[m-users.] (P ; not P)

[Julian Fondren]

Hello list,

Mercury infers that the following is nondet:

   :- func fizzbuzz(int) = string.
   fizzbuzz(N) = R :-
       (
           fizz(N), buzz(N),           R = "FizzBuzz"
       ;
           fizz(N), not buzz(N),       R = "Fizz"
       ;
           not fizz(N), buzz(N),       R = "Buzz"
       ;
           not fizz(N), not buzz(N),   R = string.from_int(N)
       ).

Where fizz/1 and buzz/1 are just semidet predicates:

   :- pred fizz(int::in) is semidet.
   :- pred buzz(int::in) is semidet.

Is there a right way to do a disjunction over semidet predicates like
this, or is turning them into the bools the best you can do if you
want a disjunction over all the cases?

   ( if fizz(N) then Fizz = yes else Fizz = no ),
   ( if buzz(N) then Buzz = yes else Buzz = no ),
   (
       Fizz = yes, Buzz = yes, R = "FizzBuzz"
   ;
       ...
   )