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

(0) Obligation:

Clauses:

goal :- ','(lte(X, s(s(s(s(0))))), even(X)).
lte(0, Y) :- !.
lte(X, s(Y)) :- ','(p(X, P), lte(P, Y)).
even(0).
even(s(s(X))) :- even(X).
p(0, 0).
p(s(X), X).

Query: goal()

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: goal\n\n", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: goal, SCOPE: 1, CLAUSE: 0\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(lte(X1, s(s(s(s(0))))), even(X1))\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(lte(X1, s(s(s(s(0))))), even(X1)), SCOPE: 2, CLAUSE: 1 | ','(lte(X1, s(s(s(s(0))))), even(X1)), SCOPE: 2, CLAUSE: 2\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(!_2, even(0)) | ','(lte(X1, s(s(s(s(0))))), even(X1)), SCOPE: 2, CLAUSE: 2\n\nX1 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: even(0)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: even(0), SCOPE: 3, CLAUSE: 3 | even(0), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: even(0), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: even(0), SCOPE: 3, CLAUSE: 4\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 13 [arrowhead = none ];
13 -> 2;

14 [label="ONLY EVAL with clause\ngoal :- ','(lte(X1, s(s(s(s(0))))), even(X1)).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 14 [arrowhead = none ];
14 -> 3;

15 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 15 [arrowhead = none ];
15 -> 4;

16 [label="ONLY EVAL with clause\nlte(0, X4) :- !_2.\nand substitutionX1 -> 0,\nX4 -> s(s(s(s(0))))", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 16 [arrowhead = none ];
16 -> 5;

17 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 17 [arrowhead = none ];
17 -> 6;

18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
6 -> 18 [arrowhead = none ];
18 -> 7;

19 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
7 -> 19 [arrowhead = none ];
19 -> 8;

20 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
7 -> 20 [arrowhead = none ];
20 -> 9;

21 [label="ONLY EVAL with clause\neven(0).\nand substitution", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 21 [arrowhead = none ];
21 -> 10;

22 [label="BACKTRACK\nfor clause: even(s(s(X))) :- even(X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 22 [arrowhead = none ];
22 -> 12;

23 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 23 [arrowhead = none ];
23 -> 11;

}

(2) Obligation:

Triples:


Clauses:


Afs:

(3) TPisEmptyProof (EQUIVALENT transformation)

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

(4) YES