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, X).
add(X, Y, s(Z)) :- ','(\+(isZero(Y)), ','(p(Y, P), add(X, P, Z))).
p(0, 0).
p(s(X), X).
isZero(0).

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: true | 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\nX2 is free\nadd(T1, T2, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: add(T1, 0, T3), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(\\+(isZero(0)), ','(p(0, X9), add(T10, X9, T11)))\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(call(isZero(0)), ','(!_2, fail)) | ','(p(0, X9), add(T10, X9, T11))\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(isZero(0), ','(!_2, fail)) | ?_3 | ','(p(0, X9), add(T10, X9, T11))\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(isZero(0), ','(!_2, fail)), SCOPE: 4, CLAUSE: 4 | ?_4 | ?_3 | ','(p(0, X9), add(T10, X9, T11))\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(!_2, fail) | ?_4 | ?_3 | ','(p(0, X9), add(T10, X9, T11))\n\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: fail\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(\\+(isZero(T16)), ','(p(T16, X16), add(T18, X16, T19)))\n\nT16 is ground\nX2 is free\nX16 is free\nadd(T1, T16, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(call(isZero(T16)), ','(!_5, fail)) | ','(p(T16, X16), add(T18, X16, T19))\n\nT16 is ground\nX2 is free\nX16 is free\nadd(T1, T16, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(isZero(T16), ','(!_5, fail)) | ?_6 | ','(p(T16, X16), add(T18, X16, T19))\n\nT16 is ground\nX2 is free\nX16 is free\nadd(T1, T16, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(isZero(T16), ','(!_5, fail)), SCOPE: 7, CLAUSE: 4 | ?_7 | ?_6 | ','(p(T16, X16), add(T18, X16, T19))\n\nT16 is ground\nX2 is free\nX16 is free\nadd(T1, T16, T3) !=? add(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(!_5, fail) | ?_7 | ?_6 | ','(p(0, X16), add(T18, X16, T19))\n\nX16 is free", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ?_7 | ?_6 | ','(p(T16, X16), add(T18, X16, T19))\n\nT16 is ground\nX2 is free\nX16 is free\nadd(T1, T16, T3) !=? add(X2, 0, X2)\nisZero(T16) !=? isZero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: fail\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ?_6 | ','(p(T16, X16), add(T18, X16, T19))\n\nT16 is ground\nX2 is free\nX16 is free\nadd(T1, T16, T3) !=? add(X2, 0, X2)\nisZero(T16) !=? isZero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(p(T16, X16), add(T18, X16, T19))\n\nT16 is ground\nX2 is free\nX16 is free\nadd(T1, T16, T3) !=? add(X2, 0, X2)\nisZero(T16) !=? isZero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(p(T16, X16), add(T18, X16, T19)), SCOPE: 8, CLAUSE: 2 | ','(p(T16, X16), add(T18, X16, T19)), SCOPE: 8, CLAUSE: 3\n\nT16 is ground\nX2 is free\nX16 is free\nadd(T1, T16, T3) !=? add(X2, 0, X2)\nisZero(T16) !=? isZero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: ','(p(T16, X16), add(T18, X16, T19)), SCOPE: 8, CLAUSE: 3\n\nT16 is ground\nX2 is free\nX16 is free\nadd(T1, T16, T3) !=? add(X2, 0, X2)\nisZero(T16) !=? isZero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: add(T18, T24, T19)\n\nT24 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 29 [arrowhead = none ];
29 -> 2;

30 [label="EVAL with clause\nadd(X2, 0, X2).\nand substitutionT1 -> T5,\nX2 -> T5,\nT2 -> 0,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 30 [arrowhead = none ];
30 -> 3;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 31 [arrowhead = none ];
31 -> 4;

32 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 32 [arrowhead = none ];
32 -> 5;

33 [label="EVAL with clause\nadd(X13, X14, s(X15)) :- ','(\\+(isZero(X14)), ','(p(X14, X16), add(X13, X16, X15))).\nand substitutionT1 -> T18,\nX13 -> T18,\nT2 -> T16,\nX14 -> T16,\nX15 -> T19,\nT3 -> s(T19),\nT15 -> T18,\nT17 -> T19", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 33 [arrowhead = none ];
33 -> 14;

34 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 34 [arrowhead = none ];
34 -> 15;

35 [label="EVAL with clause\nadd(X6, X7, s(X8)) :- ','(\\+(isZero(X7)), ','(p(X7, X9), add(X6, X9, X8))).\nand substitutionT1 -> T10,\nX6 -> T10,\nX7 -> 0,\nX8 -> T11,\nT3 -> s(T11),\nT8 -> T10,\nT9 -> T11", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 35 [arrowhead = none ];
35 -> 6;

