Left Termination of the query pattern
fl(g,a,g)
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,g)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: fl(T1, T2, T3)\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: fl(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | fl(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: true | fl([], T2, 0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: fl(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT3 is ground\nfl(T1, T2, T3) !=? fl([], [], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: fl([], T2, 0), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="B: ','(append(T8, X13, T12), fl(T9, X13, T11))\n\nT8 is ground\nT9 is ground\nT11 is ground\nX13 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: ','(append(T8, X13, T12), fl(T9, X13, T11)), SCOPE: 2, CLAUSE: 2 | ','(append(T8, X13, T12), fl(T9, X13, T11)), SCOPE: 2, CLAUSE: 3\n\nT8 is ground\nT9 is ground\nT11 is ground\nX13 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(append(T8, X13, T12), fl(T9, X13, T11)), SCOPE: 2, CLAUSE: 2\n\nT8 is ground\nT9 is ground\nT11 is ground\nX13 is free", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: ','(append(T8, X13, T12), fl(T9, X13, T11)), SCOPE: 2, CLAUSE: 3\n\nT8 is ground\nT9 is ground\nT11 is ground\nX13 is free", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: fl(T9, T18, T11)\n\nT9 is ground\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: ','(append(T26, X38, T28), fl(T9, X38, T11))\n\nT9 is ground\nT11 is ground\nT26 is ground\nX38 is free", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 43 [arrowhead = none ];
43 -> 5;

44 [label="EVAL with clause\nfl([], [], 0).\nand substitutionT1 -> [],\nT2 -> [],\nT3 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 44 [arrowhead = none ];
44 -> 10;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 45 [arrowhead = none ];
45 -> 13;

46 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 46 [arrowhead = none ];
46 -> 16;

47 [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 -> T11,\nT3 -> s(T11),\nT10 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 47 [arrowhead = none ];
47 -> 20;

48 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 48 [arrowhead = none ];
48 -> 24;

49 [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];
16 -> 49 [arrowhead = none ];
49 -> 18;

50 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 50 [arrowhead = none ];
50 -> 26;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
26 -> 51 [arrowhead = none ];
51 -> 28;

52 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
26 -> 52 [arrowhead = none ];
52 -> 29;

53 [label="EVAL with clause\nappend([], X22, X22).\nand substitutionT8 -> [],\nX13 -> T18,\nX22 -> T18,\nT12 -> T18,\nX23 -> T18,\nT17 -> T18", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 53 [arrowhead = none ];
53 -> 32;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
28 -> 54 [arrowhead = none ];
54 -> 33;

55 [label="EVAL with clause\nappend(.(X34, X35), X36, .(X34, X37)) :- append(X35, X36, X37).\nand substitutionX34 -> T25,\nX35 -> T26,\nT8 -> .(T25, T26),\nX13 -> X38,\nX36 -> X38,\nX37 -> T28,\nT12 -> .(T25, T28),\nT27 -> T28", fontsize=14, style = filled, fillcolor = lightblue];
29 -> 55 [arrowhead = none ];
55 -> 41;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
29 -> 56 [arrowhead = none ];
56 -> 42;

57 [label="INSTANCE with matching:\nT1 -> T9\nT2 -> T18\nT3 -> T11", fontsize=14, style = filled, fillcolor = lightgrey];
32 -> 57 [arrowhead = none , style=dashed];
57 -> 1[style=dashed];

58 [label="INSTANCE with matching:\nT8 -> T26\nX13 -> X38\nT12 -> T28", fontsize=14, style = filled, fillcolor = lightgrey];
41 -> 58 [arrowhead = none , style=dashed];
58 -> 20[style=dashed];

