(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).
Queries:
gopher(g,a).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:gopher1(cons(cons(cons(T34, T35), T36), T37), T39) :- gopher1(cons(T34, cons(T35, cons(T36, T37))), T39).
Clauses:
gopherc1(nil, nil).
gopherc1(cons(nil, T4), cons(nil, T4)).
gopherc1(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23))).
gopherc1(cons(cons(cons(T34, T35), T36), T37), T39) :- gopherc1(cons(T34, cons(T35, cons(T36, T37))), T39).
Afs:
gopher1(x1, x2) = gopher1(x1)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:gopher1_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
GOPHER1_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → U1_GA(T34, T35, T36, T37, T39, gopher1_in_ga(cons(T34, cons(T35, cons(T36, T37))), T39))
GOPHER1_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → GOPHER1_IN_GA(cons(T34, cons(T35, cons(T36, T37))), T39)
R is empty.
The argument filtering Pi contains the following mapping:
gopher1_in_ga(x1, x2) = gopher1_in_ga(x1)
cons(x1, x2) = cons(x1, x2)
GOPHER1_IN_GA(x1, x2) = GOPHER1_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:
GOPHER1_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → U1_GA(T34, T35, T36, T37, T39, gopher1_in_ga(cons(T34, cons(T35, cons(T36, T37))), T39))
GOPHER1_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → GOPHER1_IN_GA(cons(T34, cons(T35, cons(T36, T37))), T39)
R is empty.
The argument filtering Pi contains the following mapping:
gopher1_in_ga(x1, x2) = gopher1_in_ga(x1)
cons(x1, x2) = cons(x1, x2)
GOPHER1_IN_GA(x1, x2) = GOPHER1_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:
GOPHER1_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → GOPHER1_IN_GA(cons(T34, cons(T35, cons(T36, T37))), T39)
R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x1, x2)
GOPHER1_IN_GA(x1, x2) = GOPHER1_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:
GOPHER1_IN_GA(cons(cons(cons(T34, T35), T36), T37)) → GOPHER1_IN_GA(cons(T34, cons(T35, cons(T36, T37))))
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:
GOPHER1_IN_GA(cons(cons(cons(T34, T35), T36), T37)) → GOPHER1_IN_GA(cons(T34, cons(T35, cons(T36, T37))))
Used ordering: Polynomial interpretation [POLO]:
POL(GOPHER1_IN_GA(x1)) = 2·x1
POL(cons(x1, x2)) = 2 + 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.