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

(0) Obligation:

Clauses:

p(X, Y) :- ','(q(X, Y), r(X)).
q(a, 0).
q(X, s(Y)) :- q(X, Y).
r(b) :- r(b).

Query: p(g,g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="B: p(T1, T2)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: p(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="A: ','(q(T5, T6), r(T5))\n\nT5 is ground\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(q(T5, T6), r(T5)), SCOPE: 2, CLAUSE: 1 | ','(q(T5, T6), r(T5)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(q(T5, T6), r(T5)), SCOPE: 2, CLAUSE: 1\n\nT5 is ground\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(q(T5, T6), r(T5)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: r(a)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: r(a), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(q(T11, T12), r(T11))\n\nT11 is ground\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 13 [arrowhead = none ];
13 -> 2;

14 [label="ONLY EVAL with clause\np(X3, X4) :- ','(q(X3, X4), r(X3)).\nand substitutionT1 -> T5,\nX3 -> T5,\nT2 -> T6,\nX4 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 14 [arrowhead = none ];
14 -> 3;

15 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 15 [arrowhead = none ];
15 -> 4;

16 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 16 [arrowhead = none ];
16 -> 5;

17 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 17 [arrowhead = none ];
17 -> 6;

18 [label="EVAL with clause\nq(a, 0).\nand substitutionT5 -> a,\nT6 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 18 [arrowhead = none ];
18 -> 7;

19 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 19 [arrowhead = none ];
19 -> 8;

20 [label="EVAL with clause\nq(X9, s(X10)) :- q(X9, X10).\nand substitutionT5 -> T11,\nX9 -> T11,\nX10 -> T12,\nT6 -> s(T12)", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 20 [arrowhead = none ];
20 -> 11;

21 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 21 [arrowhead = none ];
21 -> 12;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 22 [arrowhead = none ];
22 -> 9;

23 [label="BACKTRACK\nfor clause: r(b) :- r(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 23 [arrowhead = none ];
23 -> 10;

24 [label="INSTANCE with matching:\nT5 -> T11\nT6 -> T12", fontsize=14, style = filled, fillcolor = lightgrey];
11 -> 24 [arrowhead = none , style=dashed];
24 -> 3[style=dashed];

}

(2) Obligation:

Triples:

pA(X1, s(X2)) :- pA(X1, X2).
pB(X1, X2) :- pA(X1, X2).

Clauses:

qcA(X1, s(X2)) :- qcA(X1, X2).

Afs:

pB(x1, x2)  =  pB(x1, x2)

(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:
pB_in: (b,b)
pA_in: (b,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

PB_IN_GG(X1, X2) → U2_GG(X1, X2, pA_in_gg(X1, X2))
PB_IN_GG(X1, X2) → PA_IN_GG(X1, X2)
PA_IN_GG(X1, s(X2)) → U1_GG(X1, X2, pA_in_gg(X1, X2))
PA_IN_GG(X1, s(X2)) → PA_IN_GG(X1, X2)

R is empty.
Pi is empty.
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:

PB_IN_GG(X1, X2) → U2_GG(X1, X2, pA_in_gg(X1, X2))
PB_IN_GG(X1, X2) → PA_IN_GG(X1, X2)
PA_IN_GG(X1, s(X2)) → U1_GG(X1, X2, pA_in_gg(X1, X2))
PA_IN_GG(X1, s(X2)) → PA_IN_GG(X1, X2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

PA_IN_GG(X1, s(X2)) → PA_IN_GG(X1, X2)

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

(7) PiDPToQDPProof (EQUIVALENT 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:

PA_IN_GG(X1, s(X2)) → PA_IN_GG(X1, X2)

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:

• PA_IN_GG(X1, s(X2)) → PA_IN_GG(X1, X2)
  The graph contains the following edges 1 >= 1, 2 > 2

(10) YES