- #1

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Anyway, I would be useful to be able to prove that:

([itex]P_b[/itex] and not [itex]P_1[/itex]) and ([itex]P_b[/tex] and not [itex]P_2[/tex] and ...

as P1, P2, P3 ...

equals Pb and not (P1, P2, P3 ...)

I don't know if this is the proper way to express this, but I hope you will get my point. If not ask. And please correct me.