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, Z) :- ','(!, eq(Y, Z)).
app(X, Y, .(H, Z)) :- ','(head(X, H), ','(tail(X, T), app(T, Y, Z))).
head([], X1).
head(.(X, X2), X).
tail([], []).
tail(.(X3, Xs), Xs).
eq(X, X).

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 | ?_1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(!_1, eq(T6, T7)) | app([], T2, T3), SCOPE: 1, CLAUSE: 1 | ?_1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: app(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX6 is free; \nX7 is free; \napp(T1, T2, T3) !=? app([], X6, X7)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: eq(T6, T7)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: eq(T6, T7), SCOPE: 2, CLAUSE: 6\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(head(T9, T13), ','(tail(T9, X18), app(X18, T14, T15)))\n\nT9 is ground\nX6 is free; \nX7 is free; \nX18 is free; \napp(T9, T2, T3) !=? app([], X6, X7)", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\nX6 is free\nX7 is free; \nX14 is free; \nX15 is free; \nX16 is free; \nX17 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(head(T9, T13), ','(tail(T9, X18), app(X18, T14, T15))), SCOPE: 3, CLAUSE: 2 | ','(head(T9, T13), ','(tail(T9, X18), app(X18, T14, T15))), SCOPE: 3, CLAUSE: 3\n\nT9 is ground\nX6 is free; \nX7 is free; \nX18 is free; \napp(T9, T2, T3) !=? app([], X6, X7)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(head(T9, T13), ','(tail(T9, X18), app(X18, T14, T15))), SCOPE: 3, CLAUSE: 3\n\nT9 is ground\nX6 is free; \nX7 is free; \nX18 is free; \nX20 is free; \napp(T9, T2, T3) !=? app([], X6, X7)\nhead(T9, T13) !=? head([], X20)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(tail(.(T16, T17), X18), app(X18, T18, T19))\n\nT16 is ground\nT17 is ground\nX6 is free; \nX7 is free; \nX18 is free; \nX20 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\nX6 is free\nX7 is free; \nX20 is free; \nX23 is free; \nX24 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(tail(.(T16, T17), X18), app(X18, T18, T19)), SCOPE: 4, CLAUSE: 4 | ','(tail(.(T16, T17), X18), app(X18, T18, T19)), SCOPE: 4, CLAUSE: 5\n\nT16 is ground\nT17 is ground\nX18 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(tail(.(T16, T17), X18), app(X18, T18, T19)), SCOPE: 4, CLAUSE: 5\n\nT16 is ground\nT17 is ground\nX18 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: app(T21, T18, T19)\n\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 19 [arrowhead = none ];
19 -> 2;

20 [label="EVAL with clause\napp([], X6, X7) :- ','(!, eq(X6, X7)).\nand substitution\nT1 -> []\nT2 -> T6\nX6 -> T6\nT3 -> T7\nX7 -> T7\nT4 -> T6\nT5 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 20 [arrowhead = none ];
20 -> 3;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 21 [arrowhead = none ];
21 -> 4;

22 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 22 [arrowhead = none ];
22 -> 5;

23 [label="EVAL with clause\napp(X14, X15, .(X16, X17)) :- ','(head(X14, X16), ','(tail(X14, X18), app(X18, X15, X17))).\nand substitution\nT1 -> T9\nX14 -> T9\nT2 -> T14\nX15 -> T14\nX16 -> T13\nX17 -> T15\nT3 -> .(T13, T15)\nT11 -> T13\nT10 -> T14\nT12 -> T15", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 23 [arrowhead = none ];
23 -> 10;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 24 [arrowhead = none ];
24 -> 11;

25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 25 [arrowhead = none ];
25 -> 6;

26 [label="EVAL with clause\neq(X9, X9).\nand substitution\nT6 -> T8\nX9 -> T8\nT7 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 26 [arrowhead = none ];
26 -> 7;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 27 [arrowhead = none ];
27 -> 8;

28 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 28 [arrowhead = none ];
28 -> 9;

29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 29 [arrowhead = none ];
29 -> 12;

30 [label="BACKTRACK\nfor clause: head([], X1)\nwith clash: (app(T9, T2, T3), app([], X6, X7))", fontsize=14, style = filled, fillcolor = white];
12 -> 30 [arrowhead = none ];
30 -> 13;

31 [label="EVAL with clause\nhead(.(X23, X24), X23).\nand substitution\nX23 -> T16\nX24 -> T17\nT9 -> .(T16, T17)\nT13 -> T16\nT14 -> T18\nT15 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 31 [arrowhead = none ];
31 -> 14;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 32 [arrowhead = none ];
32 -> 15;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 33 [arrowhead = none ];
33 -> 16;

