(0) Obligation:

Clauses:

fl([], [], 0).
fl(.(E, X), R, s(Z)) :- ','(append(E, Y, R), fl(X, Y, Z)).
append([], X, X).
append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).

Queries:

fl(g,a,g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

p7([], T18, T18, T9, T11) :- fl1(T9, T18, T11).
p7(.(T25, T26), X43, .(T25, T28), T9, T11) :- p7(T26, X43, T28, T9, T11).
fl1(.(T8, T9), T12, s(T11)) :- p7(T8, X13, T12, T9, T11).

Clauses:

flc1([], [], 0).
flc1(.(T8, T9), T12, s(T11)) :- qc7(T8, X13, T12, T9, T11).
qc7([], T18, T18, T9, T11) :- flc1(T9, T18, T11).
qc7(.(T25, T26), X43, .(T25, T28), T9, T11) :- qc7(T26, X43, T28, T9, T11).

Afs:

fl1(x1, x2, x3)  =  fl1(x1, x3)

(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:
fl1_in: (b,f,b)
p7_in: (b,f,f,b,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

FL1_IN_GAG(.(T8, T9), T12, s(T11)) → U3_GAG(T8, T9, T12, T11, p7_in_gaagg(T8, X13, T12, T9, T11))
FL1_IN_GAG(.(T8, T9), T12, s(T11)) → P7_IN_GAAGG(T8, X13, T12, T9, T11)
P7_IN_GAAGG([], T18, T18, T9, T11) → U1_GAAGG(T18, T9, T11, fl1_in_gag(T9, T18, T11))
P7_IN_GAAGG([], T18, T18, T9, T11) → FL1_IN_GAG(T9, T18, T11)
P7_IN_GAAGG(.(T25, T26), X43, .(T25, T28), T9, T11) → U2_GAAGG(T25, T26, X43, T28, T9, T11, p7_in_gaagg(T26, X43, T28, T9, T11))
P7_IN_GAAGG(.(T25, T26), X43, .(T25, T28), T9, T11) → P7_IN_GAAGG(T26, X43, T28, T9, T11)

R is empty.
The argument filtering Pi contains the following mapping:
fl1_in_gag(x1, x2, x3)  =  fl1_in_gag(x1, x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
p7_in_gaagg(x1, x2, x3, x4, x5)  =  p7_in_gaagg(x1, x4, x5)
[]  =  []
FL1_IN_GAG(x1, x2, x3)  =  FL1_IN_GAG(x1, x3)
U3_GAG(x1, x2, x3, x4, x5)  =  U3_GAG(x1, x2, x4, x5)
P7_IN_GAAGG(x1, x2, x3, x4, x5)  =  P7_IN_GAAGG(x1, x4, x5)
U1_GAAGG(x1, x2, x3, x4)  =  U1_GAAGG(x2, x3, x4)
U2_GAAGG(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAAGG(x1, x2, x5, x6, x7)

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:

FL1_IN_GAG(.(T8, T9), T12, s(T11)) → U3_GAG(T8, T9, T12, T11, p7_in_gaagg(T8, X13, T12, T9, T11))
FL1_IN_GAG(.(T8, T9), T12, s(T11)) → P7_IN_GAAGG(T8, X13, T12, T9, T11)
P7_IN_GAAGG([], T18, T18, T9, T11) → U1_GAAGG(T18, T9, T11, fl1_in_gag(T9, T18, T11))
P7_IN_GAAGG([], T18, T18, T9, T11) → FL1_IN_GAG(T9, T18, T11)
P7_IN_GAAGG(.(T25, T26), X43, .(T25, T28), T9, T11) → U2_GAAGG(T25, T26, X43, T28, T9, T11, p7_in_gaagg(T26, X43, T28, T9, T11))
P7_IN_GAAGG(.(T25, T26), X43, .(T25, T28), T9, T11) → P7_IN_GAAGG(T26, X43, T28, T9, T11)

R is empty.
The argument filtering Pi contains the following mapping:
fl1_in_gag(x1, x2, x3)  =  fl1_in_gag(x1, x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
p7_in_gaagg(x1, x2, x3, x4, x5)  =  p7_in_gaagg(x1, x4, x5)
[]  =  []
FL1_IN_GAG(x1, x2, x3)  =  FL1_IN_GAG(x1, x3)
U3_GAG(x1, x2, x3, x4, x5)  =  U3_GAG(x1, x2, x4, x5)
P7_IN_GAAGG(x1, x2, x3, x4, x5)  =  P7_IN_GAAGG(x1, x4, x5)
U1_GAAGG(x1, x2, x3, x4)  =  U1_GAAGG(x2, x3, x4)
U2_GAAGG(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAAGG(x1, x2, x5, x6, x7)

We have to consider all (P,R,Pi)-chains

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

FL1_IN_GAG(.(T8, T9), T12, s(T11)) → P7_IN_GAAGG(T8, X13, T12, T9, T11)
P7_IN_GAAGG([], T18, T18, T9, T11) → FL1_IN_GAG(T9, T18, T11)
P7_IN_GAAGG(.(T25, T26), X43, .(T25, T28), T9, T11) → P7_IN_GAAGG(T26, X43, T28, T9, T11)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
[]  =  []
FL1_IN_GAG(x1, x2, x3)  =  FL1_IN_GAG(x1, x3)
P7_IN_GAAGG(x1, x2, x3, x4, x5)  =  P7_IN_GAAGG(x1, x4, x5)

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:

FL1_IN_GAG(.(T8, T9), s(T11)) → P7_IN_GAAGG(T8, T9, T11)
P7_IN_GAAGG([], T9, T11) → FL1_IN_GAG(T9, T11)
P7_IN_GAAGG(.(T25, T26), T9, T11) → P7_IN_GAAGG(T26, T9, T11)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • P7_IN_GAAGG([], T9, T11) → FL1_IN_GAG(T9, T11)
    The graph contains the following edges 2 >= 1, 3 >= 2

  • P7_IN_GAAGG(.(T25, T26), T9, T11) → P7_IN_GAAGG(T26, T9, T11)
    The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3

  • FL1_IN_GAG(.(T8, T9), s(T11)) → P7_IN_GAAGG(T8, T9, T11)
    The graph contains the following edges 1 > 1, 1 > 2, 2 > 3

(10) YES