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

(0) Obligation:

Clauses:

append(.(H, X), Y, .(X, Z)) :- append(X, Y, Z).
append([], Y, Y).

Query: append(g,g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: append(T1, T2, T3)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: append(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | append(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: append(T9, T10, T12) | append(.(T8, T9), T10, T3), SCOPE: 1, CLAUSE: 1\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: append(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nT2 is ground\nX5 is free\nX6 is free\nX7 is free\nX8 is free\nappend(T1, T2, T3) !=? append(.(X5, X6), X7, .(X6, X8))", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: append(T9, T10, T12), SCOPE: 2, CLAUSE: 0 | append(T9, T10, T12), SCOPE: 2, CLAUSE: 1 | ?_2 | append(.(T8, T9), T10, T3), SCOPE: 1, CLAUSE: 1\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: append(T9, T10, T12), SCOPE: 2, CLAUSE: 0\n\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: append(T9, T10, T12), SCOPE: 2, CLAUSE: 1 | ?_2 | append(.(T8, T9), T10, T3), SCOPE: 1, CLAUSE: 1\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: append(T30, T31, T33)\n\nT30 is ground\nT31 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: append(T9, T10, T12), SCOPE: 2, CLAUSE: 1\n\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: ?_2 | append(.(T8, T9), T10, T3), SCOPE: 1, CLAUSE: 1\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: append(.(T8, T9), T10, T3), SCOPE: 1, CLAUSE: 1\n\nT8 is ground\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="43: \n\n", 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 -> 5;

46 [label="EVAL with clause\nappend(.(X5, X6), X7, .(X6, X8)) :- append(X6, X7, X8).\nand substitutionX5 -> T8,\nX6 -> T9,\nT1 -> .(T8, T9),\nT2 -> T10,\nX7 -> T10,\nX8 -> T12,\nT3 -> .(T9, T12),\nT11 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 46 [arrowhead = none ];
46 -> 9;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 47 [arrowhead = none ];
47 -> 12;

48 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 48 [arrowhead = none ];
48 -> 14;

49 [label="EVAL with clause\nappend([], X40, X40).\nand substitutionT1 -> [],\nT2 -> T44,\nX40 -> T44,\nT3 -> T44", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 49 [arrowhead = none ];
49 -> 39;

50 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 50 [arrowhead = none ];
50 -> 43;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 51 [arrowhead = none ];
51 -> 18;

52 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 52 [arrowhead = none ];
52 -> 20;

53 [label="EVAL with clause\nappend(.(X25, X26), X27, .(X26, X28)) :- append(X26, X27, X28).\nand substitutionX25 -> T29,\nX26 -> T30,\nT9 -> .(T29, T30),\nT10 -> T31,\nX27 -> T31,\nX28 -> T33,\nT12 -> .(T30, T33),\nT32 -> T33", fontsize=14, style = filled, fillcolor = lightblue];
18 -> 53 [arrowhead = none ];
53 -> 23;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 54 [arrowhead = none ];
54 -> 24;

55 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
20 -> 55 [arrowhead = none ];
55 -> 25;

56 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
20 -> 56 [arrowhead = none ];
56 -> 26;

57 [label="INSTANCE with matching:\nT1 -> T30\nT2 -> T31\nT3 -> T33", fontsize=14, style = filled, fillcolor = lightgrey];
23 -> 57 [arrowhead = none , style=dashed];
57 -> 2[style=dashed];

58 [label="EVAL with clause\nappend([], X37, X37).\nand substitutionT9 -> [],\nT10 -> T42,\nX37 -> T42,\nT12 -> T42", fontsize=14, style = filled, fillcolor = lightblue];
25 -> 58 [arrowhead = none ];
58 -> 30;

59 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 59 [arrowhead = none ];
59 -> 31;

60 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 60 [arrowhead = none ];
60 -> 33;

61 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 61 [arrowhead = none ];
61 -> 32;

62 [label="BACKTRACK\nfor clause: append([], Y, Y)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
33 -> 62 [arrowhead = none ];
62 -> 34;

63 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
39 -> 63 [arrowhead = none ];
63 -> 44;

}

(2) Obligation:

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

appendA_in_gga(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → U1_gga(T8, T29, T30, T31, T33, appendA_in_gga(T30, T31, T33))
appendA_in_gga(.(T8, []), T42, .([], T42)) → appendA_out_gga(.(T8, []), T42, .([], T42))
appendA_in_gga([], T44, T44) → appendA_out_gga([], T44, T44)
U1_gga(T8, T29, T30, T31, T33, appendA_out_gga(T30, T31, T33)) → appendA_out_gga(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33)))

