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

(0) Obligation:

Clauses:

factorial1(N, F) :- ','(>(N, 0), ','(is(N1, -(N, 1)), ','(factorial1(N1, F1), is(F, *(N, F1))))).
factorial1(0, 1).
factorial2(N, F) :- factorial2(0, N, 1, F).
factorial2(I, N, T, F) :- ','(<(I, N), ','(is(I1, +(I, 1)), ','(is(T1, *(T, I1)), factorial2(I1, N, T1, F)))).
factorial2(N, N, F, F).
factorial3(N, F) :- factorial3(N, 1, F).
factorial3(N, T, F) :- ','(>(N, 0), ','(is(T1, *(T, N)), ','(is(N1, -(N, 1)), factorial3(N1, T1, F)))).
factorial3(0, F, F).

Query: factorial(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: factorial(T2, T3)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="UNDEFINED ERROR", fontsize=14, style = filled, fillcolor = lightgreen];
1 -> 3 [arrowhead = none ];
3 -> 2;

}

(2) Obligation:

Triples:


Clauses:


Afs:

(3) TPisEmptyProof (EQUIVALENT transformation)

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

(4) YES