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([], 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).

Query: app(g,a,a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

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: ','(!_1, eq(T8, T9)) | 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\nX6 is free\nX7 is free\napp(T1, T2, T3) !=? app([], X6, X7)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: eq(T8, T9)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: eq(T8, T9), 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\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(head(T17, T21), ','(tail(T17, X19), app(X19, T22, T23)))\n\nT17 is ground\nX6 is free\nX7 is free\nX19 is free\napp(T17, T2, T3) !=? app([], X6, X7)", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(head(T17, T21), ','(tail(T17, X19), app(X19, T22, T23))), SCOPE: 3, CLAUSE: 2 | ','(head(T17, T21), ','(tail(T17, X19), app(X19, T22, T23))), SCOPE: 3, CLAUSE: 3\n\nT17 is ground\nX6 is free\nX7 is free\nX19 is free\napp(T17, T2, T3) !=? app([], X6, X7)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(head(T17, T21), ','(tail(T17, X19), app(X19, T22, T23))), SCOPE: 3, CLAUSE: 3\n\nT17 is ground\nX6 is free\nX7 is free\nX19 is free\napp(T17, T2, T3) !=? app([], X6, X7)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(tail(.(T29, T30), X19), app(X19, T31, T32))\n\nT29 is ground\nT30 is ground\nX19 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(tail(.(T29, T30), X19), app(X19, T31, T32)), SCOPE: 4, CLAUSE: 4 | ','(tail(.(T29, T30), X19), app(X19, T31, T32)), SCOPE: 4, CLAUSE: 5\n\nT29 is ground\nT30 is ground\nX19 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(tail(.(T29, T30), X19), app(X19, T31, T32)), SCOPE: 4, CLAUSE: 5\n\nT29 is ground\nT30 is ground\nX19 is free", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: app(T42, T31, T32)\n\nT42 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) :- ','(!_1, eq(X6, X7)).\nand substitutionT1 -> [],\nT2 -> T8,\nX6 -> T8,\nT3 -> T9,\nX7 -> T9,\nT6 -> T8,\nT7 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 20 [arrowhead = none ];
20 -> 3;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
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(X15, X16, .(X17, X18)) :- ','(head(X15, X17), ','(tail(X15, X19), app(X19, X16, X18))).\nand substitutionT1 -> T17,\nX15 -> T17,\nT2 -> T22,\nX16 -> T22,\nX17 -> T21,\nX18 -> T23,\nT3 -> .(T21, T23),\nT19 -> T21,\nT18 -> T22,\nT20 -> T23", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 23 [arrowhead = none ];
23 -> 10;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
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(X10, X10).\nand substitutionT8 -> T12,\nX10 -> T12,\nT9 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 26 [arrowhead = none ];
26 -> 7;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
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(T17, T2, T3), app([], X6, X7))", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 30 [arrowhead = none ];
30 -> 13;

31 [label="EVAL with clause\nhead(.(X25, X26), X25).\nand substitutionX25 -> T29,\nX26 -> T30,\nT17 -> .(T29, T30),\nT21 -> T29,\nT22 -> T31,\nT23 -> T32", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 31 [arrowhead = none ];
31 -> 14;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
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 = lightgreen];
16 -> 34 [arrowhead = none ];
34 -> 17;

35 [label="ONLY EVAL with clause\ntail(.(X35, X36), X36).\nand substitutionT29 -> T41,\nX35 -> T41,\nT30 -> T42,\nX36 -> T42,\nX19 -> T42", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 35 [arrowhead = none ];
35 -> 18;

36 [label="INSTANCE with matching:\nT1 -> T42\nT2 -> T31\nT3 -> T32", fontsize=14, style = filled, fillcolor = lightgrey];
18 -> 36 [arrowhead = none , style=dashed];
36 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

appA([], T12, T12).
appA(.(T41, T42), T31, .(T41, T32)) :- appA(T42, T31, T32).

