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

(0) Obligation:

Clauses:

append(X, Y, Z) :- ','(=(X, []), ','(!, =(Y, Z))).
append(X, Y, Z) :- ','(=(X, .(H, Xs)), ','(!, ','(=(Z, .(H, Zs)), append(Xs, Y, Zs)))).
=(X, X).

Query: append(g,a,a)

(1) BuiltinConflictTransformerProof (EQUIVALENT transformation)

Renamed defined predicates conflicting with built-in predicates [PROLOG].

(2) Obligation:

Clauses:

append(X, Y, Z) :- ','(user_defined_=(X, []), ','(!, user_defined_=(Y, Z))).
append(X, Y, Z) :- ','(user_defined_=(X, .(H, Xs)), ','(!, ','(user_defined_=(Z, .(H, Zs)), append(Xs, Y, Zs)))).
user_defined_=(X, X).

Query: append(g,a,a)

(3) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: append(T1, T2, T3)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: append(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | append(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(user_defined_=(T7, []), ','(!_1, user_defined_=(T10, T11))) | append(T7, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(user_defined_=(T7, []), ','(!_1, user_defined_=(T10, T11))), SCOPE: 2, CLAUSE: 2 | ?_2 | append(T7, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(!_1, user_defined_=(T10, T11)) | ?_2 | append([], T2, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ?_2 | append(T7, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT7 is ground\nX9 is free\nuser_defined_=(T7, []) !=? user_defined_=(X9, X9)", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: user_defined_=(T10, T11)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: user_defined_=(T10, T11), SCOPE: 3, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: append(T7, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT7 is ground\nX9 is free\nuser_defined_=(T7, []) !=? user_defined_=(X9, X9)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(user_defined_=(T26, .(X25, X26)), ','(!_1, ','(user_defined_=(T29, .(X25, X27)), append(X26, T30, X27))))\n\nT26 is ground\nX9 is free\nX25 is free\nX26 is free\nX27 is free\nuser_defined_=(T26, []) !=? user_defined_=(X9, X9)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(user_defined_=(T26, .(X25, X26)), ','(!_1, ','(user_defined_=(T29, .(X25, X27)), append(X26, T30, X27)))), SCOPE: 4, CLAUSE: 2\n\nT26 is ground\nX9 is free\nX25 is free\nX26 is free\nX27 is free\nuser_defined_=(T26, []) !=? user_defined_=(X9, X9)", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(!_1, ','(user_defined_=(T29, .(T38, X27)), append(T39, T30, X27)))\n\nT38 is ground\nT39 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(user_defined_=(T29, .(T38, X27)), append(T39, T30, X27))\n\nT38 is ground\nT39 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(user_defined_=(T29, .(T38, X27)), append(T39, T30, X27)), SCOPE: 5, CLAUSE: 2\n\nT38 is ground\nT39 is ground\nX27 is free", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: append(T39, T51, T52)\n\nT39 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 21 [arrowhead = none ];
21 -> 2;

22 [label="ONLY EVAL with clause\nappend(X4, X5, X6) :- ','(user_defined_=(X4, []), ','(!_1, user_defined_=(X5, X6))).\nand substitutionT1 -> T7,\nX4 -> T7,\nT2 -> T10,\nX5 -> T10,\nT3 -> T11,\nX6 -> T11,\nT8 -> T10,\nT9 -> T11", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 22 [arrowhead = none ];
22 -> 3;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 23 [arrowhead = none ];
23 -> 4;

24 [label="EVAL with clause\nuser_defined_=(X9, X9).\nand substitutionT7 -> [],\nX9 -> [],\nT14 -> []", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 24 [arrowhead = none ];
24 -> 5;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 25 [arrowhead = none ];
25 -> 6;

26 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 26 [arrowhead = none ];
26 -> 7;

27 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 27 [arrowhead = none ];
27 -> 12;

28 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 28 [arrowhead = none ];
28 -> 8;

29 [label="EVAL with clause\nuser_defined_=(X12, X12).\nand substitutionT10 -> T17,\nX12 -> T17,\nT11 -> T17", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 29 [arrowhead = none ];
29 -> 9;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 30 [arrowhead = none ];
30 -> 10;

31 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 31 [arrowhead = none ];
31 -> 11;

32 [label="ONLY EVAL with clause\nappend(X22, X23, X24) :- ','(user_defined_=(X22, .(X25, X26)), ','(!_1, ','(user_defined_=(X24, .(X25, X27)), append(X26, X23, X27)))).\nand substitutionT7 -> T26,\nX22 -> T26,\nT2 -> T30,\nX23 -> T30,\nT3 -> T29,\nX24 -> T29,\nT28 -> T29,\nT27 -> T30", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 32 [arrowhead = none ];
32 -> 13;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 33 [arrowhead = none ];
33 -> 14;

34 [label="EVAL with clause\nuser_defined_=(X30, X30).\nand substitutionT26 -> .(T38, T39),\nX30 -> .(T38, T39),\nX25 -> T38,\nX26 -> T39,\nT37 -> .(T38, T39)", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 34 [arrowhead = none ];
34 -> 15;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 35 [arrowhead = none ];
35 -> 16;

36 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 36 [arrowhead = none ];
36 -> 17;

37 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 37 [arrowhead = none ];
37 -> 18;

38 [label="EVAL with clause\nuser_defined_=(X37, X37).\nand substitutionT29 -> .(T49, T52),\nX37 -> .(T49, T52),\nT38 -> T49,\nX27 -> T52,\nT48 -> .(T49, T52),\nT30 -> T51,\nT50 -> T52", fontsize=14, style = filled, fillcolor = lightblue];
18 -> 38 [arrowhead = none ];
38 -> 19;

39 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 39 [arrowhead = none ];
39 -> 20;

40 [label="INSTANCE with matching:\nT1 -> T39\nT2 -> T51\nT3 -> T52", fontsize=14, style = filled, fillcolor = lightgrey];
19 -> 40 [arrowhead = none , style=dashed];
40 -> 1[style=dashed];

}

(4) Obligation:

Clauses:

appendA([], T17, T17).
appendA(.(T49, T39), T51, .(T49, T52)) :- appendA(T39, T51, T52).

