quick boolean algebra question

[Stephen Tu]

another quick question:

can somebody explain how
O1 = !P1!P0 + !P1!I + !P0!I

reduces to:
O1 = !(P1P0+P1I+P0I)

, where ! denotes NOT

i'm sure it's pretty simple but i dont see it

[Stephen Tu]

Willem Jan van Vliet wrote:
> !p1!p0 + !p1!i + !p0!i
> = !(p1 + p0) + !(p1 + i) + !(p0 + i) de Morgan's Law
> = ![(p1 + p0)(p1 + i)] + !(p0 + i) de Morgan's Law
> = ![(p1 + p0)(p1 + i)(p0 + i)] de Morgan's Law
> = ![(p1p1 + p1i + p0p1 + p0i)(p0 + i)]
> = ![(p1 + p1i + p0p1 + p0i)(p0 + i)] idempotency
> = ![p1p0 + p1p0i + p0p0p1 + p0p0i + p1i + p1ii + p0p1i + p0ii]
> = ![p1p0 + p1p0i + p0i + p1i] idempotency
> = ![p1p0(1 + i) + p0i + p1i] distributive
> = ![p1p0(1) + p0i + p1i] law of 1's
> = !(p1p0 + p0i + p1i) identity
>

thanks dude. last night i got down all the way to the distributive step,
but for some reason (1+i) was not equaling (1) in my head :(

i'll blame it on the lateness :)