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

(0) Obligation:

Clauses:

sum(X, 0, X).
sum(X, s(Y), s(Z)) :- sum(X, Y, Z).

Query: sum(a,a,g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: sum(T1, T2, T3)\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: sum(T1, T2, T3), SCOPE: 1, CLAUSE: 0 | sum(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | sum(T1, T2, T5), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: sum(T1, T2, T3), SCOPE: 1, CLAUSE: 1\n\nT3 is ground\nX2 is free\nsum(T1, T2, T3) !=? sum(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: sum(T1, T2, T5), SCOPE: 1, CLAUSE: 1\n\nT5 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: sum(T12, T13, T11)\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: sum(T12, T13, T11), SCOPE: 2, CLAUSE: 0 | sum(T12, T13, T11), SCOPE: 2, CLAUSE: 1\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: sum(T12, T13, T11), SCOPE: 2, CLAUSE: 0\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: sum(T12, T13, T11), SCOPE: 2, CLAUSE: 1\n\nT11 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: sum(T28, T29, T27)\n\nT27 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: sum(T38, T39, T37)\n\nT37 is ground\nX2 is free\nsum(T1, T2, s(T37)) !=? sum(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: sum(T38, T39, T37), SCOPE: 3, CLAUSE: 0 | sum(T38, T39, T37), SCOPE: 3, CLAUSE: 1\n\nT37 is ground\nX2 is free\nsum(T1, T2, s(T37)) !=? sum(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: sum(T38, T39, T37), SCOPE: 3, CLAUSE: 0\n\nT37 is ground\nX2 is free\nsum(T1, T2, s(T37)) !=? sum(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: sum(T38, T39, T37), SCOPE: 3, CLAUSE: 1\n\nT37 is ground\nX2 is free\nsum(T1, T2, s(T37)) !=? sum(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: sum(T54, T55, T53)\n\nT53 is ground\nX2 is free\nsum(T1, T2, s(s(T53))) !=? sum(X2, 0, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 26 [arrowhead = none ];
26 -> 2;

27 [label="EVAL with clause\nsum(X2, 0, X2).\nand substitutionT1 -> T5,\nX2 -> T5,\nT2 -> 0,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 27 [arrowhead = none ];
27 -> 3;

28 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 28 [arrowhead = none ];
28 -> 4;

29 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 29 [arrowhead = none ];
29 -> 5;

30 [label="EVAL with clause\nsum(X28, s(X29), s(X30)) :- sum(X28, X29, X30).\nand substitutionT1 -> T38,\nX28 -> T38,\nX29 -> T39,\nT2 -> s(T39),\nX30 -> T37,\nT3 -> s(T37),\nT35 -> T38,\nT36 -> T39", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 30 [arrowhead = none ];
30 -> 16;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 31 [arrowhead = none ];
31 -> 17;

32 [label="EVAL with clause\nsum(X6, s(X7), s(X8)) :- sum(X6, X7, X8).\nand substitutionT1 -> T12,\nX6 -> T12,\nX7 -> T13,\nT2 -> s(T13),\nX8 -> T11,\nT5 -> s(T11),\nT9 -> T12,\nT10 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 32 [arrowhead = none ];
32 -> 6;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 33 [arrowhead = none ];
33 -> 7;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 34 [arrowhead = none ];
34 -> 8;

35 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 35 [arrowhead = none ];
35 -> 9;

36 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 36 [arrowhead = none ];
36 -> 10;

37 [label="EVAL with clause\nsum(X13, 0, X13).\nand substitutionT12 -> T18,\nX13 -> T18,\nT13 -> 0,\nT11 -> T18", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 37 [arrowhead = none ];
37 -> 11;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 38 [arrowhead = none ];
38 -> 12;

39 [label="EVAL with clause\nsum(X20, s(X21), s(X22)) :- sum(X20, X21, X22).\nand substitutionT12 -> T28,\nX20 -> T28,\nX21 -> T29,\nT13 -> s(T29),\nX22 -> T27,\nT11 -> s(T27),\nT25 -> T28,\nT26 -> T29", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 39 [arrowhead = none ];
39 -> 14;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 40 [arrowhead = none ];
40 -> 15;

