Andre Poenitz <[EMAIL PROTECTED]> writes:
| 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
yes... tired I guess... and since we are on this topic, I have
forgotten the name of this transformation, can you remind me?
I found it: DeMorgan's Law.
| 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.
You are right... back to the original question... was my adding of
!a && ... wrong?
--
Lgb
