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) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: q\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: q, SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: p(s(s(0)))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: 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];
5 [label="5: p(s(s(0))), SCOPE: 2, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: p(s(s(0))), SCOPE: 2, CLAUSE: 2 | p(s(s(0))), SCOPE: 2, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: p(s(0))\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: 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];
9 [label="9: p(s(0)), SCOPE: 3, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: p(s(0)), SCOPE: 3, CLAUSE: 2 | p(s(0)), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: p(0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: 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];
13 [label="13: p(0), SCOPE: 4, CLAUSE: 2 | p(0), SCOPE: 4, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: p(0), SCOPE: 4, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: p(0), SCOPE: 4, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: p(s(0)), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: p(s(s(0))), SCOPE: 2, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 23 [arrowhead = none ];
23 -> 2;

24 [label="ONLY EVAL with clause\nq :- p(s(s(0))).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 24 [arrowhead = none ];
24 -> 3;

25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 25 [arrowhead = none ];
25 -> 4;

26 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 26 [arrowhead = none ];
26 -> 5;

27 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 27 [arrowhead = none ];
27 -> 6;

28 [label="ONLY EVAL with clause\np(s(X7)) :- p(X7).\nand substitutionX7 -> s(0)", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 28 [arrowhead = none ];
28 -> 7;

29 [label="BACKTRACK\nfor clause: p(0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 29 [arrowhead = none ];
29 -> 21;

30 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 30 [arrowhead = none ];
30 -> 8;

31 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 31 [arrowhead = none ];
31 -> 9;

32 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 32 [arrowhead = none ];
32 -> 10;

33 [label="ONLY EVAL with clause\np(s(X15)) :- p(X15).\nand substitutionX15 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 33 [arrowhead = none ];
33 -> 11;

34 [label="BACKTRACK\nfor clause: p(0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 34 [arrowhead = none ];
34 -> 19;

35 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 35 [arrowhead = none ];
35 -> 12;

36 [label="BACKTRACK\nfor clause: p(s(X)) :- p(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 36 [arrowhead = none ];
36 -> 13;

37 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
13 -> 37 [arrowhead = none ];
37 -> 14;

38 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
13 -> 38 [arrowhead = none ];
38 -> 15;

39 [label="ONLY EVAL with clause\np(0).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 39 [arrowhead = none ];
39 -> 16;

40 [label="BACKTRACK\nfor clause: p(s(s(s(0)))) :- p(s(s(s(0))))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 40 [arrowhead = none ];
40 -> 18;

41 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 41 [arrowhead = none ];
41 -> 17;

42 [label="BACKTRACK\nfor clause: p(s(s(s(0)))) :- p(s(s(s(0))))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 42 [arrowhead = none ];
42 -> 20;

43 [label="BACKTRACK\nfor clause: p(s(s(s(0)))) :- p(s(s(s(0))))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 43 [arrowhead = none ];
43 -> 22;

}

(2) Obligation:

Triples:


Clauses:


Afs:

(3) TPisEmptyProof (EQUIVALENT transformation)

There are no more dependency triples. Hence, the dependency triple problem trivially terminates.

(4) YES