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

(0) Obligation:

Clauses:

add(X, 0, Y) :- ','(!, eq(X, Y)).
add(X, Y, s(Z)) :- ','(p(Y, P), add(X, P, Z)).
p(0, 0).
p(s(X), X).
eq(X, X).

Query: add(a,g,a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: add(T1, T2, T3)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: add(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | add(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(!_1, eq(T8, T9)) | add(T1, 0, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: add(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT2 is ground\nX3 is free\nX4 is free\nadd(T1, T2, T3) !=? add(X3, 0, X4)", 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: 4\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: ','(p(T17, X14), add(T19, X14, T20))\n\nT17 is ground\nX3 is free\nX4 is free\nX14 is free\nadd(T1, T17, T3) !=? add(X3, 0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(p(T17, X14), add(T19, X14, T20)), SCOPE: 3, CLAUSE: 2 | ','(p(T17, X14), add(T19, X14, T20)), SCOPE: 3, CLAUSE: 3\n\nT17 is ground\nX3 is free\nX4 is free\nX14 is free\nadd(T1, T17, T3) !=? add(X3, 0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(T17, X14), add(T19, X14, T20)), SCOPE: 3, CLAUSE: 3\n\nT17 is ground\nX3 is free\nX4 is free\nX14 is free\nadd(T1, T17, T3) !=? add(X3, 0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: add(T19, T23, T20)\n\nT23 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 16 [arrowhead = none ];
16 -> 2;

17 [label="EVAL with clause\nadd(X3, 0, X4) :- ','(!_1, eq(X3, X4)).\nand substitutionT1 -> T8,\nX3 -> T8,\nT2 -> 0,\nT3 -> T9,\nX4 -> T9,\nT6 -> T8,\nT7 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 17 [arrowhead = none ];
17 -> 3;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 18 [arrowhead = none ];
18 -> 4;

19 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 19 [arrowhead = none ];
19 -> 5;

20 [label="EVAL with clause\nadd(X11, X12, s(X13)) :- ','(p(X12, X14), add(X11, X14, X13)).\nand substitutionT1 -> T19,\nX11 -> T19,\nT2 -> T17,\nX12 -> T17,\nX13 -> T20,\nT3 -> s(T20),\nT16 -> T19,\nT18 -> T20", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 20 [arrowhead = none ];
20 -> 10;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 21 [arrowhead = none ];
21 -> 11;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="EVAL with clause\neq(X7, X7).\nand substitutionT8 -> T12,\nX7 -> T12,\nT9 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 23 [arrowhead = none ];
23 -> 7;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 24 [arrowhead = none ];
24 -> 8;

25 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 25 [arrowhead = none ];
25 -> 9;

26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 26 [arrowhead = none ];
26 -> 12;

27 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (add(T1, T17, T3), add(X3, 0, X4))", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 27 [arrowhead = none ];
27 -> 13;

28 [label="EVAL with clause\np(s(X17), X17).\nand substitutionX17 -> T23,\nT17 -> s(T23),\nX14 -> T23", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 28 [arrowhead = none ];
28 -> 14;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 29 [arrowhead = none ];
29 -> 15;

30 [label="INSTANCE with matching:\nT1 -> T19\nT2 -> T23\nT3 -> T20", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 30 [arrowhead = none , style=dashed];
30 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

addA(T12, 0, T12).
addA(T19, s(T23), s(T20)) :- addA(T19, T23, T20).

Query: addA(a,g,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:
addA_in: (f,b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

addA_in_aga(T12, 0, T12) → addA_out_aga(T12, 0, T12)
addA_in_aga(T19, s(T23), s(T20)) → U1_aga(T19, T23, T20, addA_in_aga(T19, T23, T20))
U1_aga(T19, T23, T20, addA_out_aga(T19, T23, T20)) → addA_out_aga(T19, s(T23), s(T20))

