Left Termination of the query pattern app_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

app([], Y, Y).
app(X, Y, .(H, Z)) :- ','(no(empty(X)), ','(head(X, H), ','(tail(X, T), app(T, Y, Z)))).
head([], X1).
head(.(X, X2), X).
tail([], []).
tail(.(X3, Xs), Xs).
empty([]).
no(X) :- ','(X, ','(!, failure(a))).
no(X4).
failure(b).

Queries:

app(g,a,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: app(T1, T2, T3)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: app(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | app(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: app(T1, T2, T3), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: app(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\nX7 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(no(empty(T5)), ','(head(T5, T9), ','(tail(T5, X16), app(X16, T10, T11))))\n\nT5 is ground\nX16 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\nX12 is free\nX13 is free; \nX14 is free; \nX15 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(no(empty(T5)), ','(head(T5, T9), ','(tail(T5, X16), app(X16, T10, T11)))), SCOPE: 2, CLAUSE: 7 | ','(no(empty(T5)), ','(head(T5, T9), ','(tail(T5, X16), app(X16, T10, T11)))), SCOPE: 2, CLAUSE: 8 | ?_2\n\nT5 is ground\nX16 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(empty(T12), ','(!_2, ','(failure(a), ','(head(T12, T9), ','(tail(T12, X16), app(X16, T10, T11)))))) | ','(no(empty(T12)), ','(head(T12, T9), ','(tail(T12, X16), app(X16, T10, T11)))), SCOPE: 2, CLAUSE: 8 | ?_2\n\nT12 is ground\nX16 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(empty(T12), ','(!_2, ','(failure(a), ','(head(T12, T9), ','(tail(T12, X16), app(X16, T10, T11)))))), SCOPE: 3, CLAUSE: 6 | ','(no(empty(T12)), ','(head(T12, T9), ','(tail(T12, X16), app(X16, T10, T11)))), SCOPE: 2, CLAUSE: 8 | ?_2\n\nT12 is ground\nX16 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(!_2, ','(failure(a), ','(head([], T9), ','(tail([], X16), app(X16, T10, T11))))) | ','(no(empty([])), ','(head([], T9), ','(tail([], X16), app(X16, T10, T11)))), SCOPE: 2, CLAUSE: 8 | ?_2\n\nX16 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(no(empty(T12)), ','(head(T12, T9), ','(tail(T12, X16), app(X16, T10, T11)))), SCOPE: 2, CLAUSE: 8\n\nT12 is ground\nX16 is free; \nempty(T12) !=? empty([])", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(failure(a), ','(head([], T9), ','(tail([], X16), app(X16, T10, T11))))\n\nX16 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(failure(a), ','(head([], T9), ','(tail([], X16), app(X16, T10, T11)))), SCOPE: 4, CLAUSE: 9\n\nX16 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(head(T13, T9), ','(tail(T13, X16), app(X16, T10, T11)))\n\nT13 is ground\nX16 is free; \nempty(T13) !=? empty([])", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(head(T13, T9), ','(tail(T13, X16), app(X16, T10, T11))), SCOPE: 5, CLAUSE: 2 | ','(head(T13, T9), ','(tail(T13, X16), app(X16, T10, T11))), SCOPE: 5, CLAUSE: 3\n\nT13 is ground\nX16 is free; \nempty(T13) !=? empty([])", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(head(T13, T9), ','(tail(T13, X16), app(X16, T10, T11))), SCOPE: 5, CLAUSE: 3\n\nT13 is ground\nX16 is free; \nX22 is free; \nempty(T13) !=? empty([])\nhead(T13, T9) !=? head([], X22)", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(tail(.(T14, T15), X16), app(X16, T16, T17))\n\nT14 is ground\nT15 is ground\nX16 is free; \nX22 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\nX22 is free\nX25 is free; \nX26 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ','(tail(.(T14, T15), X16), app(X16, T16, T17)), SCOPE: 6, CLAUSE: 4 | ','(tail(.(T14, T15), X16), app(X16, T16, T17)), SCOPE: 6, CLAUSE: 5\n\nT14 is ground\nT15 is ground\nX16 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(tail(.(T14, T15), X16), app(X16, T16, T17)), SCOPE: 6, CLAUSE: 5\n\nT14 is ground\nT15 is ground\nX16 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: app(T19, T16, T17)\n\nT19 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 26 [arrowhead = none ];
26 -> 2;

