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

(0) Obligation:

Clauses:

q(X) :- ','(not_zero(X), ','(p(X, Y), q(Y))).
p(0, 0).
p(s(X), X).
zero(0).
not_zero(X) :- ','(zero(X), ','(!, failure(a))).
not_zero(X1).
failure(b).

Query: q(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: q(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: q(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(not_zero(T3), ','(p(T3, X4), q(X4)))\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(not_zero(T3), ','(p(T3, X4), q(X4))), SCOPE: 2, CLAUSE: 4 | ','(not_zero(T3), ','(p(T3, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(','(zero(T6), ','(!_2, failure(a))), ','(p(T6, X4), q(X4))) | ','(not_zero(T6), ','(p(T6, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(','(zero(T6), ','(!_2, failure(a))), ','(p(T6, X4), q(X4))), SCOPE: 3, CLAUSE: 3 | ?_3 | ','(not_zero(T6), ','(p(T6, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(','(!_2, failure(a)), ','(p(0, X4), q(X4))) | ?_3 | ','(not_zero(0), ','(p(0, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ?_3 | ','(not_zero(T6), ','(p(T6, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX4 is free\nzero(T6) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(failure(a), ','(p(0, X4), q(X4)))\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(failure(a), ','(p(0, X4), q(X4))), SCOPE: 4, CLAUSE: 6\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(not_zero(T6), ','(p(T6, X4), q(X4))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX4 is free\nzero(T6) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(T9, X4), q(X4))\n\nT9 is ground\nX4 is free\nzero(T9) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(p(T9, X4), q(X4)), SCOPE: 5, CLAUSE: 1 | ','(p(T9, X4), q(X4)), SCOPE: 5, CLAUSE: 2\n\nT9 is ground\nX4 is free\nzero(T9) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(p(T9, X4), q(X4)), SCOPE: 5, CLAUSE: 2\n\nT9 is ground\nX4 is free\nzero(T9) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: q(T12)\n\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 18 [arrowhead = none ];
18 -> 2;

19 [label="ONLY EVAL with clause\nq(X3) :- ','(not_zero(X3), ','(p(X3, X4), q(X4))).\nand substitutionT1 -> T3,\nX3 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 19 [arrowhead = none ];
19 -> 3;

20 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 20 [arrowhead = none ];
20 -> 4;

21 [label="ONLY EVAL with clause\nnot_zero(X7) :- ','(zero(X7), ','(!_2, failure(a))).\nand substitutionT3 -> T6,\nX7 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 21 [arrowhead = none ];
21 -> 5;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="EVAL with clause\nzero(0).\nand substitutionT6 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 23 [arrowhead = none ];
23 -> 7;

24 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 24 [arrowhead = none ];
24 -> 8;

25 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 25 [arrowhead = none ];
25 -> 9;

26 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 26 [arrowhead = none ];
26 -> 12;

27 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
9 -> 27 [arrowhead = none ];
27 -> 10;

28 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 28 [arrowhead = none ];
28 -> 11;

29 [label="ONLY EVAL with clause\nnot_zero(X16).\nand substitutionT6 -> T9,\nX16 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 29 [arrowhead = none ];
29 -> 13;

30 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
13 -> 30 [arrowhead = none ];
30 -> 14;

31 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (zero(T9), zero(0))", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 31 [arrowhead = none ];
31 -> 15;

32 [label="EVAL with clause\np(s(X22), X22).\nand substitutionX22 -> T12,\nT9 -> s(T12),\nX4 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 32 [arrowhead = none ];
32 -> 16;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 33 [arrowhead = none ];
33 -> 17;

34 [label="INSTANCE with matching:\nT1 -> T12", fontsize=14, style = filled, fillcolor = lightgrey];
16 -> 34 [arrowhead = none , style=dashed];
34 -> 1[style=dashed];

}

(2) Obligation:

Triples:

qA(s(X1)) :- qA(X1).

Clauses:

qcA(s(X1)) :- qcA(X1).

Afs:

qA(x1)  =  qA(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:
qA_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

QA_IN_G(s(X1)) → U1_G(X1, qA_in_g(X1))
QA_IN_G(s(X1)) → QA_IN_G(X1)

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:

QA_IN_G(s(X1)) → U1_G(X1, qA_in_g(X1))
QA_IN_G(s(X1)) → QA_IN_G(X1)

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 1 less node.

(6) Obligation:

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

QA_IN_G(s(X1)) → QA_IN_G(X1)

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:

QA_IN_G(s(X1)) → QA_IN_G(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:

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

(10) YES