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

(0) Obligation:

Clauses:

q(X) :- ','(grammar(G), ','(-(terminal, string(X)), produces(G, X))).
grammar(S) :- ','(equal(S, or(N, and(a, and(S, b)))), equal(N, or(c, and(c, N)))).
equal(X, X).
-(terminal, string(nil)).
-(terminal, string(and(A, X))) :- ','(terminal(A), -(terminal, string(X))).
terminal(a).
terminal(b).
terminal(c).
produces(G, X) :- gives(G, X, nil).
gives(or(P, Q), X, Y) :- gives(P, X, Y).
gives(or(P, Q), X, Y) :- gives(Q, X, Y).
gives(and(P, Q), X, Z) :- ','(gives(P, X, Y), gives(Q, Y, Z)).
gives(A, and(A, X), X) :- terminal(A).

Query: q(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: q(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: q(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(grammar(X3), ','(-(terminal, string(T3)), produces(X3, T3)))\n\nT3 is ground\nX3 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(grammar(X3), ','(-(terminal, string(T3)), produces(X3, T3))), SCOPE: 2, CLAUSE: 1\n\nT3 is ground\nX3 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(','(equal(X11, or(X10, and(a, and(X11, b)))), equal(X10, or(c, and(c, X10)))), ','(-(terminal, string(T3)), produces(X11, T3)))\n\nT3 is ground\nX11 is free\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ','(','(equal(X11, or(X10, and(a, and(X11, b)))), equal(X10, or(c, and(c, X10)))), ','(-(terminal, string(T3)), produces(X11, T3))), SCOPE: 3, CLAUSE: 2\n\nT3 is ground\nX11 is free\nX10 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 8 [arrowhead = none ];
8 -> 2;

9 [label="ONLY EVAL with clause\nq(X2) :- ','(grammar(X3), ','(-(terminal, string(X2)), produces(X3, X2))).\nand substitutionT1 -> T3,\nX2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 9 [arrowhead = none ];
9 -> 3;

10 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 10 [arrowhead = none ];
10 -> 4;

11 [label="ONLY EVAL with clause\ngrammar(X9) :- ','(equal(X9, or(X10, and(a, and(X9, b)))), equal(X10, or(c, and(c, X10)))).\nand substitutionX3 -> X11,\nX9 -> X11", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 11 [arrowhead = none ];
11 -> 5;

12 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 12 [arrowhead = none ];
12 -> 6;

13 [label="BACKTRACK\nfor clause: equal(X, X)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 13 [arrowhead = none ];
13 -> 7;

}

(2) Obligation:

Triples:


Clauses:


Afs:

(3) TPisEmptyProof (EQUIVALENT transformation)

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

(4) YES