Left Termination of the query pattern append_in_3(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).

Queries:

append(g,a,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

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 | ?_1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(=(T4, []), ','(!_1, =(T7, T8))) | append(T4, T2, T3), SCOPE: 1, CLAUSE: 1 | ?_1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(=(T4, []), ','(!_1, =(T7, T8))), SCOPE: 2, CLAUSE: 2 | append(T4, T2, T3), SCOPE: 1, CLAUSE: 1 | ?_1\n\nT4 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(!_1, =(T7, T8)) | append([], T2, T3), SCOPE: 1, CLAUSE: 1 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: append(T4, T2, T3), SCOPE: 1, CLAUSE: 1 | ?_1\n\nT4 is ground\nX8 is free; \n=(T4, []) !=? =(X8, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: =(T7, T8)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: =(T7, T8), 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\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(=(T11, .(X17, X18)), ','(!_1, ','(=(T14, .(X17, X19)), append(X18, T15, X19)))) | ?_1\n\nT11 is ground\nX8 is free; \nX17 is free; \nX18 is free; \nX19 is free; \n=(T11, []) !=? =(X8, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(=(T11, .(X17, X18)), ','(!_1, ','(=(T14, .(X17, X19)), append(X18, T15, X19)))), SCOPE: 4, CLAUSE: 2 | ?_1\n\nT11 is ground\nX8 is free; \nX17 is free; \nX18 is free; \nX19 is free; \n=(T11, []) !=? =(X8, X8)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(!_1, ','(=(T14, .(T17, X19)), append(T18, T15, X19))) | ?_1\n\nT17 is ground\nT18 is ground\nX8 is free; \nX19 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\nX8 is free\nX17 is free; \nX18 is free; \nX21 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(=(T14, .(T17, X19)), append(T18, T15, X19))\n\nT17 is ground\nT18 is ground\nX19 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(=(T14, .(T17, X19)), append(T18, T15, X19)), SCOPE: 5, CLAUSE: 2\n\nT17 is ground\nT18 is ground\nX19 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: append(T18, T22, T23)\n\nT18 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\nX19 is free\nX23 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 20 [arrowhead = none ];
20 -> 2;

21 [label="EVAL with clause\nappend(X4, X5, X6) :- ','(=(X4, []), ','(!, =(X5, X6))).\nand substitution\nT1 -> T4\nX4 -> T4\nT2 -> T7\nX5 -> T7\nT3 -> T8\nX6 -> T8\nT5 -> T7\nT6 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 21 [arrowhead = none ];
21 -> 3;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 22 [arrowhead = none ];
22 -> 4;

23 [label="EVAL with clause\n=(X8, X8).\nand substitution\nT4 -> []\nX8 -> []\nT9 -> []", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 23 [arrowhead = none ];
23 -> 5;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 24 [arrowhead = none ];
24 -> 6;

25 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 25 [arrowhead = none ];
25 -> 7;

26 [label="EVAL with clause\nappend(X14, X15, X16) :- ','(=(X14, .(X17, X18)), ','(!, ','(=(X16, .(X17, X19)), append(X18, X15, X19)))).\nand substitution\nT4 -> T11\nX14 -> T11\nT2 -> T15\nX15 -> T15\nT3 -> T14\nX16 -> T14\nT13 -> T14\nT12 -> T15", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 26 [arrowhead = none ];
26 -> 12;

27 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 27 [arrowhead = none ];
27 -> 8;

28 [label="EVAL with clause\n=(X10, X10).\nand substitution\nT7 -> T10\nX10 -> T10\nT8 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 28 [arrowhead = none ];
28 -> 9;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 29 [arrowhead = none ];
29 -> 10;

30 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 30 [arrowhead = none ];
30 -> 11;

31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 31 [arrowhead = none ];
31 -> 13;

32 [label="EVAL with clause\n=(X21, X21).\nand substitution\nT11 -> .(T17, T18)\nX21 -> .(T17, T18)\nX17 -> T17\nX18 -> T18\nT16 -> .(T17, T18)", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 32 [arrowhead = none ];
32 -> 14;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 33 [arrowhead = none ];
33 -> 15;

34 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 34 [arrowhead = none ];
34 -> 16;

35 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
16 -> 35 [arrowhead = none ];
35 -> 17;

36 [label="EVAL with clause\n=(X23, X23).\nand substitution\nT14 -> .(T20, T23)\nX23 -> .(T20, T23)\nT17 -> T20\nX19 -> T23\nT19 -> .(T20, T23)\nT15 -> T22\nT21 -> T23", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 36 [arrowhead = none ];
36 -> 18;

37 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 37 [arrowhead = none ];
37 -> 19;

38 [label="INSTANCE with matching:\nT1 -> T18\nT2 -> T22\nT3 -> T23", fontsize=14, style = filled, fillcolor = lightgrey];
18 -> 38 [arrowhead = none , style=dashed];
38 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

append1([], T10, T10).
append1(.(T20, T18), T22, .(T20, T23)) :- append1(T18, T22, T23).