34 [label="BACKTRACK\nfor clause: tail([], [])because of non-unification", fontsize=14, style = filled, fillcolor = white];
16 -> 34 [arrowhead = none ];
34 -> 17;

35 [label="EVAL with clause\ntail(.(X27, X28), X28).\nand substitution\nT16 -> T20\nX27 -> T20\nT17 -> T21\nX28 -> T21\nX18 -> T21", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 35 [arrowhead = none ];
35 -> 18;

36 [label="INSTANCE with matching:\nT1 -> T21\nT2 -> T18\nT3 -> T19", fontsize=14, style = filled, fillcolor = lightgrey];
18 -> 36 [arrowhead = none , style=dashed];
36 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

app1([], T8, T8).
app1(.(T20, T21), T18, .(T20, T19)) :- app1(T21, T18, T19).

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([], T8, T8) → app1_out_gaa([], T8, T8)
app1_in_gaa(.(T20, T21), T18, .(T20, T19)) → U1_gaa(T20, T21, T18, T19, app1_in_gaa(T21, T18, T19))
U1_gaa(T20, T21, T18, T19, app1_out_gaa(T21, T18, T19)) → app1_out_gaa(.(T20, T21), T18, .(T20, T19))

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, 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:

app1_in_gaa([], T8, T8) → app1_out_gaa([], T8, T8)
app1_in_gaa(.(T20, T21), T18, .(T20, T19)) → U1_gaa(T20, T21, T18, T19, app1_in_gaa(T21, T18, T19))
U1_gaa(T20, T21, T18, T19, app1_out_gaa(T21, T18, T19)) → app1_out_gaa(.(T20, T21), T18, .(T20, T19))

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, 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:

APP1_IN_GAA(.(T20, T21), T18, .(T20, T19)) → U1_GAA(T20, T21, T18, T19, app1_in_gaa(T21, T18, T19))
APP1_IN_GAA(.(T20, T21), T18, .(T20, T19)) → APP1_IN_GAA(T21, T18, T19)

The TRS R consists of the following rules:

app1_in_gaa([], T8, T8) → app1_out_gaa([], T8, T8)
app1_in_gaa(.(T20, T21), T18, .(T20, T19)) → U1_gaa(T20, T21, T18, T19, app1_in_gaa(T21, T18, T19))
U1_gaa(T20, T21, T18, T19, app1_out_gaa(T21, T18, T19)) → app1_out_gaa(.(T20, T21), T18, .(T20, T19))

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, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APP1_IN_GAA(x1, x2, x3)  =  APP1_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:

APP1_IN_GAA(.(T20, T21), T18, .(T20, T19)) → U1_GAA(T20, T21, T18, T19, app1_in_gaa(T21, T18, T19))
APP1_IN_GAA(.(T20, T21), T18, .(T20, T19)) → APP1_IN_GAA(T21, T18, T19)

The TRS R consists of the following rules:

app1_in_gaa([], T8, T8) → app1_out_gaa([], T8, T8)
app1_in_gaa(.(T20, T21), T18, .(T20, T19)) → U1_gaa(T20, T21, T18, T19, app1_in_gaa(T21, T18, T19))
U1_gaa(T20, T21, T18, T19, app1_out_gaa(T21, T18, T19)) → app1_out_gaa(.(T20, T21), T18, .(T20, T19))

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, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APP1_IN_GAA(x1, x2, x3)  =  APP1_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:

APP1_IN_GAA(.(T20, T21), T18, .(T20, T19)) → APP1_IN_GAA(T21, T18, T19)

The TRS R consists of the following rules:

app1_in_gaa([], T8, T8) → app1_out_gaa([], T8, T8)
app1_in_gaa(.(T20, T21), T18, .(T20, T19)) → U1_gaa(T20, T21, T18, T19, app1_in_gaa(T21, T18, T19))
U1_gaa(T20, T21, T18, T19, app1_out_gaa(T21, T18, T19)) → app1_out_gaa(.(T20, T21), T18, .(T20, T19))

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, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(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(.(T20, T21), T18, .(T20, T19)) → APP1_IN_GAA(T21, T18, T19)

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(.(T20, T21)) → APP1_IN_GAA(T21)

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(.(T20, T21)) → APP1_IN_GAA(T21)
  The graph contains the following edges 1 > 1

(14) TRUE