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?
a - pit is first par
b - rit is first row
!a && !b -> !(a || b)
remember that && and || is shortcur and why a was added was to avoid
the evaluation of b.
so in !(a || b)
a is false then b is tested -> error
with (!a && !b)
a is false (thus true) and b is never tested.
I guess the (!a && !b) -> !(a || b) transmformation only holds if a
and b are independant. That is not the case here.
--
Lgb
