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

(0) Obligation:

Clauses:

p(s(X), X).
plus(0, Y, Y).
plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z)).

Query: plus(g,a,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: plus(T1, T2, T3)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: plus(T1, T2, T3), SCOPE: 1, CLAUSE: 1 | plus(T1, T2, T3), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | plus(0, T2, T3), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: plus(T1, T2, T3), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nX2 is free\nplus(T1, T2, T3) !=? plus(0, X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: plus(0, T2, T3), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(p(s(T9), X12), plus(X12, T12, T13))\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(p(s(T9), X12), plus(X12, T12, T13)), SCOPE: 2, CLAUSE: 0\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: plus(T18, T12, T13)\n\nT18 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 11 [arrowhead = none ];
11 -> 2;

12 [label="EVAL with clause\nplus(0, X2, X2).\nand substitutionT1 -> 0,\nT2 -> T5,\nX2 -> T5,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 12 [arrowhead = none ];
12 -> 3;

13 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 13 [arrowhead = none ];
13 -> 4;

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

15 [label="EVAL with clause\nplus(s(X9), X10, s(X11)) :- ','(p(s(X9), X12), plus(X12, X10, X11)).\nand substitutionX9 -> T9,\nT1 -> s(T9),\nT2 -> T12,\nX10 -> T12,\nX11 -> T13,\nT3 -> s(T13),\nT10 -> T12,\nT11 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 15 [arrowhead = none ];
15 -> 7;

16 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 16 [arrowhead = none ];
16 -> 8;

17 [label="BACKTRACK\nfor clause: plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 17 [arrowhead = none ];
17 -> 6;

18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 18 [arrowhead = none ];
18 -> 9;

19 [label="ONLY EVAL with clause\np(s(X17), X17).\nand substitutionT9 -> T18,\nX17 -> T18,\nX12 -> T18", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 19 [arrowhead = none ];
19 -> 10;

20 [label="INSTANCE with matching:\nT1 -> T18\nT2 -> T12\nT3 -> T13", fontsize=14, style = filled, fillcolor = lightgrey];
10 -> 20 [arrowhead = none , style=dashed];
20 -> 1[style=dashed];

}

(2) Obligation:

Triples:

plusA(s(X1), X2, s(X3)) :- plusA(X1, X2, X3).

Clauses:

pluscA(0, X1, X1).
pluscA(s(X1), X2, s(X3)) :- pluscA(X1, X2, X3).

Afs:

plusA(x1, x2, x3)  =  plusA(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
plusA_in: (b,f,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

PLUSA_IN_GAA(s(X1), X2, s(X3)) → U1_GAA(X1, X2, X3, plusA_in_gaa(X1, X2, X3))
PLUSA_IN_GAA(s(X1), X2, s(X3)) → PLUSA_IN_GAA(X1, X2, X3)

R is empty.
The argument filtering Pi contains the following mapping:
plusA_in_gaa(x1, x2, x3)  =  plusA_in_gaa(x1)
s(x1)  =  s(x1)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

PLUSA_IN_GAA(s(X1), X2, s(X3)) → U1_GAA(X1, X2, X3, plusA_in_gaa(X1, X2, X3))
PLUSA_IN_GAA(s(X1), X2, s(X3)) → PLUSA_IN_GAA(X1, X2, X3)

R is empty.
The argument filtering Pi contains the following mapping:
plusA_in_gaa(x1, x2, x3)  =  plusA_in_gaa(x1)
s(x1)  =  s(x1)
PLUSA_IN_GAA(x1, x2, x3)  =  PLUSA_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(6) Obligation:

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

PLUSA_IN_GAA(s(X1), X2, s(X3)) → PLUSA_IN_GAA(X1, X2, X3)

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

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

(7) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(8) Obligation:

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

PLUSA_IN_GAA(s(X1)) → PLUSA_IN_GAA(X1)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(9) 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:

• PLUSA_IN_GAA(s(X1)) → PLUSA_IN_GAA(X1)
  The graph contains the following edges 1 > 1

(10) YES