}

(2) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

flA_in_gag([], [], 0) → flA_out_gag([], [], 0)
flA_in_gag(.(T8, T9), T12, s(T11)) → U1_gag(T8, T9, T12, T11, pB_in_gaagg(T8, X13, T12, T9, T11))
pB_in_gaagg([], T18, T18, T9, T11) → U2_gaagg(T18, T9, T11, flA_in_gag(T9, T18, T11))
U2_gaagg(T18, T9, T11, flA_out_gag(T9, T18, T11)) → pB_out_gaagg([], T18, T18, T9, T11)
pB_in_gaagg(.(T25, T26), X38, .(T25, T28), T9, T11) → U3_gaagg(T25, T26, X38, T28, T9, T11, pB_in_gaagg(T26, X38, T28, T9, T11))
U3_gaagg(T25, T26, X38, T28, T9, T11, pB_out_gaagg(T26, X38, T28, T9, T11)) → pB_out_gaagg(.(T25, T26), X38, .(T25, T28), T9, T11)
U1_gag(T8, T9, T12, T11, pB_out_gaagg(T8, X13, T12, T9, T11)) → flA_out_gag(.(T8, T9), T12, s(T11))

The argument filtering Pi contains the following mapping:
flA_in_gag(x1, x2, x3)  =  flA_in_gag(x1, x3)
[]  =  []
0  =  0
flA_out_gag(x1, x2, x3)  =  flA_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1_gag(x1, x2, x3, x4, x5)  =  U1_gag(x1, x2, x4, x5)
pB_in_gaagg(x1, x2, x3, x4, x5)  =  pB_in_gaagg(x1, x4, x5)
U2_gaagg(x1, x2, x3, x4)  =  U2_gaagg(x2, x3, x4)
pB_out_gaagg(x1, x2, x3, x4, x5)  =  pB_out_gaagg(x1, x2, x3, x4, x5)
U3_gaagg(x1, x2, x3, x4, x5, x6, x7)  =  U3_gaagg(x1, x2, x5, x6, x7)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

FLA_IN_GAG(.(T8, T9), T12, s(T11)) → U1_GAG(T8, T9, T12, T11, pB_in_gaagg(T8, X13, T12, T9, T11))
FLA_IN_GAG(.(T8, T9), T12, s(T11)) → PB_IN_GAAGG(T8, X13, T12, T9, T11)
PB_IN_GAAGG([], T18, T18, T9, T11) → U2_GAAGG(T18, T9, T11, flA_in_gag(T9, T18, T11))
PB_IN_GAAGG([], T18, T18, T9, T11) → FLA_IN_GAG(T9, T18, T11)
PB_IN_GAAGG(.(T25, T26), X38, .(T25, T28), T9, T11) → U3_GAAGG(T25, T26, X38, T28, T9, T11, pB_in_gaagg(T26, X38, T28, T9, T11))
PB_IN_GAAGG(.(T25, T26), X38, .(T25, T28), T9, T11) → PB_IN_GAAGG(T26, X38, T28, T9, T11)

The TRS R consists of the following rules:

flA_in_gag([], [], 0) → flA_out_gag([], [], 0)
flA_in_gag(.(T8, T9), T12, s(T11)) → U1_gag(T8, T9, T12, T11, pB_in_gaagg(T8, X13, T12, T9, T11))
pB_in_gaagg([], T18, T18, T9, T11) → U2_gaagg(T18, T9, T11, flA_in_gag(T9, T18, T11))
U2_gaagg(T18, T9, T11, flA_out_gag(T9, T18, T11)) → pB_out_gaagg([], T18, T18, T9, T11)
pB_in_gaagg(.(T25, T26), X38, .(T25, T28), T9, T11) → U3_gaagg(T25, T26, X38, T28, T9, T11, pB_in_gaagg(T26, X38, T28, T9, T11))
U3_gaagg(T25, T26, X38, T28, T9, T11, pB_out_gaagg(T26, X38, T28, T9, T11)) → pB_out_gaagg(.(T25, T26), X38, .(T25, T28), T9, T11)
U1_gag(T8, T9, T12, T11, pB_out_gaagg(T8, X13, T12, T9, T11)) → flA_out_gag(.(T8, T9), T12, s(T11))

