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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: fl(T1, T2, T3)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
6 [label="6: 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];
11 [label="11: true | fl([], T2, T3), 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\nfl(T1, T2, T3) !=? fl([], [], 0)", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: fl([], T2, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [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];
25 [label="25: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: ','(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];
30 [label="30: ','(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];
31 [label="31: ','(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];
34 [label="34: fl(T9, T19, T20)\n\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="43: ','(append(T28, X38, T30), fl(T9, X38, T31))\n\nT9 is ground\nT28 is ground\nX38 is free", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="44: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
45 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 45 [arrowhead = none ];
45 -> 6;

46 [label="EVAL with clause\nfl([], [], 0).\nand substitutionT1 -> [],\nT2 -> [],\nT3 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 46 [arrowhead = none ];
46 -> 11;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 47 [arrowhead = none ];
47 -> 13;

48 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 48 [arrowhead = none ];
48 -> 16;

49 [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];
13 -> 49 [arrowhead = none ];
49 -> 23;

50 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 50 [arrowhead = none ];
50 -> 25;

51 [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 -> 51 [arrowhead = none ];
51 -> 19;

52 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
23 -> 52 [arrowhead = none ];
52 -> 27;

53 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 53 [arrowhead = none ];
53 -> 30;

54 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 54 [arrowhead = none ];
54 -> 31;

55 [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];
30 -> 55 [arrowhead = none ];
55 -> 34;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 56 [arrowhead = none ];
56 -> 35;

57 [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];
31 -> 57 [arrowhead = none ];
57 -> 43;

58 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
31 -> 58 [arrowhead = none ];
58 -> 44;

59 [label="INSTANCE with matching:\nT1 -> T9\nT2 -> T19\nT3 -> T20", fontsize=14, style = filled, fillcolor = lightgrey];
34 -> 59 [arrowhead = none , style=dashed];
59 -> 2[style=dashed];

60 [label="INSTANCE with matching:\nT8 -> T28\nX13 -> X38\nT12 -> T30\nT13 -> T31", fontsize=14, style = filled, fillcolor = lightgrey];
43 -> 60 [arrowhead = none , style=dashed];
60 -> 23[style=dashed];

}

(2) Obligation:

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

flA_in_gaa([], [], 0) → flA_out_gaa([], [], 0)
flA_in_gaa(.(T8, T9), T12, s(T13)) → U1_gaa(T8, T9, T12, T13, pB_in_gaaga(T8, X13, T12, T9, T13))
pB_in_gaaga([], T19, T19, T9, T20) → U2_gaaga(T19, T9, T20, flA_in_gaa(T9, T19, T20))
U2_gaaga(T19, T9, T20, flA_out_gaa(T9, T19, T20)) → pB_out_gaaga([], T19, T19, T9, T20)
pB_in_gaaga(.(T27, T28), X38, .(T27, T30), T9, T31) → U3_gaaga(T27, T28, X38, T30, T9, T31, pB_in_gaaga(T28, X38, T30, T9, T31))
U3_gaaga(T27, T28, X38, T30, T9, T31, pB_out_gaaga(T28, X38, T30, T9, T31)) → pB_out_gaaga(.(T27, T28), X38, .(T27, T30), T9, T31)
U1_gaa(T8, T9, T12, T13, pB_out_gaaga(T8, X13, T12, T9, T13)) → flA_out_gaa(.(T8, T9), T12, s(T13))

The argument filtering Pi contains the following mapping:
flA_in_gaa(x1, x2, x3)  =  flA_in_gaa(x1)
[]  =  []
flA_out_gaa(x1, x2, x3)  =  flA_out_gaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
pB_in_gaaga(x1, x2, x3, x4, x5)  =  pB_in_gaaga(x1, x4)
U2_gaaga(x1, x2, x3, x4)  =  U2_gaaga(x2, x4)
pB_out_gaaga(x1, x2, x3, x4, x5)  =  pB_out_gaaga(x1, x2, x3, x4, x5)
U3_gaaga(x1, x2, x3, x4, x5, x6, x7)  =  U3_gaaga(x1, x2, x5, x7)
s(x1)  =  s(x1)

(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_GAA(.(T8, T9), T12, s(T13)) → U1_GAA(T8, T9, T12, T13, pB_in_gaaga(T8, X13, T12, T9, T13))
FLA_IN_GAA(.(T8, T9), T12, s(T13)) → PB_IN_GAAGA(T8, X13, T12, T9, T13)
PB_IN_GAAGA([], T19, T19, T9, T20) → U2_GAAGA(T19, T9, T20, flA_in_gaa(T9, T19, T20))
PB_IN_GAAGA([], T19, T19, T9, T20) → FLA_IN_GAA(T9, T19, T20)
PB_IN_GAAGA(.(T27, T28), X38, .(T27, T30), T9, T31) → U3_GAAGA(T27, T28, X38, T30, T9, T31, pB_in_gaaga(T28, X38, T30, T9, T31))
PB_IN_GAAGA(.(T27, T28), X38, .(T27, T30), T9, T31) → PB_IN_GAAGA(T28, X38, T30, T9, T31)