27 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 27 [arrowhead = none ];
27 -> 3;

28 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 28 [arrowhead = none ];
28 -> 4;

29 [label="EVAL with clause\napp([], X7, X7).\nand substitution\nT1 -> []\nT2 -> T4\nX7 -> T4\nT3 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 29 [arrowhead = none ];
29 -> 5;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 30 [arrowhead = none ];
30 -> 6;

31 [label="EVAL with clause\napp(X12, X13, .(X14, X15)) :- ','(no(empty(X12)), ','(head(X12, X14), ','(tail(X12, X16), app(X16, X13, X15)))).\nand substitution\nT1 -> T5\nX12 -> T5\nT2 -> T10\nX13 -> T10\nX14 -> T9\nX15 -> T11\nT3 -> .(T9, T11)\nT7 -> T9\nT6 -> T10\nT8 -> T11", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 31 [arrowhead = none ];
31 -> 8;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 32 [arrowhead = none ];
32 -> 9;

33 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 33 [arrowhead = none ];
33 -> 7;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 34 [arrowhead = none ];
34 -> 10;

35 [label="EVAL with clause\nno(X18) :- ','(X18, ','(!, failure(a))).\nand substitution\nT5 -> T12\nX18 -> empty(T12)", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 35 [arrowhead = none ];
35 -> 11;

36 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 36 [arrowhead = none ];
36 -> 12;

37 [label="EVAL with clause\nempty([]).\nand substitution\nT12 -> []", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 37 [arrowhead = none ];
37 -> 13;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 38 [arrowhead = none ];
38 -> 14;

39 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 39 [arrowhead = none ];
39 -> 15;

40 [label="EVAL with clause\nno(X20).\nand substitution\nT12 -> T13\nX20 -> empty(T13)", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 40 [arrowhead = none ];
40 -> 18;

41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 41 [arrowhead = none ];
41 -> 16;

42 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = white];
16 -> 42 [arrowhead = none ];
42 -> 17;

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
18 -> 43 [arrowhead = none ];
43 -> 19;

44 [label="BACKTRACK\nfor clause: head([], X1)\nwith clash: (empty(T13), empty([]))", fontsize=14, style = filled, fillcolor = white];
19 -> 44 [arrowhead = none ];
44 -> 20;

45 [label="EVAL with clause\nhead(.(X25, X26), X25).\nand substitution\nX25 -> T14\nX26 -> T15\nT13 -> .(T14, T15)\nT9 -> T14\nT10 -> T16\nT11 -> T17", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 45 [arrowhead = none ];
45 -> 21;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 46 [arrowhead = none ];
46 -> 22;

47 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
21 -> 47 [arrowhead = none ];
47 -> 23;

48 [label="BACKTRACK\nfor clause: tail([], [])because of non-unification", fontsize=14, style = filled, fillcolor = white];
23 -> 48 [arrowhead = none ];
48 -> 24;

49 [label="EVAL with clause\ntail(.(X29, X30), X30).\nand substitution\nT14 -> T18\nX29 -> T18\nT15 -> T19\nX30 -> T19\nX16 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 49 [arrowhead = none ];
49 -> 25;

50 [label="INSTANCE with matching:\nT1 -> T19\nT2 -> T16\nT3 -> T17", fontsize=14, style = filled, fillcolor = lightgrey];
25 -> 50 [arrowhead = none , style=dashed];
50 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

app1([], T4, T4).
app1(.(T18, T19), T16, .(T18, T17)) :- app1(T19, T16, T17).