Query: appendA(g,a,a)

(5) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
appendA_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

appendA_in_gaa([], T17, T17) → appendA_out_gaa([], T17, T17)
appendA_in_gaa(.(T49, T39), T51, .(T49, T52)) → U1_gaa(T49, T39, T51, T52, appendA_in_gaa(T39, T51, T52))
U1_gaa(T49, T39, T51, T52, appendA_out_gaa(T39, T51, T52)) → appendA_out_gaa(.(T49, T39), T51, .(T49, T52))

The argument filtering Pi contains the following mapping:
appendA_in_gaa(x1, x2, x3)  =  appendA_in_gaa(x1)
[]  =  []
appendA_out_gaa(x1, x2, x3)  =  appendA_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(6) Obligation:

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

appendA_in_gaa([], T17, T17) → appendA_out_gaa([], T17, T17)
appendA_in_gaa(.(T49, T39), T51, .(T49, T52)) → U1_gaa(T49, T39, T51, T52, appendA_in_gaa(T39, T51, T52))
U1_gaa(T49, T39, T51, T52, appendA_out_gaa(T39, T51, T52)) → appendA_out_gaa(.(T49, T39), T51, .(T49, T52))

The argument filtering Pi contains the following mapping:
appendA_in_gaa(x1, x2, x3)  =  appendA_in_gaa(x1)
[]  =  []
appendA_out_gaa(x1, x2, x3)  =  appendA_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)

(7) 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_GAA(.(T49, T39), T51, .(T49, T52)) → U1_GAA(T49, T39, T51, T52, appendA_in_gaa(T39, T51, T52))
APPENDA_IN_GAA(.(T49, T39), T51, .(T49, T52)) → APPENDA_IN_GAA(T39, T51, T52)

The TRS R consists of the following rules:

appendA_in_gaa([], T17, T17) → appendA_out_gaa([], T17, T17)
appendA_in_gaa(.(T49, T39), T51, .(T49, T52)) → U1_gaa(T49, T39, T51, T52, appendA_in_gaa(T39, T51, T52))
U1_gaa(T49, T39, T51, T52, appendA_out_gaa(T39, T51, T52)) → appendA_out_gaa(.(T49, T39), T51, .(T49, T52))

The argument filtering Pi contains the following mapping:
appendA_in_gaa(x1, x2, x3)  =  appendA_in_gaa(x1)
[]  =  []
appendA_out_gaa(x1, x2, x3)  =  appendA_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APPENDA_IN_GAA(x1, x2, x3)  =  APPENDA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x5)

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

(8) Obligation:

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

APPENDA_IN_GAA(.(T49, T39), T51, .(T49, T52)) → U1_GAA(T49, T39, T51, T52, appendA_in_gaa(T39, T51, T52))
APPENDA_IN_GAA(.(T49, T39), T51, .(T49, T52)) → APPENDA_IN_GAA(T39, T51, T52)

The TRS R consists of the following rules:

appendA_in_gaa([], T17, T17) → appendA_out_gaa([], T17, T17)
appendA_in_gaa(.(T49, T39), T51, .(T49, T52)) → U1_gaa(T49, T39, T51, T52, appendA_in_gaa(T39, T51, T52))
U1_gaa(T49, T39, T51, T52, appendA_out_gaa(T39, T51, T52)) → appendA_out_gaa(.(T49, T39), T51, .(T49, T52))

The argument filtering Pi contains the following mapping:
appendA_in_gaa(x1, x2, x3)  =  appendA_in_gaa(x1)
[]  =  []
appendA_out_gaa(x1, x2, x3)  =  appendA_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APPENDA_IN_GAA(x1, x2, x3)  =  APPENDA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x5)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Obligation:

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

APPENDA_IN_GAA(.(T49, T39), T51, .(T49, T52)) → APPENDA_IN_GAA(T39, T51, T52)

The TRS R consists of the following rules:

appendA_in_gaa([], T17, T17) → appendA_out_gaa([], T17, T17)
appendA_in_gaa(.(T49, T39), T51, .(T49, T52)) → U1_gaa(T49, T39, T51, T52, appendA_in_gaa(T39, T51, T52))
U1_gaa(T49, T39, T51, T52, appendA_out_gaa(T39, T51, T52)) → appendA_out_gaa(.(T49, T39), T51, .(T49, T52))

The argument filtering Pi contains the following mapping:
appendA_in_gaa(x1, x2, x3)  =  appendA_in_gaa(x1)
[]  =  []
appendA_out_gaa(x1, x2, x3)  =  appendA_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APPENDA_IN_GAA(x1, x2, x3)  =  APPENDA_IN_GAA(x1)

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

(11) UsableRulesProof (EQUIVALENT transformation)

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

(12) Obligation:

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

APPENDA_IN_GAA(.(T49, T39), T51, .(T49, T52)) → APPENDA_IN_GAA(T39, T51, T52)

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

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

(13) PiDPToQDPProof (SOUND transformation)

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

(14) Obligation:

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

APPENDA_IN_GAA(.(T49, T39)) → APPENDA_IN_GAA(T39)

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

(15) 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_GAA(.(T49, T39)) → APPENDA_IN_GAA(T39)
  The graph contains the following edges 1 > 1

(16) YES