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

(0) Obligation:

Clauses:

dyck(X) :- ','(grammar(G), ','(-(terminal, string(X)), produces(G, X))).
grammar(S) :- =(S, or(e, or(and(S, S), and(o, and(S, c))))).
equal(X, X).
-(terminal, string(nil)).
-(terminal, string(and(A, X))) :- ','(terminal(A), -(terminal, string(X))).
terminal(o).
terminal(c).
terminal(e).
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: dyck(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: dyck(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: dyck(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: ','(=(X9, or(e, or(and(X9, X9), and(o, and(X9, c))))), ','(-(terminal, string(T3)), produces(X9, T3)))\n\nT3 is ground\nX9 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 7 [arrowhead = none ];
7 -> 2;

8 [label="ONLY EVAL with clause\ndyck(X2) :- ','(grammar(X3), ','(-(terminal, string(X2)), produces(X3, X2))).\nand substitutionT1 -> T3,\nX2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 8 [arrowhead = none ];
8 -> 3;

9 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 9 [arrowhead = none ];
9 -> 4;

10 [label="ONLY EVAL with clause\ngrammar(X8) :- =(X8, or(e, or(and(X8, X8), and(o, and(X8, c))))).\nand substitutionX3 -> X9,\nX8 -> X9", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 10 [arrowhead = none ];
10 -> 5;

11 [label="UNIFY-FAIL\nbecause of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 11 [arrowhead = none ];
11 -> 6;

}

(2) Obligation:

Triples:


Clauses:


Afs:

(3) TPisEmptyProof (EQUIVALENT transformation)

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

(4) YES