in_ga(
x1)
empty =
empty
countstack1_out_ga(
x1,
x2) =
countstack1_out_ga(
x2)
push(
x1,
x2) =
push(
x1,
x2)
nil =
nil
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
cons(
x1,
x2) =
cons(
x1,
x2)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
s(
x1) =
s(
x1)
COUNTSTACK1_IN_GA(
x1,
x2) =
COUNTSTACK1_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x3)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x5)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
COUNTSTACK1_IN_GA(push(nil, T10), T5) → U1_GA(T10, T5, countstack1_in_ga(T10, T5))
COUNTSTACK1_IN_GA(push(nil, T10), T5) → COUNTSTACK1_IN_GA(T10, T5)
COUNTSTACK1_IN_GA(push(cons(T24, T25), T26), s(T13)) → U2_GA(T24, T25, T26, T13, countstack1_in_ga(push(T24, push(T25, T26)), T13))
COUNTSTACK1_IN_GA(push(cons(T24, T25), T26), s(T13)) → COUNTSTACK1_IN_GA(push(T24, push(T25, T26)), T13)
The TRS R consists of the following rules:
countstack1_in_ga(empty, 0) → countstack1_out_ga(empty, 0)
countstack1_in_ga(push(nil, T10), T5) → U1_ga(T10, T5, countstack1_in_ga(T10, T5))
countstack1_in_ga(push(cons(T24, T25), T26), s(T13)) → U2_ga(T24, T25, T26, T13, countstack1_in_ga(push(T24, push(T25, T26)), T13))
U2_ga(T24, T25, T26, T13, countstack1_out_ga(push(T24, push(T25, T26)), T13)) → countstack1_out_ga(push(cons(T24, T25), T26), s(T13))
U1_ga(T10, T5, countstack1_out_ga(T10, T5)) → countstack1_out_ga(push(nil, T10), T5)
The argument filtering Pi contains the following mapping:
countstack1_in_ga(
x1,
x2) =
countstack1_in_ga(
x1)
empty =
empty
countstack1_out_ga(
x1,
x2) =
countstack1_out_ga(
x2)
push(
x1,
x2) =
push(
x1,
x2)
nil =
nil
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
cons(
x1,
x2) =
cons(
x1,
x2)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
s(
x1) =
s(
x1)
COUNTSTACK1_IN_GA(
x1,
x2) =
COUNTSTACK1_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x3)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x5)
We have to consider all (P,R,Pi)-chains
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
COUNTSTACK1_IN_GA(push(cons(T24, T25), T26), s(T13)) → COUNTSTACK1_IN_GA(push(T24, push(T25, T26)), T13)
COUNTSTACK1_IN_GA(push(nil, T10), T5) → COUNTSTACK1_IN_GA(T10, T5)
The TRS R consists of the following rules:
countstack1_in_ga(empty, 0) → countstack1_out_ga(empty, 0)
countstack1_in_ga(push(nil, T10), T5) → U1_ga(T10, T5, countstack1_in_ga(T10, T5))
countstack1_in_ga(push(cons(T24, T25), T26), s(T13)) → U2_ga(T24, T25, T26, T13, countstack1_in_ga(push(T24, push(T25, T26)), T13))
U2_ga(T24, T25, T26, T13, countstack1_out_ga(push(T24, push(T25, T26)), T13)) → countstack1_out_ga(push(cons(T24, T25), T26), s(T13))
U1_ga(T10, T5, countstack1_out_ga(T10, T5)) → countstack1_out_ga(push(nil, T10), T5)
The argument filtering Pi contains the following mapping:
countstack1_in_ga(
x1,
x2) =
countstack1_in_ga(
x1)
empty =
empty
countstack1_out_ga(
x1,
x2) =
countstack1_out_ga(
x2)
push(
x1,
x2) =
push(
x1,
x2)
nil =
nil
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
cons(
x1,
x2) =
cons(
x1,
x2)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
s(
x1) =
s(
x1)
COUNTSTACK1_IN_GA(
x1,
x2) =
COUNTSTACK1_IN_GA(
x1)
We have to consider all (P,R,Pi)-chains
(9) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(10) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
COUNTSTACK1_IN_GA(push(cons(T24, T25), T26), s(T13)) → COUNTSTACK1_IN_GA(push(T24, push(T25, T26)), T13)
COUNTSTACK1_IN_GA(push(nil, T10), T5) → COUNTSTACK1_IN_GA(T10, T5)
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)
COUNTSTACK1_IN_GA(
x1,
x2) =
COUNTSTACK1_IN_GA(
x1)
We have to consider all (P,R,Pi)-chains
(11) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COUNTSTACK1_IN_GA(push(cons(T24, T25), T26)) → COUNTSTACK1_IN_GA(push(T24, push(T25, T26)))
COUNTSTACK1_IN_GA(push(nil, T10)) → COUNTSTACK1_IN_GA(T10)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(13) 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:
COUNTSTACK1_IN_GA(push(cons(T24, T25), T26)) → COUNTSTACK1_IN_GA(push(T24, push(T25, T26)))
COUNTSTACK1_IN_GA(push(nil, T10)) → COUNTSTACK1_IN_GA(T10)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(COUNTSTACK1_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
(14) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(15) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(16) TRUE