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, X).
add(X, Y, s(Z)) :- ','(no(zero(Y)), ','(p(Y, P), add(X, P, Z))).
p(0, 0).
p(s(X), X).
zero(0).
no(X) :- ','(X, ','(!, failure(a))).
no(X).
failure(b).

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\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: add(T1, T2, T3), SCOPE: 1, CLAUSE: 0\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: add(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\nX3 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(no(zero(T6)), ','(p(T6, X10), add(T8, X10, T9)))\n\nT6 is ground\nX10 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\nX7 is free\nX8 is free; \nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(no(zero(T6)), ','(p(T6, X10), add(T8, X10, T9))), SCOPE: 2, CLAUSE: 5 | ','(no(zero(T6)), ','(p(T6, X10), add(T8, X10, T9))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT6 is ground\nX10 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(zero(T10), ','(!_2, ','(failure(a), ','(p(T10, X10), add(T8, X10, T9))))) | ','(no(zero(T10)), ','(p(T10, X10), add(T8, X10, T9))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT10 is ground\nX10 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(zero(T10), ','(!_2, ','(failure(a), ','(p(T10, X10), add(T8, X10, T9))))), SCOPE: 3, CLAUSE: 4 | ','(no(zero(T10)), ','(p(T10, X10), add(T8, X10, T9))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nT10 is ground\nX10 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(!_2, ','(failure(a), ','(p(0, X10), add(T8, X10, T9)))) | ','(no(zero(0)), ','(p(0, X10), add(T8, X10, T9))), SCOPE: 2, CLAUSE: 6 | ?_2\n\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(no(zero(T10)), ','(p(T10, X10), add(T8, X10, T9))), SCOPE: 2, CLAUSE: 6\n\nT10 is ground\nX10 is free; \nzero(T10) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(failure(a), ','(p(0, X10), add(T8, X10, T9)))\n\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(failure(a), ','(p(0, X10), add(T8, X10, T9))), SCOPE: 4, CLAUSE: 7\n\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(p(T11, X10), add(T8, X10, T9))\n\nT11 is ground\nX10 is free; \nzero(T11) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(p(T11, X10), add(T8, X10, T9)), SCOPE: 5, CLAUSE: 2 | ','(p(T11, X10), add(T8, X10, T9)), SCOPE: 5, CLAUSE: 3\n\nT11 is ground\nX10 is free; \nzero(T11) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(p(T11, X10), add(T8, X10, T9)), SCOPE: 5, CLAUSE: 3\n\nT11 is ground\nX10 is free; \nzero(T11) !=? zero(0)\np(T11, X10) !=? p(0, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: add(T8, T12, T9)\n\nT12 is ground\nX10 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\nX10 is free\nX16 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 23 [arrowhead = none ];
23 -> 2;

24 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 24 [arrowhead = none ];
24 -> 3;

25 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 25 [arrowhead = none ];
25 -> 4;

26 [label="EVAL with clause\nadd(X3, 0, X3).\nand substitution\nT1 -> T4\nX3 -> T4\nT2 -> 0\nT3 -> T4", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 26 [arrowhead = none ];
26 -> 5;

27 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 27 [arrowhead = none ];
27 -> 6;

28 [label="EVAL with clause\nadd(X7, X8, s(X9)) :- ','(no(zero(X8)), ','(p(X8, X10), add(X7, X10, X9))).\nand substitution\nT1 -> T8\nX7 -> T8\nT2 -> T6\nX8 -> T6\nX9 -> T9\nT3 -> s(T9)\nT5 -> T8\nT7 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 28 [arrowhead = none ];
28 -> 8;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 29 [arrowhead = none ];
29 -> 9;

30 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 30 [arrowhead = none ];
30 -> 7;

31 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 31 [arrowhead = none ];
31 -> 10;

