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Boolean algebra question

Owen · 2009-01-28 12:09:09

My head is wrecked, I've got the following statement : X=A'BCD + AB'CD + ABC'D + ABCD' + ABCD I've reduced this to : X=CD(A'B+AB') + AB(C+D) Can this be reduced any further using boolean algebra? I've been googling lots, but most examples are only using ABC, not ABCD, and I'm stumped.

28-01-2009 12:09pm

#1

Owen

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My head is wrecked, I've got the following statement :

X=A'BCD + AB'CD + ABC'D + ABCD' + ABCD

I've reduced this to :

X=CD(A'B+AB') + AB(C+D)

Can this be reduced any further using boolean algebra? I've been googling lots, but most examples are only using ABC, not ABCD, and I'm stumped.

0

Comments

#2 28-01-2009 4:53pm

nialo

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i think you can reduce it to CD + AB(C +D) as A'B + AB' is 1.

or

CD(A'B + AB') + AB(C'D + CD') + ABCD
CD(1) + AB(1) + ABCD

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#3 28-01-2009 4:55pm

Owen

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nialo wrote: »

A'B + AB' is 1.

Oh, well, that's where I've been screwing up! Thanks Nialo

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#4 28-01-2009 4:59pm

nialo

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Dont take that as golden. but i think its right. in simpler terms A + A' is 1 from what ive read. so would stand to reason it applys to AB as well.

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#5 28-01-2009 6:17pm

nobodythere

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Karnaugh Map:
   CD: 00  01  11  10
AB
00     
01              1
11          1   1   1
10              1

Best reduction is
ABC + ABD + BCD + ACD
= BC(A+D) + AD(B+C)

#6 29-01-2009 1:29pm

Cpt Tremendous

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Divide by zero

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