On Tue, Nov 25, 2003 at 11:13:40AM +0100, Lars Gullik Bj�nnes wrote:
> Alfredo Braunstein <[EMAIL PROTECTED]> writes:
>
> | Alfredo Braunstein wrote:
> >
> >> But it should be "!(pit_ is the first par && rit_ is the first row) and
> >> ... i.e. (!pit is not the first par || !rit is the first row) and ...
> | without this ^^^
> >
> | i.e. !(a && b) == (!a || !b)
>
> Is it?
Yes:
!(a && b) !a || !b
a == 0, b == 0: a. Wrong. Stop. Negate. True !a. W. S. N. T
a == 0, b == 1: a. Wrong. Stop. Negate. True !a. W. S. N. T
a == 1, b == 0: a. Good. b Wrong. Negate. True a Good. N. b. W. N. Or T
a == 1, b == 1: a. Good. b good. Negate. False a Good. N. b. G. N. Or F
> I guess the (!a && !b) -> !(a || b) transmformation only holds if a
> and b are independant. That is not the case here.
It also holds for short-cut evaluation.
Andre'