The argument filtering Pi contains the following mapping:
flA_in_gag(x1, x2, x3)  =  flA_in_gag(x1, x3)
[]  =  []
0  =  0
flA_out_gag(x1, x2, x3)  =  flA_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1_gag(x1, x2, x3, x4, x5)  =  U1_gag(x1, x2, x4, x5)
pB_in_gaagg(x1, x2, x3, x4, x5)  =  pB_in_gaagg(x1, x4, x5)
U2_gaagg(x1, x2, x3, x4)  =  U2_gaagg(x2, x3, x4)
pB_out_gaagg(x1, x2, x3, x4, x5)  =  pB_out_gaagg(x1, x2, x3, x4, x5)
U3_gaagg(x1, x2, x3, x4, x5, x6, x7)  =  U3_gaagg(x1, x2, x5, x6, x7)
FLA_IN_GAG(x1, x2, x3)  =  FLA_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5)  =  U1_GAG(x1, x2, x4, x5)
PB_IN_GAAGG(x1, x2, x3, x4, x5)  =  PB_IN_GAAGG(x1, x4, x5)
U2_GAAGG(x1, x2, x3, x4)  =  U2_GAAGG(x2, x3, x4)
U3_GAAGG(x1, x2, x3, x4, x5, x6, x7)  =  U3_GAAGG(x1, x2, x5, x6, x7)

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

(4) Obligation:

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

FLA_IN_GAG(.(T8, T9), T12, s(T11)) → U1_GAG(T8, T9, T12, T11, pB_in_gaagg(T8, X13, T12, T9, T11))
FLA_IN_GAG(.(T8, T9), T12, s(T11)) → PB_IN_GAAGG(T8, X13, T12, T9, T11)
PB_IN_GAAGG([], T18, T18, T9, T11) → U2_GAAGG(T18, T9, T11, flA_in_gag(T9, T18, T11))
PB_IN_GAAGG([], T18, T18, T9, T11) → FLA_IN_GAG(T9, T18, T11)
PB_IN_GAAGG(.(T25, T26), X38, .(T25, T28), T9, T11) → U3_GAAGG(T25, T26, X38, T28, T9, T11, pB_in_gaagg(T26, X38, T28, T9, T11))
PB_IN_GAAGG(.(T25, T26), X38, .(T25, T28), T9, T11) → PB_IN_GAAGG(T26, X38, T28, T9, T11)

The TRS R consists of the following rules:

flA_in_gag([], [], 0) → flA_out_gag([], [], 0)
flA_in_gag(.(T8, T9), T12, s(T11)) → U1_gag(T8, T9, T12, T11, pB_in_gaagg(T8, X13, T12, T9, T11))
pB_in_gaagg([], T18, T18, T9, T11) → U2_gaagg(T18, T9, T11, flA_in_gag(T9, T18, T11))
U2_gaagg(T18, T9, T11, flA_out_gag(T9, T18, T11)) → pB_out_gaagg([], T18, T18, T9, T11)
pB_in_gaagg(.(T25, T26), X38, .(T25, T28), T9, T11) → U3_gaagg(T25, T26, X38, T28, T9, T11, pB_in_gaagg(T26, X38, T28, T9, T11))
U3_gaagg(T25, T26, X38, T28, T9, T11, pB_out_gaagg(T26, X38, T28, T9, T11)) → pB_out_gaagg(.(T25, T26), X38, .(T25, T28), T9, T11)
U1_gag(T8, T9, T12, T11, pB_out_gaagg(T8, X13, T12, T9, T11)) → flA_out_gag(.(T8, T9), T12, s(T11))

The argument filtering Pi contains the following mapping:
flA_in_gag(x1, x2, x3)  =  flA_in_gag(x1, x3)
[]  =  []
0  =  0
flA_out_gag(x1, x2, x3)  =  flA_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1_gag(x1, x2, x3, x4, x5)  =  U1_gag(x1, x2, x4, x5)
pB_in_gaagg(x1, x2, x3, x4, x5)  =  pB_in_gaagg(x1, x4, x5)
U2_gaagg(x1, x2, x3, x4)  =  U2_gaagg(x2, x3, x4)
pB_out_gaagg(x1, x2, x3, x4, x5)  =  pB_out_gaagg(x1, x2, x3, x4, x5)
U3_gaagg(x1, x2, x3, x4, x5, x6, x7)  =  U3_gaagg(x1, x2, x5, x6, x7)
FLA_IN_GAG(x1, x2, x3)  =  FLA_IN_GAG(x1, x3)
U1_GAG(x1, x2, x3, x4, x5)  =  U1_GAG(x1, x2, x4, x5)
PB_IN_GAAGG(x1, x2, x3, x4, x5)  =  PB_IN_GAAGG(x1, x4, x5)
U2_GAAGG(x1, x2, x3, x4)  =  U2_GAAGG(x2, x3, x4)
U3_GAAGG(x1, x2, x3, x4, x5, x6, x7)  =  U3_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:

