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

(0) Obligation:

Clauses:

q :- p(s(s(0))).
p(s(X)) :- p(X).
p(0).
p(s(s(s(0)))) :- p(s(s(s(0)))).

Query: q()

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
3 [label="A: q\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
6 [label="6: q, SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: p(s(s(0)))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: p(s(s(0))), SCOPE: 2, CLAUSE: 1 | p(s(s(0))), SCOPE: 2, CLAUSE: 2 | p(s(s(0))), SCOPE: 2, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: p(s(s(0))), SCOPE: 2, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: p(s(s(0))), SCOPE: 2, CLAUSE: 2 | p(s(s(0))), SCOPE: 2, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: p(s(0))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: p(s(0)), SCOPE: 3, CLAUSE: 1 | p(s(0)), SCOPE: 3, CLAUSE: 2 | p(s(0)), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: p(s(0)), SCOPE: 3, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: p(s(0)), SCOPE: 3, CLAUSE: 2 | p(s(0)), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: p(0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
53 [label="53: p(0), SCOPE: 4, CLAUSE: 1 | p(0), SCOPE: 4, CLAUSE: 2 | p(0), SCOPE: 4, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
56 [label="56: p(0), SCOPE: 4, CLAUSE: 2 | p(0), SCOPE: 4, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
58 [label="58: p(0), SCOPE: 4, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
59 [label="59: p(0), SCOPE: 4, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
60 [label="60: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
61 [label="61: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
62 [label="62: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
63 [label="63: p(s(0)), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
64 [label="64: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
65 [label="65: p(s(s(0))), SCOPE: 2, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
66 [label="66: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
67 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 67 [arrowhead = none ];
67 -> 6;

68 [label="ONLY EVAL with clause\nq :- p(s(s(0))).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 68 [arrowhead = none ];
68 -> 8;

69 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 69 [arrowhead = none ];
69 -> 11;

70 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 70 [arrowhead = none ];
70 -> 13;

71 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
11 -> 71 [arrowhead = none ];
71 -> 15;

72 [label="ONLY EVAL with clause\np(s(X7)) :- p(X7).\nand substitutionX7 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 72 [arrowhead = none ];
72 -> 17;

73 [label="BACKTRACK\nfor clause: p(0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 73 [arrowhead = none ];
73 -> 65;

74 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 74 [arrowhead = none ];
74 -> 19;

75 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
19 -> 75 [arrowhead = none ];
75 -> 23;

76 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
19 -> 76 [arrowhead = none ];
76 -> 24;

77 [label="ONLY EVAL with clause\np(s(X15)) :- p(X15).\nand substitutionX15 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
23 -> 77 [arrowhead = none ];
77 -> 28;

78 [label="BACKTRACK\nfor clause: p(0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
24 -> 78 [arrowhead = none ];
78 -> 63;

79 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
28 -> 79 [arrowhead = none ];
79 -> 53;

80 [label="BACKTRACK\nfor clause: p(s(X)) :- p(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
53 -> 80 [arrowhead = none ];
80 -> 56;

81 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
56 -> 81 [arrowhead = none ];
81 -> 58;

82 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
56 -> 82 [arrowhead = none ];
82 -> 59;

83 [label="ONLY EVAL with clause\np(0).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
58 -> 83 [arrowhead = none ];
83 -> 60;

84 [label="BACKTRACK\nfor clause: p(s(s(s(0)))) :- p(s(s(s(0))))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
59 -> 84 [arrowhead = none ];
84 -> 62;

85 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
60 -> 85 [arrowhead = none ];
85 -> 61;

86 [label="BACKTRACK\nfor clause: p(s(s(s(0)))) :- p(s(s(s(0))))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
63 -> 86 [arrowhead = none ];
86 -> 64;

87 [label="BACKTRACK\nfor clause: p(s(s(s(0)))) :- p(s(s(s(0))))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
65 -> 87 [arrowhead = none ];
87 -> 66;

}

(2) Obligation:

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

qA_in_ → qA_out_

Pi is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
P is empty.
The TRS R consists of the following rules:

qA_in_ → qA_out_

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

(4) Obligation:

Pi DP problem:
P is empty.
The TRS R consists of the following rules:

qA_in_ → qA_out_

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

(5) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,R,Pi) chain.

(6) YES