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

(0) Obligation:

Clauses:

e(X) :- ','(=(Y, .(e, .(o, Y))), c(Y, X)).
c(.(e, X1), 0).
c(.(X2, X), s(Y)) :- c(X, Y).

Query: e(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: e(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: e(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(=(X5, .(e, .(o, X5))), c(X5, T3))\n\nT3 is ground\nX5 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 5 [arrowhead = none ];
5 -> 2;

6 [label="ONLY EVAL with clause\ne(X4) :- ','(=(X5, .(e, .(o, X5))), c(X5, X4)).\nand substitutionT1 -> T3,\nX4 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 6 [arrowhead = none ];
6 -> 3;

7 [label="UNIFY-FAIL\nbecause of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 7 [arrowhead = none ];
7 -> 4;

}

(2) Obligation:

Triples:


Clauses:


Afs:

(3) TPisEmptyProof (EQUIVALENT transformation)

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

(4) YES