FLA_IN_GAG(.(T8, T9), T12, s(T11)) → PB_IN_GAAGG(T8, X13, T12, T9, T11)
PB_IN_GAAGG([], T18, T18, T9, T11) → FLA_IN_GAG(T9, T18, T11)
PB_IN_GAAGG(.(T25, T26), X38, .(T25, T28), T9, T11) → PB_IN_GAAGG(T26, X38, T28, T9, T11)

The TRS R consists of the following rules:

flA_in_gag([], [], 0) → flA_out_gag([], [], 0)
flA_in_gag(.(T8, T9), T12, s(T11)) → U1_gag(T8, T9, T12, T11, pB_in_gaagg(T8, X13, T12, T9, T11))
pB_in_gaagg([], T18, T18, T9, T11) → U2_gaagg(T18, T9, T11, flA_in_gag(T9, T18, T11))
U2_gaagg(T18, T9, T11, flA_out_gag(T9, T18, T11)) → pB_out_gaagg([], T18, T18, T9, T11)
pB_in_gaagg(.(T25, T26), X38, .(T25, T28), T9, T11) → U3_gaagg(T25, T26, X38, T28, T9, T11, pB_in_gaagg(T26, X38, T28, T9, T11))
U3_gaagg(T25, T26, X38, T28, T9, T11, pB_out_gaagg(T26, X38, T28, T9, T11)) → pB_out_gaagg(.(T25, T26), X38, .(T25, T28), T9, T11)
U1_gag(T8, T9, T12, T11, pB_out_gaagg(T8, X13, T12, T9, T11)) → flA_out_gag(.(T8, T9), T12, s(T11))

The argument filtering Pi contains the following mapping:
flA_in_gag(x1, x2, x3)  =  flA_in_gag(x1, x3)
[]  =  []
0  =  0
flA_out_gag(x1, x2, x3)  =  flA_out_gag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1_gag(x1, x2, x3, x4, x5)  =  U1_gag(x1, x2, x4, x5)
pB_in_gaagg(x1, x2, x3, x4, x5)  =  pB_in_gaagg(x1, x4, x5)
U2_gaagg(x1, x2, x3, x4)  =  U2_gaagg(x2, x3, x4)
pB_out_gaagg(x1, x2, x3, x4, x5)  =  pB_out_gaagg(x1, x2, x3, x4, x5)
U3_gaagg(x1, x2, x3, x4, x5, x6, x7)  =  U3_gaagg(x1, x2, x5, x6, x7)
FLA_IN_GAG(x1, x2, x3)  =  FLA_IN_GAG(x1, x3)
PB_IN_GAAGG(x1, x2, x3, x4, x5)  =  PB_IN_GAAGG(x1, x4, x5)

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

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

FLA_IN_GAG(.(T8, T9), T12, s(T11)) → PB_IN_GAAGG(T8, X13, T12, T9, T11)
PB_IN_GAAGG([], T18, T18, T9, T11) → FLA_IN_GAG(T9, T18, T11)
PB_IN_GAAGG(.(T25, T26), X38, .(T25, T28), T9, T11) → PB_IN_GAAGG(T26, X38, T28, T9, T11)

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

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

(9) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(10) Obligation:

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

FLA_IN_GAG(.(T8, T9), s(T11)) → PB_IN_GAAGG(T8, T9, T11)
PB_IN_GAAGG([], T9, T11) → FLA_IN_GAG(T9, T11)
PB_IN_GAAGG(.(T25, T26), T9, T11) → PB_IN_GAAGG(T26, T9, T11)

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

(11) 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_GAAGG([], T9, T11) → FLA_IN_GAG(T9, T11)
  The graph contains the following edges 2 >= 1, 3 >= 2

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

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

(12) YES