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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: q(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
6 [label="6: q(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(not_zero(T3), ','(p(T3, X4), q(X4)))\n\nT3 is ground\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(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];
17 [label="17: ','(','(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];
20 [label="20: ','(','(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];
23 [label="23: ','(','(!_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];
24 [label="24: ?_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];
26 [label="26: ','(failure(a), ','(p(0, X4), q(X4)))\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: ','(failure(a), ','(p(0, X4), q(X4))), SCOPE: 4, CLAUSE: 6\n\nX4 is free", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: ','(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];
38 [label="38: ','(p(T9, X4), q(X4))\n\nT9 is ground\nX4 is free\nzero(T9) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="43: ','(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];
45 [label="45: ','(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];
49 [label="49: q(T12)\n\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
50 [label="50: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
51 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 51 [arrowhead = none ];
51 -> 6;

52 [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];
6 -> 52 [arrowhead = none ];
52 -> 7;

53 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 53 [arrowhead = none ];
53 -> 13;

54 [label="ONLY EVAL with clause\nnot_zero(X7) :- ','(zero(X7), ','(!_2, failure(a))).\nand substitutionT3 -> T6,\nX7 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 54 [arrowhead = none ];
54 -> 17;

55 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 55 [arrowhead = none ];
55 -> 20;

56 [label="EVAL with clause\nzero(0).\nand substitutionT6 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 56 [arrowhead = none ];
56 -> 23;

57 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 57 [arrowhead = none ];
57 -> 24;

58 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
23 -> 58 [arrowhead = none ];
58 -> 26;

59 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 59 [arrowhead = none ];
59 -> 36;

60 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
26 -> 60 [arrowhead = none ];
60 -> 32;

61 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
32 -> 61 [arrowhead = none ];
61 -> 35;

62 [label="ONLY EVAL with clause\nnot_zero(X16).\nand substitutionT6 -> T9,\nX16 -> T9", fontsize=14, style = filled, fillcolor = lightblue];
36 -> 62 [arrowhead = none ];
62 -> 38;

63 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
38 -> 63 [arrowhead = none ];
63 -> 43;

64 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (zero(T9), zero(0))", fontsize=14, style = filled, fillcolor = lightgreen];
43 -> 64 [arrowhead = none ];
64 -> 45;

65 [label="EVAL with clause\np(s(X22), X22).\nand substitutionX22 -> T12,\nT9 -> s(T12),\nX4 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
45 -> 65 [arrowhead = none ];
65 -> 49;

66 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
45 -> 66 [arrowhead = none ];
66 -> 50;

67 [label="INSTANCE with matching:\nT1 -> T12", fontsize=14, style = filled, fillcolor = lightgrey];
49 -> 67 [arrowhead = none , style=dashed];
67 -> 2[style=dashed];

}

(2) Obligation:

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

qA_in_g(s(T12)) → U1_g(T12, qA_in_g(T12))
U1_g(T12, qA_out_g(T12)) → qA_out_g(s(T12))

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:

QA_IN_G(s(T12)) → U1_G(T12, qA_in_g(T12))
QA_IN_G(s(T12)) → QA_IN_G(T12)

The TRS R consists of the following rules:

qA_in_g(s(T12)) → U1_g(T12, qA_in_g(T12))
U1_g(T12, qA_out_g(T12)) → qA_out_g(s(T12))

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:

QA_IN_G(s(T12)) → U1_G(T12, qA_in_g(T12))
QA_IN_G(s(T12)) → QA_IN_G(T12)

The TRS R consists of the following rules:

qA_in_g(s(T12)) → U1_g(T12, qA_in_g(T12))
U1_g(T12, qA_out_g(T12)) → qA_out_g(s(T12))

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(T12)) → QA_IN_G(T12)

The TRS R consists of the following rules:

qA_in_g(s(T12)) → U1_g(T12, qA_in_g(T12))
U1_g(T12, qA_out_g(T12)) → qA_out_g(s(T12))

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:

QA_IN_G(s(T12)) → QA_IN_G(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:

QA_IN_G(s(T12)) → QA_IN_G(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:

• QA_IN_G(s(T12)) → QA_IN_G(T12)
  The graph contains the following edges 1 > 1

(12) YES