The TRS R consists of the following rules:

flA_in_gaa([], [], 0) → flA_out_gaa([], [], 0)
flA_in_gaa(.(T8, T9), T12, s(T13)) → U1_gaa(T8, T9, T12, T13, pB_in_gaaga(T8, X13, T12, T9, T13))
pB_in_gaaga([], T19, T19, T9, T20) → U2_gaaga(T19, T9, T20, flA_in_gaa(T9, T19, T20))
U2_gaaga(T19, T9, T20, flA_out_gaa(T9, T19, T20)) → pB_out_gaaga([], T19, T19, T9, T20)
pB_in_gaaga(.(T27, T28), X38, .(T27, T30), T9, T31) → U3_gaaga(T27, T28, X38, T30, T9, T31, pB_in_gaaga(T28, X38, T30, T9, T31))
U3_gaaga(T27, T28, X38, T30, T9, T31, pB_out_gaaga(T28, X38, T30, T9, T31)) → pB_out_gaaga(.(T27, T28), X38, .(T27, T30), T9, T31)
U1_gaa(T8, T9, T12, T13, pB_out_gaaga(T8, X13, T12, T9, T13)) → flA_out_gaa(.(T8, T9), T12, s(T13))

The argument filtering Pi contains the following mapping:
flA_in_gaa(x1, x2, x3)  =  flA_in_gaa(x1)
[]  =  []
flA_out_gaa(x1, x2, x3)  =  flA_out_gaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
pB_in_gaaga(x1, x2, x3, x4, x5)  =  pB_in_gaaga(x1, x4)
U2_gaaga(x1, x2, x3, x4)  =  U2_gaaga(x2, x4)
pB_out_gaaga(x1, x2, x3, x4, x5)  =  pB_out_gaaga(x1, x2, x3, x4, x5)
U3_gaaga(x1, x2, x3, x4, x5, x6, x7)  =  U3_gaaga(x1, x2, x5, x7)
s(x1)  =  s(x1)
FLA_IN_GAA(x1, x2, x3)  =  FLA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, x5)
PB_IN_GAAGA(x1, x2, x3, x4, x5)  =  PB_IN_GAAGA(x1, x4)
U2_GAAGA(x1, x2, x3, x4)  =  U2_GAAGA(x2, x4)
U3_GAAGA(x1, x2, x3, x4, x5, x6, x7)  =  U3_GAAGA(x1, x2, x5, 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_GAA(.(T8, T9), T12, s(T13)) → U1_GAA(T8, T9, T12, T13, pB_in_gaaga(T8, X13, T12, T9, T13))
FLA_IN_GAA(.(T8, T9), T12, s(T13)) → PB_IN_GAAGA(T8, X13, T12, T9, T13)
PB_IN_GAAGA([], T19, T19, T9, T20) → U2_GAAGA(T19, T9, T20, flA_in_gaa(T9, T19, T20))
PB_IN_GAAGA([], T19, T19, T9, T20) → FLA_IN_GAA(T9, T19, T20)
PB_IN_GAAGA(.(T27, T28), X38, .(T27, T30), T9, T31) → U3_GAAGA(T27, T28, X38, T30, T9, T31, pB_in_gaaga(T28, X38, T30, T9, T31))
PB_IN_GAAGA(.(T27, T28), X38, .(T27, T30), T9, T31) → PB_IN_GAAGA(T28, X38, T30, T9, T31)

The TRS R consists of the following rules:

flA_in_gaa([], [], 0) → flA_out_gaa([], [], 0)
flA_in_gaa(.(T8, T9), T12, s(T13)) → U1_gaa(T8, T9, T12, T13, pB_in_gaaga(T8, X13, T12, T9, T13))
pB_in_gaaga([], T19, T19, T9, T20) → U2_gaaga(T19, T9, T20, flA_in_gaa(T9, T19, T20))
U2_gaaga(T19, T9, T20, flA_out_gaa(T9, T19, T20)) → pB_out_gaaga([], T19, T19, T9, T20)
pB_in_gaaga(.(T27, T28), X38, .(T27, T30), T9, T31) → U3_gaaga(T27, T28, X38, T30, T9, T31, pB_in_gaaga(T28, X38, T30, T9, T31))
U3_gaaga(T27, T28, X38, T30, T9, T31, pB_out_gaaga(T28, X38, T30, T9, T31)) → pB_out_gaaga(.(T27, T28), X38, .(T27, T30), T9, T31)
U1_gaa(T8, T9, T12, T13, pB_out_gaaga(T8, X13, T12, T9, T13)) → flA_out_gaa(.(T8, T9), T12, s(T13))

