Discussion:
quick boolean algebra question
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Stephen Tu
2008-08-13 09:08:06 UTC
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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
2008-08-13 20:16:50 UTC
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!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 :)

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