41 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 41 [arrowhead = none ];
41 -> 13;

42 [label="INSTANCE with matching:\nT1 -> T28\nT2 -> T29\nT3 -> T27", fontsize=14, style = filled, fillcolor = lightgrey];
14 -> 42 [arrowhead = none , style=dashed];
42 -> 1[style=dashed];

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
16 -> 43 [arrowhead = none ];
43 -> 18;

44 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
18 -> 44 [arrowhead = none ];
44 -> 19;

45 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
18 -> 45 [arrowhead = none ];
45 -> 20;

46 [label="EVAL with clause\nsum(X35, 0, X35).\nand substitutionT38 -> T44,\nX35 -> T44,\nT39 -> 0,\nT37 -> T44", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 46 [arrowhead = none ];
46 -> 21;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 47 [arrowhead = none ];
47 -> 22;

48 [label="EVAL with clause\nsum(X42, s(X43), s(X44)) :- sum(X42, X43, X44).\nand substitutionT38 -> T54,\nX42 -> T54,\nX43 -> T55,\nT39 -> s(T55),\nX44 -> T53,\nT37 -> s(T53),\nT51 -> T54,\nT52 -> T55", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 48 [arrowhead = none ];
48 -> 24;

49 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 49 [arrowhead = none ];
49 -> 25;

50 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 50 [arrowhead = none ];
50 -> 23;

51 [label="INSTANCE with matching:\nT1 -> T54\nT2 -> T55\nT3 -> T53", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 51 [arrowhead = none , style=dashed];
51 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

sumA(T5, 0, T5).
sumA(T18, s(0), s(T18)).
sumA(T28, s(s(T29)), s(s(T27))) :- sumA(T28, T29, T27).
sumA(T44, s(0), s(T44)).
sumA(T54, s(s(T55)), s(s(T53))) :- sumA(T54, T55, T53).

Query: sumA(a,a,g)

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

sumA_in_aag(T5, 0, T5) → sumA_out_aag(T5, 0, T5)
sumA_in_aag(T18, s(0), s(T18)) → sumA_out_aag(T18, s(0), s(T18))
sumA_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, T27, sumA_in_aag(T28, T29, T27))
sumA_in_aag(T54, s(s(T55)), s(s(T53))) → U2_aag(T54, T55, T53, sumA_in_aag(T54, T55, T53))
U2_aag(T54, T55, T53, sumA_out_aag(T54, T55, T53)) → sumA_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sumA_out_aag(T28, T29, T27)) → sumA_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sumA_in_aag(x1, x2, x3)  =  sumA_in_aag(x3)
sumA_out_aag(x1, x2, x3)  =  sumA_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(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:

sumA_in_aag(T5, 0, T5) → sumA_out_aag(T5, 0, T5)
sumA_in_aag(T18, s(0), s(T18)) → sumA_out_aag(T18, s(0), s(T18))
sumA_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, T27, sumA_in_aag(T28, T29, T27))
sumA_in_aag(T54, s(s(T55)), s(s(T53))) → U2_aag(T54, T55, T53, sumA_in_aag(T54, T55, T53))
U2_aag(T54, T55, T53, sumA_out_aag(T54, T55, T53)) → sumA_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sumA_out_aag(T28, T29, T27)) → sumA_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sumA_in_aag(x1, x2, x3)  =  sumA_in_aag(x3)
sumA_out_aag(x1, x2, x3)  =  sumA_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(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:

SUMA_IN_AAG(T28, s(s(T29)), s(s(T27))) → U1_AAG(T28, T29, T27, sumA_in_aag(T28, T29, T27))
SUMA_IN_AAG(T28, s(s(T29)), s(s(T27))) → SUMA_IN_AAG(T28, T29, T27)
SUMA_IN_AAG(T54, s(s(T55)), s(s(T53))) → U2_AAG(T54, T55, T53, sumA_in_aag(T54, T55, T53))

