Left Termination of the query pattern
fl(g,a,a)
w.r.t. the given Prolog program could successfully be proven:

(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,a,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: fl(T1, T2, T3)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: fl(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | fl(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | fl([], T2, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: fl(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nfl(T1, T2, T3) !=? fl([], [], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: fl([], T2, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="B: ','(append(T8, X13, T12), fl(T9, X13, T13))\n\nT8 is ground\nT9 is ground\nX13 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(append(T8, X13, T12), fl(T9, X13, T13)), SCOPE: 2, CLAUSE: 2 | ','(append(T8, X13, T12), fl(T9, X13, T13)), SCOPE: 2, CLAUSE: 3\n\nT8 is ground\nT9 is ground\nX13 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(append(T8, X13, T12), fl(T9, X13, T13)), SCOPE: 2, CLAUSE: 2\n\nT8 is ground\nT9 is ground\nX13 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(append(T8, X13, T12), fl(T9, X13, T13)), SCOPE: 2, CLAUSE: 3\n\nT8 is ground\nT9 is ground\nX13 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: fl(T9, T19, T20)\n\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(append(T28, X38, T30), fl(T9, X38, T31))\n\nT9 is ground\nT28 is ground\nX38 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 16 [arrowhead = none ];
16 -> 2;

17 [label="EVAL with clause\nfl([], [], 0).\nand substitutionT1 -> [],\nT2 -> [],\nT3 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 17 [arrowhead = none ];
17 -> 3;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 18 [arrowhead = none ];
18 -> 4;

19 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 19 [arrowhead = none ];
19 -> 5;

20 [label="EVAL with clause\nfl(.(X9, X10), X11, s(X12)) :- ','(append(X9, X13, X11), fl(X10, X13, X12)).\nand substitutionX9 -> T8,\nX10 -> T9,\nT1 -> .(T8, T9),\nT2 -> T12,\nX11 -> T12,\nX12 -> T13,\nT3 -> s(T13),\nT10 -> T12,\nT11 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 20 [arrowhead = none ];
20 -> 7;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 21 [arrowhead = none ];
21 -> 8;

22 [label="BACKTRACK\nfor clause: fl(.(E, X), R, s(Z)) :- ','(append(E, Y, R), fl(X, Y, Z))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 23 [arrowhead = none ];
23 -> 9;

24 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 24 [arrowhead = none ];
24 -> 10;

25 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 25 [arrowhead = none ];
25 -> 11;

26 [label="EVAL with clause\nappend([], X22, X22).\nand substitutionT8 -> [],\nX13 -> T19,\nX22 -> T19,\nT12 -> T19,\nX23 -> T19,\nT18 -> T19,\nT13 -> T20", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 26 [arrowhead = none ];
26 -> 12;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 27 [arrowhead = none ];
27 -> 13;

28 [label="EVAL with clause\nappend(.(X34, X35), X36, .(X34, X37)) :- append(X35, X36, X37).\nand substitutionX34 -> T27,\nX35 -> T28,\nT8 -> .(T27, T28),\nX13 -> X38,\nX36 -> X38,\nX37 -> T30,\nT12 -> .(T27, T30),\nT29 -> T30,\nT13 -> T31", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 28 [arrowhead = none ];
28 -> 14;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 29 [arrowhead = none ];
29 -> 15;

30 [label="INSTANCE with matching:\nT1 -> T9\nT2 -> T19\nT3 -> T20", fontsize=14, style = filled, fillcolor = lightgrey];
12 -> 30 [arrowhead = none , style=dashed];
30 -> 1[style=dashed];

31 [label="INSTANCE with matching:\nT8 -> T28\nX13 -> X38\nT12 -> T30\nT13 -> T31", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 31 [arrowhead = none , style=dashed];
31 -> 7[style=dashed];

}

(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)