The argument filtering Pi contains the following mapping:
appendA_in_gga(x1, x2, x3)  =  appendA_in_gga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6)  =  U1_gga(x1, x2, x3, x4, x6)
[]  =  []
appendA_out_gga(x1, x2, x3)  =  appendA_out_gga(x1, x2, x3)

(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:

APPENDA_IN_GGA(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → U1_GGA(T8, T29, T30, T31, T33, appendA_in_gga(T30, T31, T33))
APPENDA_IN_GGA(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → APPENDA_IN_GGA(T30, T31, T33)

The TRS R consists of the following rules:

appendA_in_gga(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → U1_gga(T8, T29, T30, T31, T33, appendA_in_gga(T30, T31, T33))
appendA_in_gga(.(T8, []), T42, .([], T42)) → appendA_out_gga(.(T8, []), T42, .([], T42))
appendA_in_gga([], T44, T44) → appendA_out_gga([], T44, T44)
U1_gga(T8, T29, T30, T31, T33, appendA_out_gga(T30, T31, T33)) → appendA_out_gga(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33)))

The argument filtering Pi contains the following mapping:
appendA_in_gga(x1, x2, x3)  =  appendA_in_gga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6)  =  U1_gga(x1, x2, x3, x4, x6)
[]  =  []
appendA_out_gga(x1, x2, x3)  =  appendA_out_gga(x1, x2, x3)
APPENDA_IN_GGA(x1, x2, x3)  =  APPENDA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6)  =  U1_GGA(x1, x2, x3, x4, x6)

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

(4) Obligation:

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

APPENDA_IN_GGA(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → U1_GGA(T8, T29, T30, T31, T33, appendA_in_gga(T30, T31, T33))
APPENDA_IN_GGA(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → APPENDA_IN_GGA(T30, T31, T33)

The TRS R consists of the following rules:

appendA_in_gga(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → U1_gga(T8, T29, T30, T31, T33, appendA_in_gga(T30, T31, T33))
appendA_in_gga(.(T8, []), T42, .([], T42)) → appendA_out_gga(.(T8, []), T42, .([], T42))
appendA_in_gga([], T44, T44) → appendA_out_gga([], T44, T44)
U1_gga(T8, T29, T30, T31, T33, appendA_out_gga(T30, T31, T33)) → appendA_out_gga(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33)))

The argument filtering Pi contains the following mapping:
appendA_in_gga(x1, x2, x3)  =  appendA_in_gga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6)  =  U1_gga(x1, x2, x3, x4, x6)
[]  =  []
appendA_out_gga(x1, x2, x3)  =  appendA_out_gga(x1, x2, x3)
APPENDA_IN_GGA(x1, x2, x3)  =  APPENDA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6)  =  U1_GGA(x1, x2, x3, x4, x6)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

APPENDA_IN_GGA(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → APPENDA_IN_GGA(T30, T31, T33)

The TRS R consists of the following rules:

appendA_in_gga(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → U1_gga(T8, T29, T30, T31, T33, appendA_in_gga(T30, T31, T33))
appendA_in_gga(.(T8, []), T42, .([], T42)) → appendA_out_gga(.(T8, []), T42, .([], T42))
appendA_in_gga([], T44, T44) → appendA_out_gga([], T44, T44)
U1_gga(T8, T29, T30, T31, T33, appendA_out_gga(T30, T31, T33)) → appendA_out_gga(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33)))

The argument filtering Pi contains the following mapping:
appendA_in_gga(x1, x2, x3)  =  appendA_in_gga(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6)  =  U1_gga(x1, x2, x3, x4, x6)
[]  =  []
appendA_out_gga(x1, x2, x3)  =  appendA_out_gga(x1, x2, x3)
APPENDA_IN_GGA(x1, x2, x3)  =  APPENDA_IN_GGA(x1, x2)

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:

APPENDA_IN_GGA(.(T8, .(T29, T30)), T31, .(.(T29, T30), .(T30, T33))) → APPENDA_IN_GGA(T30, T31, T33)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPENDA_IN_GGA(x1, x2, x3)  =  APPENDA_IN_GGA(x1, x2)

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:

APPENDA_IN_GGA(.(T8, .(T29, T30)), T31) → APPENDA_IN_GGA(T30, T31)

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:

• APPENDA_IN_GGA(.(T8, .(T29, T30)), T31) → APPENDA_IN_GGA(T30, T31)
  The graph contains the following edges 1 > 1, 2 >= 2

(12) YES