32 [label="EVAL with clause\nno(X12) :- ','(X12, ','(!, failure(a))).\nand substitution\nT6 -> T10\nX12 -> zero(T10)", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 32 [arrowhead = none ];
32 -> 11;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 33 [arrowhead = none ];
33 -> 12;

34 [label="EVAL with clause\nzero(0).\nand substitution\nT10 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 34 [arrowhead = none ];
34 -> 13;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 35 [arrowhead = none ];
35 -> 14;

36 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 36 [arrowhead = none ];
36 -> 15;

37 [label="EVAL with clause\nno(X14).\nand substitution\nT10 -> T11\nX14 -> zero(T11)", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 37 [arrowhead = none ];
37 -> 18;

38 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 38 [arrowhead = none ];
38 -> 16;

39 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = white];
16 -> 39 [arrowhead = none ];
39 -> 17;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
18 -> 40 [arrowhead = none ];
40 -> 19;

41 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (zero(T11), zero(0))", fontsize=14, style = filled, fillcolor = white];
19 -> 41 [arrowhead = none ];
41 -> 20;

42 [label="EVAL with clause\np(s(X16), X16).\nand substitution\nX16 -> T12\nT11 -> s(T12)\nX10 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 42 [arrowhead = none ];
42 -> 21;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 43 [arrowhead = none ];
43 -> 22;

44 [label="INSTANCE with matching:\nT1 -> T8\nT2 -> T12\nT3 -> T9", fontsize=14, style = filled, fillcolor = lightgrey];
21 -> 44 [arrowhead = none , style=dashed];
44 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

add1(T4, 0, T4).
add1(T8, s(T12), s(T9)) :- add1(T8, T12, T9).

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(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T8, s(T12), s(T9)) → U1_aga(T8, T12, T9, add1_in_aga(T8, T12, T9))
U1_aga(T8, T12, T9, add1_out_aga(T8, T12, T9)) → add1_out_aga(T8, s(T12), s(T9))

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(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T8, s(T12), s(T9)) → U1_aga(T8, T12, T9, add1_in_aga(T8, T12, T9))
U1_aga(T8, T12, T9, add1_out_aga(T8, T12, T9)) → add1_out_aga(T8, s(T12), s(T9))

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(T8, s(T12), s(T9)) → U1_AGA(T8, T12, T9, add1_in_aga(T8, T12, T9))
ADD1_IN_AGA(T8, s(T12), s(T9)) → ADD1_IN_AGA(T8, T12, T9)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T8, s(T12), s(T9)) → U1_aga(T8, T12, T9, add1_in_aga(T8, T12, T9))
U1_aga(T8, T12, T9, add1_out_aga(T8, T12, T9)) → add1_out_aga(T8, s(T12), s(T9))

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(T8, s(T12), s(T9)) → U1_AGA(T8, T12, T9, add1_in_aga(T8, T12, T9))
ADD1_IN_AGA(T8, s(T12), s(T9)) → ADD1_IN_AGA(T8, T12, T9)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T8, s(T12), s(T9)) → U1_aga(T8, T12, T9, add1_in_aga(T8, T12, T9))
U1_aga(T8, T12, T9, add1_out_aga(T8, T12, T9)) → add1_out_aga(T8, s(T12), s(T9))

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(T8, s(T12), s(T9)) → ADD1_IN_AGA(T8, T12, T9)

The TRS R consists of the following rules:

add1_in_aga(T4, 0, T4) → add1_out_aga(T4, 0, T4)
add1_in_aga(T8, s(T12), s(T9)) → U1_aga(T8, T12, T9, add1_in_aga(T8, T12, T9))
U1_aga(T8, T12, T9, add1_out_aga(T8, T12, T9)) → add1_out_aga(T8, s(T12), s(T9))

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(T8, s(T12), s(T9)) → ADD1_IN_AGA(T8, T12, T9)

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(T12)) → ADD1_IN_AGA(T12)

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(T12)) → ADD1_IN_AGA(T12)
  The graph contains the following edges 1 > 1

(14) TRUE