Left Termination of the query pattern add_in_3(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).

Queries:

add(a,g,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

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 | ?_1\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(!_1, eq(T6, T7)) | add(T1, 0, T3), SCOPE: 1, CLAUSE: 1 | ?_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(T6, T7)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: eq(T6, T7), 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\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(p(T10, X13), add(T12, X13, T13))\n\nT10 is ground\nX3 is free; \nX4 is free; \nX13 is free; \nadd(T1, T10, T3) !=? add(X3, 0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\nX3 is free\nX4 is free; \nX10 is free; \nX11 is free; \nX12 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(p(T10, X13), add(T12, X13, T13)), SCOPE: 3, CLAUSE: 2 | ','(p(T10, X13), add(T12, X13, T13)), SCOPE: 3, CLAUSE: 3\n\nT10 is ground\nX3 is free; \nX4 is free; \nX13 is free; \nadd(T1, T10, T3) !=? add(X3, 0, X4)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(T10, X13), add(T12, X13, T13)), SCOPE: 3, CLAUSE: 3\n\nT10 is ground\nX3 is free; \nX4 is free; \nX13 is free; \nadd(T1, T10, T3) !=? add(X3, 0, X4)\np(T10, X13) !=? p(0, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: add(T12, T14, T13)\n\nT14 is ground\nX3 is free; \nX4 is free; \nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\nX3 is free\nX4 is free; \nX13 is free; \nX15 is free; ", 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) :- ','(!, eq(X3, X4)).\nand substitution\nT1 -> T6\nX3 -> T6\nT2 -> 0\nT3 -> T7\nX4 -> T7\nT4 -> T6\nT5 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 17 [arrowhead = none ];
17 -> 3;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
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(X10, X11, s(X12)) :- ','(p(X11, X13), add(X10, X13, X12)).\nand substitution\nT1 -> T12\nX10 -> T12\nT2 -> T10\nX11 -> T10\nX12 -> T13\nT3 -> s(T13)\nT9 -> T12\nT11 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 20 [arrowhead = none ];
20 -> 10;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
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(X6, X6).\nand substitution\nT6 -> T8\nX6 -> T8\nT7 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 23 [arrowhead = none ];
23 -> 7;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
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, T10, T3), add(X3, 0, X4))", fontsize=14, style = filled, fillcolor = white];
12 -> 27 [arrowhead = none ];
27 -> 13;

28 [label="EVAL with clause\np(s(X15), X15).\nand substitution\nX15 -> T14\nT10 -> s(T14)\nX13 -> T14", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 28 [arrowhead = none ];
28 -> 14;

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

30 [label="INSTANCE with matching:\nT1 -> T12\nT2 -> T14\nT3 -> T13", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 30 [arrowhead = none , style=dashed];
30 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

add1(T8, 0, T8).
add1(T12, s(T14), s(T13)) :- add1(T12, T14, T13).

Queries:

add1(a,g,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:
add1_in: (f,b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

add1_in_aga(T8, 0, T8) → add1_out_aga(T8, 0, T8)
add1_in_aga(T12, s(T14), s(T13)) → U1_aga(T12, T14, T13, add1_in_aga(T12, T14, T13))
U1_aga(T12, T14, T13, add1_out_aga(T12, T14, T13)) → add1_out_aga(T12, s(T14), s(T13))

The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3)  =  add1_in_aga(x2)
0  =  0
add1_out_aga(x1, x2, x3)  =  add1_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:

add1_in_aga(T8, 0, T8) → add1_out_aga(T8, 0, T8)
add1_in_aga(T12, s(T14), s(T13)) → U1_aga(T12, T14, T13, add1_in_aga(T12, T14, T13))
U1_aga(T12, T14, T13, add1_out_aga(T12, T14, T13)) → add1_out_aga(T12, s(T14), s(T13))

The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3)  =  add1_in_aga(x2)
0  =  0
add1_out_aga(x1, x2, x3)  =  add1_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:

ADD1_IN_AGA(T12, s(T14), s(T13)) → U1_AGA(T12, T14, T13, add1_in_aga(T12, T14, T13))
ADD1_IN_AGA(T12, s(T14), s(T13)) → ADD1_IN_AGA(T12, T14, T13)

The TRS R consists of the following rules:

add1_in_aga(T8, 0, T8) → add1_out_aga(T8, 0, T8)
add1_in_aga(T12, s(T14), s(T13)) → U1_aga(T12, T14, T13, add1_in_aga(T12, T14, T13))
U1_aga(T12, T14, T13, add1_out_aga(T12, T14, T13)) → add1_out_aga(T12, s(T14), s(T13))

The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3)  =  add1_in_aga(x2)
0  =  0
add1_out_aga(x1, x2, x3)  =  add1_out_aga
s(x1)  =  s(x1)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)
ADD1_IN_AGA(x1, x2, x3)  =  ADD1_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:

ADD1_IN_AGA(T12, s(T14), s(T13)) → U1_AGA(T12, T14, T13, add1_in_aga(T12, T14, T13))
ADD1_IN_AGA(T12, s(T14), s(T13)) → ADD1_IN_AGA(T12, T14, T13)

The TRS R consists of the following rules:

add1_in_aga(T8, 0, T8) → add1_out_aga(T8, 0, T8)
add1_in_aga(T12, s(T14), s(T13)) → U1_aga(T12, T14, T13, add1_in_aga(T12, T14, T13))
U1_aga(T12, T14, T13, add1_out_aga(T12, T14, T13)) → add1_out_aga(T12, s(T14), s(T13))

The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3)  =  add1_in_aga(x2)
0  =  0
add1_out_aga(x1, x2, x3)  =  add1_out_aga
s(x1)  =  s(x1)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)
ADD1_IN_AGA(x1, x2, x3)  =  ADD1_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:

ADD1_IN_AGA(T12, s(T14), s(T13)) → ADD1_IN_AGA(T12, T14, T13)

The TRS R consists of the following rules:

add1_in_aga(T8, 0, T8) → add1_out_aga(T8, 0, T8)
add1_in_aga(T12, s(T14), s(T13)) → U1_aga(T12, T14, T13, add1_in_aga(T12, T14, T13))
U1_aga(T12, T14, T13, add1_out_aga(T12, T14, T13)) → add1_out_aga(T12, s(T14), s(T13))

The argument filtering Pi contains the following mapping:
add1_in_aga(x1, x2, x3)  =  add1_in_aga(x2)
0  =  0
add1_out_aga(x1, x2, x3)  =  add1_out_aga
s(x1)  =  s(x1)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)
ADD1_IN_AGA(x1, x2, x3)  =  ADD1_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:

ADD1_IN_AGA(T12, s(T14), s(T13)) → ADD1_IN_AGA(T12, T14, T13)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
ADD1_IN_AGA(x1, x2, x3)  =  ADD1_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:

ADD1_IN_AGA(s(T14)) → ADD1_IN_AGA(T14)

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:

• ADD1_IN_AGA(s(T14)) → ADD1_IN_AGA(T14)
  The graph contains the following edges 1 > 1

(14) TRUE