The TRS R consists of the following rules:

sumA_in_aag(T5, 0, T5) → sumA_out_aag(T5, 0, T5)
sumA_in_aag(T18, s(0), s(T18)) → sumA_out_aag(T18, s(0), s(T18))
sumA_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, T27, sumA_in_aag(T28, T29, T27))
sumA_in_aag(T54, s(s(T55)), s(s(T53))) → U2_aag(T54, T55, T53, sumA_in_aag(T54, T55, T53))
U2_aag(T54, T55, T53, sumA_out_aag(T54, T55, T53)) → sumA_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sumA_out_aag(T28, T29, T27)) → sumA_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sumA_in_aag(x1, x2, x3)  =  sumA_in_aag(x3)
sumA_out_aag(x1, x2, x3)  =  sumA_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(x4)
SUMA_IN_AAG(x1, x2, x3)  =  SUMA_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4)  =  U1_AAG(x4)
U2_AAG(x1, x2, x3, x4)  =  U2_AAG(x4)

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

(6) Obligation:

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

SUMA_IN_AAG(T28, s(s(T29)), s(s(T27))) → U1_AAG(T28, T29, T27, sumA_in_aag(T28, T29, T27))
SUMA_IN_AAG(T28, s(s(T29)), s(s(T27))) → SUMA_IN_AAG(T28, T29, T27)
SUMA_IN_AAG(T54, s(s(T55)), s(s(T53))) → U2_AAG(T54, T55, T53, sumA_in_aag(T54, T55, T53))

The TRS R consists of the following rules:

sumA_in_aag(T5, 0, T5) → sumA_out_aag(T5, 0, T5)
sumA_in_aag(T18, s(0), s(T18)) → sumA_out_aag(T18, s(0), s(T18))
sumA_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, T27, sumA_in_aag(T28, T29, T27))
sumA_in_aag(T54, s(s(T55)), s(s(T53))) → U2_aag(T54, T55, T53, sumA_in_aag(T54, T55, T53))
U2_aag(T54, T55, T53, sumA_out_aag(T54, T55, T53)) → sumA_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sumA_out_aag(T28, T29, T27)) → sumA_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sumA_in_aag(x1, x2, x3)  =  sumA_in_aag(x3)
sumA_out_aag(x1, x2, x3)  =  sumA_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(x4)
SUMA_IN_AAG(x1, x2, x3)  =  SUMA_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4)  =  U1_AAG(x4)
U2_AAG(x1, x2, x3, x4)  =  U2_AAG(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 2 less nodes.

(8) Obligation:

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

SUMA_IN_AAG(T28, s(s(T29)), s(s(T27))) → SUMA_IN_AAG(T28, T29, T27)

The TRS R consists of the following rules:

sumA_in_aag(T5, 0, T5) → sumA_out_aag(T5, 0, T5)
sumA_in_aag(T18, s(0), s(T18)) → sumA_out_aag(T18, s(0), s(T18))
sumA_in_aag(T28, s(s(T29)), s(s(T27))) → U1_aag(T28, T29, T27, sumA_in_aag(T28, T29, T27))
sumA_in_aag(T54, s(s(T55)), s(s(T53))) → U2_aag(T54, T55, T53, sumA_in_aag(T54, T55, T53))
U2_aag(T54, T55, T53, sumA_out_aag(T54, T55, T53)) → sumA_out_aag(T54, s(s(T55)), s(s(T53)))
U1_aag(T28, T29, T27, sumA_out_aag(T28, T29, T27)) → sumA_out_aag(T28, s(s(T29)), s(s(T27)))

The argument filtering Pi contains the following mapping:
sumA_in_aag(x1, x2, x3)  =  sumA_in_aag(x3)
sumA_out_aag(x1, x2, x3)  =  sumA_out_aag(x1, x2)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4)  =  U1_aag(x4)
U2_aag(x1, x2, x3, x4)  =  U2_aag(x4)
SUMA_IN_AAG(x1, x2, x3)  =  SUMA_IN_AAG(x3)

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:

SUMA_IN_AAG(T28, s(s(T29)), s(s(T27))) → SUMA_IN_AAG(T28, T29, T27)

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

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:

SUMA_IN_AAG(s(s(T27))) → SUMA_IN_AAG(T27)

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:

• SUMA_IN_AAG(s(s(T27))) → SUMA_IN_AAG(T27)
  The graph contains the following edges 1 > 1

(14) YES