another question concerning Pilog. Suppose following senseless Prolog
aT(N) :- t1(N).
aT(N) :- t2(N).
bT(N) :- t1(N) ; t2(N).
Here the predicate aT/1 may be rewritten using the ';' operator as in
bT/1. Now I have found, that Pilog offer a 'or' rule. But it seems, this
does not resemble ';' of Prolog. Suppose the Pilog translation of the
(be t1 (1))
(be t1 (2))
(be t1 (3))
(be t2 (4))
(be t2 (5))
(be t2 (6))
(be aT (@N) (t1 @N))
(be aT (@N) (t2 @N))
(be bT (@N)
(or (t1 @N) (t2 @N)))
Now calling (? (aT @N)) is running like the Prolog version and therefore
as I expected. But running (? (bT @N)) will return NIL at once.
So I would like to ask, if 'or' works as intended or if there is a bug?
Thanks in advance and ciao,