(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).

Query: fl(g,g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

(2) Obligation:

Triples:

pB([], X1, X1, X2, X3) :- flA(X2, X1, X3).
pB(.(X1, X2), X3, .(X1, X4), X5, X6) :- pB(X2, X3, X4, X5, X6).
flA(.(X1, X2), X3, s(X4)) :- pB(X1, X5, X3, X2, X4).

Clauses:

flcA([], [], 0).
flcA(.(X1, X2), X3, s(X4)) :- qcB(X1, X5, X3, X2, X4).
qcB([], X1, X1, X2, X3) :- flcA(X2, X1, X3).
qcB(.(X1, X2), X3, .(X1, X4), X5, X6) :- qcB(X2, X3, X4, X5, X6).

Afs:

flA(x1, x2, x3)  =  flA(x1, x2)

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

FLA_IN_GGA(.(X1, X2), X3, s(X4)) → U3_GGA(X1, X2, X3, X4, pB_in_gagga(X1, X5, X3, X2, X4))
FLA_IN_GGA(.(X1, X2), X3, s(X4)) → PB_IN_GAGGA(X1, X5, X3, X2, X4)
PB_IN_GAGGA([], X1, X1, X2, X3) → U1_GAGGA(X1, X2, X3, flA_in_gga(X2, X1, X3))
PB_IN_GAGGA([], X1, X1, X2, X3) → FLA_IN_GGA(X2, X1, X3)
PB_IN_GAGGA(.(X1, X2), X3, .(X1, X4), X5, X6) → U2_GAGGA(X1, X2, X3, X4, X5, X6, pB_in_gagga(X2, X3, X4, X5, X6))
PB_IN_GAGGA(.(X1, X2), X3, .(X1, X4), X5, X6) → PB_IN_GAGGA(X2, X3, X4, X5, X6)

R is empty.
The argument filtering Pi contains the following mapping:
flA_in_gga(x1, x2, x3)  =  flA_in_gga(x1, x2)
.(x1, x2)  =  .(x1, x2)
pB_in_gagga(x1, x2, x3, x4, x5)  =  pB_in_gagga(x1, x3, x4)
[]  =  []
s(x1)  =  s(x1)
FLA_IN_GGA(x1, x2, x3)  =  FLA_IN_GGA(x1, x2)
U3_GGA(x1, x2, x3, x4, x5)  =  U3_GGA(x1, x2, x3, x5)
PB_IN_GAGGA(x1, x2, x3, x4, x5)  =  PB_IN_GAGGA(x1, x3, x4)
U1_GAGGA(x1, x2, x3, x4)  =  U1_GAGGA(x1, x2, x4)
U2_GAGGA(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAGGA(x1, x2, x4, x5, 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:

FLA_IN_GGA(.(X1, X2), X3, s(X4)) → U3_GGA(X1, X2, X3, X4, pB_in_gagga(X1, X5, X3, X2, X4))
FLA_IN_GGA(.(X1, X2), X3, s(X4)) → PB_IN_GAGGA(X1, X5, X3, X2, X4)
PB_IN_GAGGA([], X1, X1, X2, X3) → U1_GAGGA(X1, X2, X3, flA_in_gga(X2, X1, X3))
PB_IN_GAGGA([], X1, X1, X2, X3) → FLA_IN_GGA(X2, X1, X3)
PB_IN_GAGGA(.(X1, X2), X3, .(X1, X4), X5, X6) → U2_GAGGA(X1, X2, X3, X4, X5, X6, pB_in_gagga(X2, X3, X4, X5, X6))
PB_IN_GAGGA(.(X1, X2), X3, .(X1, X4), X5, X6) → PB_IN_GAGGA(X2, X3, X4, X5, X6)

R is empty.
The argument filtering Pi contains the following mapping:
flA_in_gga(x1, x2, x3)  =  flA_in_gga(x1, x2)
.(x1, x2)  =  .(x1, x2)
pB_in_gagga(x1, x2, x3, x4, x5)  =  pB_in_gagga(x1, x3, x4)
[]  =  []
s(x1)  =  s(x1)
FLA_IN_GGA(x1, x2, x3)  =  FLA_IN_GGA(x1, x2)
U3_GGA(x1, x2, x3, x4, x5)  =  U3_GGA(x1, x2, x3, x5)
PB_IN_GAGGA(x1, x2, x3, x4, x5)  =  PB_IN_GAGGA(x1, x3, x4)
U1_GAGGA(x1, x2, x3, x4)  =  U1_GAGGA(x1, x2, x4)
U2_GAGGA(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAGGA(x1, x2, x4, x5, 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:

FLA_IN_GGA(.(X1, X2), X3, s(X4)) → PB_IN_GAGGA(X1, X5, X3, X2, X4)
PB_IN_GAGGA([], X1, X1, X2, X3) → FLA_IN_GGA(X2, X1, X3)
PB_IN_GAGGA(.(X1, X2), X3, .(X1, X4), X5, X6) → PB_IN_GAGGA(X2, X3, X4, X5, X6)

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

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:

FLA_IN_GGA(.(X1, X2), X3) → PB_IN_GAGGA(X1, X3, X2)
PB_IN_GAGGA([], X1, X2) → FLA_IN_GGA(X2, X1)
PB_IN_GAGGA(.(X1, X2), .(X1, X4), X5) → PB_IN_GAGGA(X2, X4, X5)

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:

  • PB_IN_GAGGA([], X1, X2) → FLA_IN_GGA(X2, X1)
    The graph contains the following edges 3 >= 1, 2 >= 2

  • PB_IN_GAGGA(.(X1, X2), .(X1, X4), X5) → PB_IN_GAGGA(X2, X4, X5)
    The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3

  • FLA_IN_GGA(.(X1, X2), X3) → PB_IN_GAGGA(X1, X3, X2)
    The graph contains the following edges 1 > 1, 2 >= 2, 1 > 3

(10) YES