Queries:

app1(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:
app1_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

app1_in_gaa([], T4, T4) → app1_out_gaa([], T4, T4)
app1_in_gaa(.(T18, T19), T16, .(T18, T17)) → U1_gaa(T18, T19, T16, T17, app1_in_gaa(T19, T16, T17))
U1_gaa(T18, T19, T16, T17, app1_out_gaa(T19, T16, T17)) → app1_out_gaa(.(T18, T19), T16, .(T18, T17))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, 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:

app1_in_gaa([], T4, T4) → app1_out_gaa([], T4, T4)
app1_in_gaa(.(T18, T19), T16, .(T18, T17)) → U1_gaa(T18, T19, T16, T17, app1_in_gaa(T19, T16, T17))
U1_gaa(T18, T19, T16, T17, app1_out_gaa(T19, T16, T17)) → app1_out_gaa(.(T18, T19), T16, .(T18, T17))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, 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:

APP1_IN_GAA(.(T18, T19), T16, .(T18, T17)) → U1_GAA(T18, T19, T16, T17, app1_in_gaa(T19, T16, T17))
APP1_IN_GAA(.(T18, T19), T16, .(T18, T17)) → APP1_IN_GAA(T19, T16, T17)

The TRS R consists of the following rules:

app1_in_gaa([], T4, T4) → app1_out_gaa([], T4, T4)
app1_in_gaa(.(T18, T19), T16, .(T18, T17)) → U1_gaa(T18, T19, T16, T17, app1_in_gaa(T19, T16, T17))
U1_gaa(T18, T19, T16, T17, app1_out_gaa(T19, T16, T17)) → app1_out_gaa(.(T18, T19), T16, .(T18, T17))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
APP1_IN_GAA(x1, x2, x3)  =  APP1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, x5)

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

(6) Obligation:

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

APP1_IN_GAA(.(T18, T19), T16, .(T18, T17)) → U1_GAA(T18, T19, T16, T17, app1_in_gaa(T19, T16, T17))
APP1_IN_GAA(.(T18, T19), T16, .(T18, T17)) → APP1_IN_GAA(T19, T16, T17)

The TRS R consists of the following rules:

app1_in_gaa([], T4, T4) → app1_out_gaa([], T4, T4)
app1_in_gaa(.(T18, T19), T16, .(T18, T17)) → U1_gaa(T18, T19, T16, T17, app1_in_gaa(T19, T16, T17))
U1_gaa(T18, T19, T16, T17, app1_out_gaa(T19, T16, T17)) → app1_out_gaa(.(T18, T19), T16, .(T18, T17))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
APP1_IN_GAA(x1, x2, x3)  =  APP1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, 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:

APP1_IN_GAA(.(T18, T19), T16, .(T18, T17)) → APP1_IN_GAA(T19, T16, T17)

The TRS R consists of the following rules:

app1_in_gaa([], T4, T4) → app1_out_gaa([], T4, T4)
app1_in_gaa(.(T18, T19), T16, .(T18, T17)) → U1_gaa(T18, T19, T16, T17, app1_in_gaa(T19, T16, T17))
U1_gaa(T18, T19, T16, T17, app1_out_gaa(T19, T16, T17)) → app1_out_gaa(.(T18, T19), T16, .(T18, T17))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
APP1_IN_GAA(x1, x2, x3)  =  APP1_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:

APP1_IN_GAA(.(T18, T19), T16, .(T18, T17)) → APP1_IN_GAA(T19, T16, T17)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APP1_IN_GAA(x1, x2, x3)  =  APP1_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:

APP1_IN_GAA(.(T18, T19)) → APP1_IN_GAA(T19)

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:

• APP1_IN_GAA(.(T18, T19)) → APP1_IN_GAA(T19)
  The graph contains the following edges 1 > 1

(14) TRUE