Queries:

append1(g,a,a).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
append1_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

append1_in_gaa([], T10, T10) → append1_out_gaa([], T10, T10)
append1_in_gaa(.(T20, T18), T22, .(T20, T23)) → U1_gaa(T20, T18, T22, T23, append1_in_gaa(T18, T22, T23))
U1_gaa(T20, T18, T22, T23, append1_out_gaa(T18, T22, T23)) → append1_out_gaa(.(T20, T18), T22, .(T20, T23))

The argument filtering Pi contains the following mapping:
append1_in_gaa(x1, x2, x3)  =  append1_in_gaa(x1)
[]  =  []
append1_out_gaa(x1, x2, x3)  =  append1_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

(4) Obligation:

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

append1_in_gaa([], T10, T10) → append1_out_gaa([], T10, T10)
append1_in_gaa(.(T20, T18), T22, .(T20, T23)) → U1_gaa(T20, T18, T22, T23, append1_in_gaa(T18, T22, T23))
U1_gaa(T20, T18, T22, T23, append1_out_gaa(T18, T22, T23)) → append1_out_gaa(.(T20, T18), T22, .(T20, T23))

The argument filtering Pi contains the following mapping:
append1_in_gaa(x1, x2, x3)  =  append1_in_gaa(x1)
[]  =  []
append1_out_gaa(x1, x2, x3)  =  append1_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)

(5) 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:

APPEND1_IN_GAA(.(T20, T18), T22, .(T20, T23)) → U1_GAA(T20, T18, T22, T23, append1_in_gaa(T18, T22, T23))
APPEND1_IN_GAA(.(T20, T18), T22, .(T20, T23)) → APPEND1_IN_GAA(T18, T22, T23)

The TRS R consists of the following rules:

append1_in_gaa([], T10, T10) → append1_out_gaa([], T10, T10)
append1_in_gaa(.(T20, T18), T22, .(T20, T23)) → U1_gaa(T20, T18, T22, T23, append1_in_gaa(T18, T22, T23))
U1_gaa(T20, T18, T22, T23, append1_out_gaa(T18, T22, T23)) → append1_out_gaa(.(T20, T18), T22, .(T20, T23))

The argument filtering Pi contains the following mapping:
append1_in_gaa(x1, x2, x3)  =  append1_in_gaa(x1)
[]  =  []
append1_out_gaa(x1, x2, x3)  =  append1_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APPEND1_IN_GAA(x1, x2, x3)  =  APPEND1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x5)

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

(6) Obligation:

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

APPEND1_IN_GAA(.(T20, T18), T22, .(T20, T23)) → U1_GAA(T20, T18, T22, T23, append1_in_gaa(T18, T22, T23))
APPEND1_IN_GAA(.(T20, T18), T22, .(T20, T23)) → APPEND1_IN_GAA(T18, T22, T23)

The TRS R consists of the following rules:

append1_in_gaa([], T10, T10) → append1_out_gaa([], T10, T10)
append1_in_gaa(.(T20, T18), T22, .(T20, T23)) → U1_gaa(T20, T18, T22, T23, append1_in_gaa(T18, T22, T23))
U1_gaa(T20, T18, T22, T23, append1_out_gaa(T18, T22, T23)) → append1_out_gaa(.(T20, T18), T22, .(T20, T23))

The argument filtering Pi contains the following mapping:
append1_in_gaa(x1, x2, x3)  =  append1_in_gaa(x1)
[]  =  []
append1_out_gaa(x1, x2, x3)  =  append1_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APPEND1_IN_GAA(x1, x2, x3)  =  APPEND1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x5)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

APPEND1_IN_GAA(.(T20, T18), T22, .(T20, T23)) → APPEND1_IN_GAA(T18, T22, T23)

The TRS R consists of the following rules:

append1_in_gaa([], T10, T10) → append1_out_gaa([], T10, T10)
append1_in_gaa(.(T20, T18), T22, .(T20, T23)) → U1_gaa(T20, T18, T22, T23, append1_in_gaa(T18, T22, T23))
U1_gaa(T20, T18, T22, T23, append1_out_gaa(T18, T22, T23)) → append1_out_gaa(.(T20, T18), T22, .(T20, T23))

The argument filtering Pi contains the following mapping:
append1_in_gaa(x1, x2, x3)  =  append1_in_gaa(x1)
[]  =  []
append1_out_gaa(x1, x2, x3)  =  append1_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APPEND1_IN_GAA(x1, x2, x3)  =  APPEND1_IN_GAA(x1)

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

APPEND1_IN_GAA(.(T20, T18), T22, .(T20, T23)) → APPEND1_IN_GAA(T18, T22, T23)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND1_IN_GAA(x1, x2, x3)  =  APPEND1_IN_GAA(x1)

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

(11) PiDPToQDPProof (SOUND transformation)

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

(12) Obligation:

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

APPEND1_IN_GAA(.(T20, T18)) → APPEND1_IN_GAA(T18)

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

(13) 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:

• APPEND1_IN_GAA(.(T20, T18)) → APPEND1_IN_GAA(T18)
  The graph contains the following edges 1 > 1

(14) TRUE