36 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 36 [arrowhead = none ];
36 -> 7;

37 [label="NOT", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 37 [arrowhead = none ];
37 -> 8;

38 [label="CALL", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 38 [arrowhead = none ];
38 -> 9;

39 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 39 [arrowhead = none ];
39 -> 10;

40 [label="ONLY EVAL with clause\nisZero(0).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 40 [arrowhead = none ];
40 -> 11;

41 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 41 [arrowhead = none ];
41 -> 12;

42 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 42 [arrowhead = none ];
42 -> 13;

43 [label="NOT", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 43 [arrowhead = none ];
43 -> 16;

44 [label="CALL", fontsize=14, style = filled, fillcolor = lightcyan];
16 -> 44 [arrowhead = none ];
44 -> 17;

45 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 45 [arrowhead = none ];
45 -> 18;

46 [label="EVAL with clause\nisZero(0).\nand substitutionT16 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
18 -> 46 [arrowhead = none ];
46 -> 19;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 47 [arrowhead = none ];
47 -> 20;

48 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 48 [arrowhead = none ];
48 -> 21;

49 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 49 [arrowhead = none ];
49 -> 23;

50 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 50 [arrowhead = none ];
50 -> 22;

51 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 51 [arrowhead = none ];
51 -> 24;

52 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
24 -> 52 [arrowhead = none ];
52 -> 25;

53 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (isZero(T16), isZero(0))", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 53 [arrowhead = none ];
53 -> 26;

54 [label="EVAL with clause\np(s(X21), X21).\nand substitutionX21 -> T24,\nT16 -> s(T24),\nX16 -> T24", fontsize=14, style = filled, fillcolor = lightblue];
26 -> 54 [arrowhead = none ];
54 -> 27;

55 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 55 [arrowhead = none ];
55 -> 28;

56 [label="INSTANCE with matching:\nT1 -> T18\nT2 -> T24\nT3 -> T19", fontsize=14, style = filled, fillcolor = lightgrey];
27 -> 56 [arrowhead = none , style=dashed];
56 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

addA(T5, 0, T5).
addA(T18, s(T24), s(T19)) :- addA(T18, T24, T19).

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(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T18, s(T24), s(T19)) → U1_aga(T18, T24, T19, addA_in_aga(T18, T24, T19))
U1_aga(T18, T24, T19, addA_out_aga(T18, T24, T19)) → addA_out_aga(T18, s(T24), s(T19))

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(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T18, s(T24), s(T19)) → U1_aga(T18, T24, T19, addA_in_aga(T18, T24, T19))
U1_aga(T18, T24, T19, addA_out_aga(T18, T24, T19)) → addA_out_aga(T18, s(T24), s(T19))

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(T18, s(T24), s(T19)) → U1_AGA(T18, T24, T19, addA_in_aga(T18, T24, T19))
ADDA_IN_AGA(T18, s(T24), s(T19)) → ADDA_IN_AGA(T18, T24, T19)

The TRS R consists of the following rules:

addA_in_aga(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T18, s(T24), s(T19)) → U1_aga(T18, T24, T19, addA_in_aga(T18, T24, T19))
U1_aga(T18, T24, T19, addA_out_aga(T18, T24, T19)) → addA_out_aga(T18, s(T24), s(T19))

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(T18, s(T24), s(T19)) → U1_AGA(T18, T24, T19, addA_in_aga(T18, T24, T19))
ADDA_IN_AGA(T18, s(T24), s(T19)) → ADDA_IN_AGA(T18, T24, T19)

The TRS R consists of the following rules:

addA_in_aga(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T18, s(T24), s(T19)) → U1_aga(T18, T24, T19, addA_in_aga(T18, T24, T19))
U1_aga(T18, T24, T19, addA_out_aga(T18, T24, T19)) → addA_out_aga(T18, s(T24), s(T19))

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(T18, s(T24), s(T19)) → ADDA_IN_AGA(T18, T24, T19)

The TRS R consists of the following rules:

addA_in_aga(T5, 0, T5) → addA_out_aga(T5, 0, T5)
addA_in_aga(T18, s(T24), s(T19)) → U1_aga(T18, T24, T19, addA_in_aga(T18, T24, T19))
U1_aga(T18, T24, T19, addA_out_aga(T18, T24, T19)) → addA_out_aga(T18, s(T24), s(T19))

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(T18, s(T24), s(T19)) → ADDA_IN_AGA(T18, T24, T19)

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(T24)) → ADDA_IN_AGA(T24)

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(T24)) → ADDA_IN_AGA(T24)
  The graph contains the following edges 1 > 1

(14) YES