(0) Obligation:
Clauses:
gopher(nil, nil).
gopher(cons(nil, Y), cons(nil, Y)).
gopher(cons(cons(U, V), W), X) :- gopher(cons(U, cons(V, W)), X).
Query: gopher(g,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
gopherA(cons(cons(cons(X1, X2), X3), X4), X5) :- gopherA(cons(X1, cons(X2, cons(X3, X4))), X5).
Clauses:
gophercA(nil, nil).
gophercA(cons(nil, X1), cons(nil, X1)).
gophercA(cons(cons(nil, X1), X2), cons(nil, cons(X1, X2))).
gophercA(cons(cons(cons(X1, X2), X3), X4), X5) :- gophercA(cons(X1, cons(X2, cons(X3, X4))), X5).
Afs:
gopherA(x1, x2) = gopherA(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:
gopherA_in: (b,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → U1_GA(X1, X2, X3, X4, X5, gopherA_in_ga(cons(X1, cons(X2, cons(X3, X4))), X5))
GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))), X5)
R is empty.
The argument filtering Pi contains the following mapping:
gopherA_in_ga(
x1,
x2) =
gopherA_in_ga(
x1)
cons(
x1,
x2) =
cons(
x1,
x2)
GOPHERA_IN_GA(
x1,
x2) =
GOPHERA_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GA(
x1,
x2,
x3,
x4,
x6)
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:
GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → U1_GA(X1, X2, X3, X4, X5, gopherA_in_ga(cons(X1, cons(X2, cons(X3, X4))), X5))
GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))), X5)
R is empty.
The argument filtering Pi contains the following mapping:
gopherA_in_ga(
x1,
x2) =
gopherA_in_ga(
x1)
cons(
x1,
x2) =
cons(
x1,
x2)
GOPHERA_IN_GA(
x1,
x2) =
GOPHERA_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GA(
x1,
x2,
x3,
x4,
x6)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4), X5) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))), X5)
R is empty.
The argument filtering Pi contains the following mapping:
cons(
x1,
x2) =
cons(
x1,
x2)
GOPHERA_IN_GA(
x1,
x2) =
GOPHERA_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:
GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4)) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))))
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
GOPHERA_IN_GA(cons(cons(cons(X1, X2), X3), X4)) → GOPHERA_IN_GA(cons(X1, cons(X2, cons(X3, X4))))
Used ordering: Knuth-Bendix order [KBO] with precedence:
cons2 > GOPHERAINGA1
and weight map:
GOPHERA_IN_GA_1=1
cons_2=0
The variable weight is 1
(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