Query: appA(g,a,a)

(3) PrologToPiTRSProof (SOUND transformation)

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

appA_in_gaa([], T12, T12) → appA_out_gaa([], T12, T12)
appA_in_gaa(.(T41, T42), T31, .(T41, T32)) → U1_gaa(T41, T42, T31, T32, appA_in_gaa(T42, T31, T32))
U1_gaa(T41, T42, T31, T32, appA_out_gaa(T42, T31, T32)) → appA_out_gaa(.(T41, T42), T31, .(T41, T32))

The argument filtering Pi contains the following mapping:
appA_in_gaa(x1, x2, x3)  =  appA_in_gaa(x1)
[]  =  []
appA_out_gaa(x1, x2, x3)  =  appA_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:

appA_in_gaa([], T12, T12) → appA_out_gaa([], T12, T12)
appA_in_gaa(.(T41, T42), T31, .(T41, T32)) → U1_gaa(T41, T42, T31, T32, appA_in_gaa(T42, T31, T32))
U1_gaa(T41, T42, T31, T32, appA_out_gaa(T42, T31, T32)) → appA_out_gaa(.(T41, T42), T31, .(T41, T32))

The argument filtering Pi contains the following mapping:
appA_in_gaa(x1, x2, x3)  =  appA_in_gaa(x1)
[]  =  []
appA_out_gaa(x1, x2, x3)  =  appA_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:

APPA_IN_GAA(.(T41, T42), T31, .(T41, T32)) → U1_GAA(T41, T42, T31, T32, appA_in_gaa(T42, T31, T32))
APPA_IN_GAA(.(T41, T42), T31, .(T41, T32)) → APPA_IN_GAA(T42, T31, T32)

The TRS R consists of the following rules:

appA_in_gaa([], T12, T12) → appA_out_gaa([], T12, T12)
appA_in_gaa(.(T41, T42), T31, .(T41, T32)) → U1_gaa(T41, T42, T31, T32, appA_in_gaa(T42, T31, T32))
U1_gaa(T41, T42, T31, T32, appA_out_gaa(T42, T31, T32)) → appA_out_gaa(.(T41, T42), T31, .(T41, T32))

The argument filtering Pi contains the following mapping:
appA_in_gaa(x1, x2, x3)  =  appA_in_gaa(x1)
[]  =  []
appA_out_gaa(x1, x2, x3)  =  appA_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APPA_IN_GAA(x1, x2, x3)  =  APPA_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:

APPA_IN_GAA(.(T41, T42), T31, .(T41, T32)) → U1_GAA(T41, T42, T31, T32, appA_in_gaa(T42, T31, T32))
APPA_IN_GAA(.(T41, T42), T31, .(T41, T32)) → APPA_IN_GAA(T42, T31, T32)

The TRS R consists of the following rules:

appA_in_gaa([], T12, T12) → appA_out_gaa([], T12, T12)
appA_in_gaa(.(T41, T42), T31, .(T41, T32)) → U1_gaa(T41, T42, T31, T32, appA_in_gaa(T42, T31, T32))
U1_gaa(T41, T42, T31, T32, appA_out_gaa(T42, T31, T32)) → appA_out_gaa(.(T41, T42), T31, .(T41, T32))

The argument filtering Pi contains the following mapping:
appA_in_gaa(x1, x2, x3)  =  appA_in_gaa(x1)
[]  =  []
appA_out_gaa(x1, x2, x3)  =  appA_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APPA_IN_GAA(x1, x2, x3)  =  APPA_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:

APPA_IN_GAA(.(T41, T42), T31, .(T41, T32)) → APPA_IN_GAA(T42, T31, T32)

The TRS R consists of the following rules:

appA_in_gaa([], T12, T12) → appA_out_gaa([], T12, T12)
appA_in_gaa(.(T41, T42), T31, .(T41, T32)) → U1_gaa(T41, T42, T31, T32, appA_in_gaa(T42, T31, T32))
U1_gaa(T41, T42, T31, T32, appA_out_gaa(T42, T31, T32)) → appA_out_gaa(.(T41, T42), T31, .(T41, T32))

The argument filtering Pi contains the following mapping:
appA_in_gaa(x1, x2, x3)  =  appA_in_gaa(x1)
[]  =  []
appA_out_gaa(x1, x2, x3)  =  appA_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
APPA_IN_GAA(x1, x2, x3)  =  APPA_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:

APPA_IN_GAA(.(T41, T42), T31, .(T41, T32)) → APPA_IN_GAA(T42, T31, T32)

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

(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:

APPA_IN_GAA(.(T41, T42)) → APPA_IN_GAA(T42)

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:

• APPA_IN_GAA(.(T41, T42)) → APPA_IN_GAA(T42)
  The graph contains the following edges 1 > 1

(14) YES