The argument filtering Pi contains the following mapping:
addA_in_aga(x1, x2, x3)  =  addA_in_aga(x2)
0  =  0
addA_out_aga(x1, x2, x3)  =  addA_out_aga
s(x1)  =  s(x1)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

addA_in_aga(T12, 0, T12) → addA_out_aga(T12, 0, T12)
addA_in_aga(T19, s(T23), s(T20)) → U1_aga(T19, T23, T20, addA_in_aga(T19, T23, T20))
U1_aga(T19, T23, T20, addA_out_aga(T19, T23, T20)) → addA_out_aga(T19, s(T23), s(T20))

The argument filtering Pi contains the following mapping:
addA_in_aga(x1, x2, x3)  =  addA_in_aga(x2)
0  =  0
addA_out_aga(x1, x2, x3)  =  addA_out_aga
s(x1)  =  s(x1)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)

(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:

ADDA_IN_AGA(T19, s(T23), s(T20)) → U1_AGA(T19, T23, T20, addA_in_aga(T19, T23, T20))
ADDA_IN_AGA(T19, s(T23), s(T20)) → ADDA_IN_AGA(T19, T23, T20)

The TRS R consists of the following rules:

addA_in_aga(T12, 0, T12) → addA_out_aga(T12, 0, T12)
addA_in_aga(T19, s(T23), s(T20)) → U1_aga(T19, T23, T20, addA_in_aga(T19, T23, T20))
U1_aga(T19, T23, T20, addA_out_aga(T19, T23, T20)) → addA_out_aga(T19, s(T23), s(T20))

The argument filtering Pi contains the following mapping:
addA_in_aga(x1, x2, x3)  =  addA_in_aga(x2)
0  =  0
addA_out_aga(x1, x2, x3)  =  addA_out_aga
s(x1)  =  s(x1)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)
ADDA_IN_AGA(x1, x2, x3)  =  ADDA_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4)  =  U1_AGA(x4)

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

(6) Obligation:

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

ADDA_IN_AGA(T19, s(T23), s(T20)) → U1_AGA(T19, T23, T20, addA_in_aga(T19, T23, T20))
ADDA_IN_AGA(T19, s(T23), s(T20)) → ADDA_IN_AGA(T19, T23, T20)

The TRS R consists of the following rules:

addA_in_aga(T12, 0, T12) → addA_out_aga(T12, 0, T12)
addA_in_aga(T19, s(T23), s(T20)) → U1_aga(T19, T23, T20, addA_in_aga(T19, T23, T20))
U1_aga(T19, T23, T20, addA_out_aga(T19, T23, T20)) → addA_out_aga(T19, s(T23), s(T20))

The argument filtering Pi contains the following mapping:
addA_in_aga(x1, x2, x3)  =  addA_in_aga(x2)
0  =  0
addA_out_aga(x1, x2, x3)  =  addA_out_aga
s(x1)  =  s(x1)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)
ADDA_IN_AGA(x1, x2, x3)  =  ADDA_IN_AGA(x2)
U1_AGA(x1, x2, x3, x4)  =  U1_AGA(x4)

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:

ADDA_IN_AGA(T19, s(T23), s(T20)) → ADDA_IN_AGA(T19, T23, T20)

The TRS R consists of the following rules:

addA_in_aga(T12, 0, T12) → addA_out_aga(T12, 0, T12)
addA_in_aga(T19, s(T23), s(T20)) → U1_aga(T19, T23, T20, addA_in_aga(T19, T23, T20))
U1_aga(T19, T23, T20, addA_out_aga(T19, T23, T20)) → addA_out_aga(T19, s(T23), s(T20))

The argument filtering Pi contains the following mapping:
addA_in_aga(x1, x2, x3)  =  addA_in_aga(x2)
0  =  0
addA_out_aga(x1, x2, x3)  =  addA_out_aga
s(x1)  =  s(x1)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)
ADDA_IN_AGA(x1, x2, x3)  =  ADDA_IN_AGA(x2)

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:

ADDA_IN_AGA(T19, s(T23), s(T20)) → ADDA_IN_AGA(T19, T23, T20)

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

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:

ADDA_IN_AGA(s(T23)) → ADDA_IN_AGA(T23)

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:

• ADDA_IN_AGA(s(T23)) → ADDA_IN_AGA(T23)
  The graph contains the following edges 1 > 1

(14) YES