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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
3 [label="A: p(T1, T2)\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: p(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT1 is ground\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="B: ','(q(T5, T6), r(T5))\n\nT5 is ground\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(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];
13 [label="13: ','(q(T5, T6), r(T5)), SCOPE: 2, CLAUSE: 1\n\nT5 is ground\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(q(T5, T6), r(T5)), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nT6 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: r(a)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: r(a), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: ','(q(T11, T12), r(T11))\n\nT11 is ground\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 32 [arrowhead = none ];
32 -> 4;

33 [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];
4 -> 33 [arrowhead = none ];
33 -> 7;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 34 [arrowhead = none ];
34 -> 11;

35 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 35 [arrowhead = none ];
35 -> 13;

36 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 36 [arrowhead = none ];
36 -> 15;

37 [label="EVAL with clause\nq(a, 0).\nand substitutionT5 -> a,\nT6 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 37 [arrowhead = none ];
37 -> 17;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
13 -> 38 [arrowhead = none ];
38 -> 19;

39 [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];
15 -> 39 [arrowhead = none ];
39 -> 28;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 40 [arrowhead = none ];
40 -> 31;

41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 41 [arrowhead = none ];
41 -> 21;

42 [label="BACKTRACK\nfor clause: r(b) :- r(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 42 [arrowhead = none ];
42 -> 23;

43 [label="INSTANCE with matching:\nT5 -> T11\nT6 -> T12", fontsize=14, style = filled, fillcolor = lightgrey];
28 -> 43 [arrowhead = none , style=dashed];
43 -> 7[style=dashed];

}

(2) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

pA_in_gg(T5, T6) → U1_gg(T5, T6, pB_in_gg(T5, T6))
pB_in_gg(T11, s(T12)) → U2_gg(T11, T12, pB_in_gg(T11, T12))
U2_gg(T11, T12, pB_out_gg(T11, T12)) → pB_out_gg(T11, s(T12))
U1_gg(T5, T6, pB_out_gg(T5, T6)) → pA_out_gg(T5, T6)

Pi is empty.

(3) 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:

PA_IN_GG(T5, T6) → U1_GG(T5, T6, pB_in_gg(T5, T6))
PA_IN_GG(T5, T6) → PB_IN_GG(T5, T6)
PB_IN_GG(T11, s(T12)) → U2_GG(T11, T12, pB_in_gg(T11, T12))
PB_IN_GG(T11, s(T12)) → PB_IN_GG(T11, T12)

The TRS R consists of the following rules:

pA_in_gg(T5, T6) → U1_gg(T5, T6, pB_in_gg(T5, T6))
pB_in_gg(T11, s(T12)) → U2_gg(T11, T12, pB_in_gg(T11, T12))
U2_gg(T11, T12, pB_out_gg(T11, T12)) → pB_out_gg(T11, s(T12))
U1_gg(T5, T6, pB_out_gg(T5, T6)) → pA_out_gg(T5, T6)

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

(4) Obligation:

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

PA_IN_GG(T5, T6) → U1_GG(T5, T6, pB_in_gg(T5, T6))
PA_IN_GG(T5, T6) → PB_IN_GG(T5, T6)
PB_IN_GG(T11, s(T12)) → U2_GG(T11, T12, pB_in_gg(T11, T12))
PB_IN_GG(T11, s(T12)) → PB_IN_GG(T11, T12)

The TRS R consists of the following rules:

pA_in_gg(T5, T6) → U1_gg(T5, T6, pB_in_gg(T5, T6))
pB_in_gg(T11, s(T12)) → U2_gg(T11, T12, pB_in_gg(T11, T12))
U2_gg(T11, T12, pB_out_gg(T11, T12)) → pB_out_gg(T11, s(T12))
U1_gg(T5, T6, pB_out_gg(T5, T6)) → pA_out_gg(T5, T6)

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:

PB_IN_GG(T11, s(T12)) → PB_IN_GG(T11, T12)

The TRS R consists of the following rules:

pA_in_gg(T5, T6) → U1_gg(T5, T6, pB_in_gg(T5, T6))
pB_in_gg(T11, s(T12)) → U2_gg(T11, T12, pB_in_gg(T11, T12))
U2_gg(T11, T12, pB_out_gg(T11, T12)) → pB_out_gg(T11, s(T12))
U1_gg(T5, T6, pB_out_gg(T5, T6)) → pA_out_gg(T5, T6)

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

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

PB_IN_GG(T11, s(T12)) → PB_IN_GG(T11, T12)

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

(9) PiDPToQDPProof (EQUIVALENT transformation)

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

(10) Obligation:

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

PB_IN_GG(T11, s(T12)) → PB_IN_GG(T11, T12)

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

(11) 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:

• PB_IN_GG(T11, s(T12)) → PB_IN_GG(T11, T12)
  The graph contains the following edges 1 >= 1, 2 > 2

(12) YES