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

(0) Obligation:

Clauses:

app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).

Query: app(g,a,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: 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: true | app([], T2, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: app(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nX2 is free\napp(T1, T2, T3) !=? app([], X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: app([], T2, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: app(T11, T14, T15)\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: app(T11, T14, T15), SCOPE: 2, CLAUSE: 0 | app(T11, T14, T15), SCOPE: 2, CLAUSE: 1\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: app(T11, T14, T15), SCOPE: 2, CLAUSE: 0\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: app(T11, T14, T15), SCOPE: 2, CLAUSE: 1\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: app(T30, T33, T34)\n\nT30 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 17 [arrowhead = none ];
17 -> 2;

18 [label="EVAL with clause\napp([], X2, X2).\nand substitutionT1 -> [],\nT2 -> T5,\nX2 -> T5,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 18 [arrowhead = none ];
18 -> 3;

19 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 19 [arrowhead = none ];
19 -> 4;

20 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 20 [arrowhead = none ];
20 -> 5;

21 [label="EVAL with clause\napp(.(X11, X12), X13, .(X11, X14)) :- app(X12, X13, X14).\nand substitutionX11 -> T10,\nX12 -> T11,\nT1 -> .(T10, T11),\nT2 -> T14,\nX13 -> T14,\nX14 -> T15,\nT3 -> .(T10, T15),\nT12 -> T14,\nT13 -> T15", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 21 [arrowhead = none ];
21 -> 7;

22 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 22 [arrowhead = none ];
22 -> 8;

23 [label="BACKTRACK\nfor clause: app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 23 [arrowhead = none ];
23 -> 6;

24 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 24 [arrowhead = none ];
24 -> 9;

25 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 25 [arrowhead = none ];
25 -> 10;

26 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 26 [arrowhead = none ];
26 -> 11;

27 [label="EVAL with clause\napp([], X19, X19).\nand substitutionT11 -> [],\nT14 -> T20,\nX19 -> T20,\nT15 -> T20", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 27 [arrowhead = none ];
27 -> 12;

28 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 28 [arrowhead = none ];
28 -> 13;

29 [label="EVAL with clause\napp(.(X28, X29), X30, .(X28, X31)) :- app(X29, X30, X31).\nand substitutionX28 -> T29,\nX29 -> T30,\nT11 -> .(T29, T30),\nT14 -> T33,\nX30 -> T33,\nX31 -> T34,\nT15 -> .(T29, T34),\nT31 -> T33,\nT32 -> T34", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 29 [arrowhead = none ];
29 -> 15;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 30 [arrowhead = none ];
30 -> 16;

31 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 31 [arrowhead = none ];
31 -> 14;

32 [label="INSTANCE with matching:\nT1 -> T30\nT2 -> T33\nT3 -> T34", fontsize=14, style = filled, fillcolor = lightgrey];
15 -> 32 [arrowhead = none , style=dashed];
32 -> 1[style=dashed];

}

(2) Obligation:

Triples:

appA(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) :- appA(X3, X4, X5).

Clauses:

appcA([], X1, X1).
appcA(.(X1, []), X2, .(X1, X2)).
appcA(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) :- appcA(X3, X4, X5).

Afs:

appA(x1, x2, x3)  =  appA(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
appA_in: (b,f,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

APPA_IN_GAA(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → U1_GAA(X1, X2, X3, X4, X5, appA_in_gaa(X3, X4, X5))
APPA_IN_GAA(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → APPA_IN_GAA(X3, X4, X5)

R is empty.
The argument filtering Pi contains the following mapping:
appA_in_gaa(x1, x2, x3)  =  appA_in_gaa(x1)
.(x1, x2)  =  .(x1, x2)
APPA_IN_GAA(x1, x2, x3)  =  APPA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x2, x3, x6)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

APPA_IN_GAA(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → U1_GAA(X1, X2, X3, X4, X5, appA_in_gaa(X3, X4, X5))
APPA_IN_GAA(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → APPA_IN_GAA(X3, X4, X5)

R is empty.
The argument filtering Pi contains the following mapping:
appA_in_gaa(x1, x2, x3)  =  appA_in_gaa(x1)
.(x1, x2)  =  .(x1, x2)
APPA_IN_GAA(x1, x2, x3)  =  APPA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x2, x3, 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:

APPA_IN_GAA(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → APPA_IN_GAA(X3, X4, X5)

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

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

(7) PiDPToQDPProof (SOUND transformation)

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

(8) Obligation:

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

APPA_IN_GAA(.(X1, .(X2, X3))) → APPA_IN_GAA(X3)

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

(9) 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:

• APPA_IN_GAA(.(X1, .(X2, X3))) → APPA_IN_GAA(X3)
  The graph contains the following edges 1 > 1

(10) YES