I have a rule that matches bc. When I encounter that in a string, I don't want to parse that string, otherwise parse anything else.
% Prolog
bc(B, C) --> [B, C], {
B = "b",
C = "c"
}.
not_bc(O) --> [O], % ?! bc(O, C).
% ?- phrase(not_bc(O), "bcdefg").
% false.
% ?- phrase(not_bc(O), "abcdefg").
% O = "a".
% ?- phrase(not_bc(O), "wxcybgz")
% O = "w".
% ?- phrase(not_bc(O), "wxybgz")
% O = "w".
Simplified version of my problem, hopefully solutions are isomorphic.
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An alternative:
process_bc(_) --> "bc", !, { fail }.
process_bc(C) --> [C].
This differs from my other solution in accepting:
?- time(phrase(process_bc(C), `b`, _)).
% 8 inferences, 0.000 CPU in 0.000 seconds (83% CPU, 387053 Lips)
C = 98.