The argument filtering Pi contains the following mapping:
flA_in_gaa(x1, x2, x3)  =  flA_in_gaa(x1)
[]  =  []
flA_out_gaa(x1, x2, x3)  =  flA_out_gaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
pB_in_gaaga(x1, x2, x3, x4, x5)  =  pB_in_gaaga(x1, x4)
U2_gaaga(x1, x2, x3, x4)  =  U2_gaaga(x2, x4)
pB_out_gaaga(x1, x2, x3, x4, x5)  =  pB_out_gaaga(x1, x2, x3, x4, x5)
U3_gaaga(x1, x2, x3, x4, x5, x6, x7)  =  U3_gaaga(x1, x2, x5, x7)
s(x1)  =  s(x1)
FLA_IN_GAA(x1, x2, x3)  =  FLA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, x5)
PB_IN_GAAGA(x1, x2, x3, x4, x5)  =  PB_IN_GAAGA(x1, x4)
U2_GAAGA(x1, x2, x3, x4)  =  U2_GAAGA(x2, x4)
U3_GAAGA(x1, x2, x3, x4, x5, x6, x7)  =  U3_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(.(T8, T9), T12, s(T13)) → PB_IN_GAAGA(T8, X13, T12, T9, T13)
PB_IN_GAAGA([], T19, T19, T9, T20) → FLA_IN_GAA(T9, T19, T20)
PB_IN_GAAGA(.(T27, T28), X38, .(T27, T30), T9, T31) → PB_IN_GAAGA(T28, X38, T30, T9, T31)

The TRS R consists of the following rules:

flA_in_gaa([], [], 0) → flA_out_gaa([], [], 0)
flA_in_gaa(.(T8, T9), T12, s(T13)) → U1_gaa(T8, T9, T12, T13, pB_in_gaaga(T8, X13, T12, T9, T13))
pB_in_gaaga([], T19, T19, T9, T20) → U2_gaaga(T19, T9, T20, flA_in_gaa(T9, T19, T20))
U2_gaaga(T19, T9, T20, flA_out_gaa(T9, T19, T20)) → pB_out_gaaga([], T19, T19, T9, T20)
pB_in_gaaga(.(T27, T28), X38, .(T27, T30), T9, T31) → U3_gaaga(T27, T28, X38, T30, T9, T31, pB_in_gaaga(T28, X38, T30, T9, T31))
U3_gaaga(T27, T28, X38, T30, T9, T31, pB_out_gaaga(T28, X38, T30, T9, T31)) → pB_out_gaaga(.(T27, T28), X38, .(T27, T30), T9, T31)
U1_gaa(T8, T9, T12, T13, pB_out_gaaga(T8, X13, T12, T9, T13)) → flA_out_gaa(.(T8, T9), T12, s(T13))

The argument filtering Pi contains the following mapping:
flA_in_gaa(x1, x2, x3)  =  flA_in_gaa(x1)
[]  =  []
flA_out_gaa(x1, x2, x3)  =  flA_out_gaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
pB_in_gaaga(x1, x2, x3, x4, x5)  =  pB_in_gaaga(x1, x4)
U2_gaaga(x1, x2, x3, x4)  =  U2_gaaga(x2, x4)
pB_out_gaaga(x1, x2, x3, x4, x5)  =  pB_out_gaaga(x1, x2, x3, x4, x5)
U3_gaaga(x1, x2, x3, x4, x5, x6, x7)  =  U3_gaaga(x1, x2, x5, x7)
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) 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_GAA(.(T8, T9), T12, s(T13)) → PB_IN_GAAGA(T8, X13, T12, T9, T13)
PB_IN_GAAGA([], T19, T19, T9, T20) → FLA_IN_GAA(T9, T19, T20)
PB_IN_GAAGA(.(T27, T28), X38, .(T27, T30), T9, T31) → PB_IN_GAAGA(T28, X38, T30, T9, T31)

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

(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_GAA(.(T8, T9)) → PB_IN_GAAGA(T8, T9)
PB_IN_GAAGA([], T9) → FLA_IN_GAA(T9)
PB_IN_GAAGA(.(T27, T28), T9) → PB_IN_GAAGA(T28, T9)

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_GAAGA([], T9) → FLA_IN_GAA(T9)
  The graph contains the following edges 2 >= 1

• PB_IN_GAAGA(.(T27, T28), T9) → PB_IN_GAAGA(T28, T9)
  The graph contains the following edges 1 > 1, 2 >= 2

• FLA_IN_GAA(.(T8, T9)) → PB_IN_GAAGA(T8, T9)
  The graph contains the following edges 1 > 1, 1 > 2

(12) YES