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

FLA_IN_GAA(.(X1, X2), X3, s(X4)) → U3_GAA(X1, X2, X3, X4, pB_in_gaaga(X1, X5, X3, X2, X4))
FLA_IN_GAA(.(X1, X2), X3, s(X4)) → PB_IN_GAAGA(X1, X5, X3, X2, X4)
PB_IN_GAAGA([], X1, X1, X2, X3) → U1_GAAGA(X1, X2, X3, flA_in_gaa(X2, X1, X3))
PB_IN_GAAGA([], X1, X1, X2, X3) → FLA_IN_GAA(X2, X1, X3)
PB_IN_GAAGA(.(X1, X2), X3, .(X1, X4), X5, X6) → U2_GAAGA(X1, X2, X3, X4, X5, X6, pB_in_gaaga(X2, X3, X4, X5, X6))
PB_IN_GAAGA(.(X1, X2), X3, .(X1, X4), X5, X6) → PB_IN_GAAGA(X2, X3, X4, X5, X6)

R is empty.
The argument filtering Pi contains the following mapping:
flA_in_gaa(x1, x2, x3)  =  flA_in_gaa(x1)
.(x1, x2)  =  .(x1, x2)
pB_in_gaaga(x1, x2, x3, x4, x5)  =  pB_in_gaaga(x1, x4)
[]  =  []
s(x1)  =  s(x1)
FLA_IN_GAA(x1, x2, x3)  =  FLA_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x1, x2, x5)
PB_IN_GAAGA(x1, x2, x3, x4, x5)  =  PB_IN_GAAGA(x1, x4)
U1_GAAGA(x1, x2, x3, x4)  =  U1_GAAGA(x2, x4)
U2_GAAGA(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAAGA(x1, x2, 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_GAA(.(X1, X2), X3, s(X4)) → U3_GAA(X1, X2, X3, X4, pB_in_gaaga(X1, X5, X3, X2, X4))
FLA_IN_GAA(.(X1, X2), X3, s(X4)) → PB_IN_GAAGA(X1, X5, X3, X2, X4)
PB_IN_GAAGA([], X1, X1, X2, X3) → U1_GAAGA(X1, X2, X3, flA_in_gaa(X2, X1, X3))
PB_IN_GAAGA([], X1, X1, X2, X3) → FLA_IN_GAA(X2, X1, X3)
PB_IN_GAAGA(.(X1, X2), X3, .(X1, X4), X5, X6) → U2_GAAGA(X1, X2, X3, X4, X5, X6, pB_in_gaaga(X2, X3, X4, X5, X6))
PB_IN_GAAGA(.(X1, X2), X3, .(X1, X4), X5, X6) → PB_IN_GAAGA(X2, X3, X4, X5, X6)

R is empty.
The argument filtering Pi contains the following mapping:
flA_in_gaa(x1, x2, x3)  =  flA_in_gaa(x1)
.(x1, x2)  =  .(x1, x2)
pB_in_gaaga(x1, x2, x3, x4, x5)  =  pB_in_gaaga(x1, x4)
[]  =  []
s(x1)  =  s(x1)
FLA_IN_GAA(x1, x2, x3)  =  FLA_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x1, x2, x5)
PB_IN_GAAGA(x1, x2, x3, x4, x5)  =  PB_IN_GAAGA(x1, x4)
U1_GAAGA(x1, x2, x3, x4)  =  U1_GAAGA(x2, x4)
U2_GAAGA(x1, x2, x3, x4, x5, x6, x7)  =  U2_GAAGA(x1, x2, 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_GAA(.(X1, X2), X3, s(X4)) → PB_IN_GAAGA(X1, X5, X3, X2, X4)
PB_IN_GAAGA([], X1, X1, X2, X3) → FLA_IN_GAA(X2, X1, X3)
PB_IN_GAAGA(.(X1, X2), X3, .(X1, X4), X5, X6) → PB_IN_GAAGA(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_GAA(x1, x2, x3)  =  FLA_IN_GAA(x1)
PB_IN_GAAGA(x1, x2, x3, x4, x5)  =  PB_IN_GAAGA(x1, 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_GAA(.(X1, X2)) → PB_IN_GAAGA(X1, X2)
PB_IN_GAAGA([], X2) → FLA_IN_GAA(X2)
PB_IN_GAAGA(.(X1, X2), X5) → PB_IN_GAAGA(X2, 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_GAAGA([], X2) → FLA_IN_GAA(X2)
  The graph contains the following edges 2 >= 1

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

• FLA_IN_GAA(.(X1, X2)) → PB_IN_GAAGA(X1, X2)
  The graph contains the following edges 1 > 1, 1 > 2

(10) YES