Simplify this Boolean Algebra.

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stumpy
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Simplify this Boolean Algebra.

Postby stumpy » Tue Dec 03, 2013 7:45 pm

Simplify this Boolean Algebra. Can use Karnaugh Maps. ('A = Not A)

'A'B'C'D+'AB'C'D+A'B'C'D+AB'C'D

I have 'C'D('A'B+'AB+A'B+AB)
which makes just 'C'D.

Is this right?

anonmeans
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Re: Simplify this Boolean Algebra.

Postby anonmeans » Tue Dec 03, 2013 11:44 pm

stumpy wrote:Simplify this Boolean Algebra.

'A'B'C'D+'AB'C'D+A'B'C'D+AB'C'D

I have 'C'D('A'B+'AB+A'B+AB) which makes just 'C'D.

Is this right?

The factorization is valid, and the summands in the parentheses cancel out pairwise. The canceled pairs equal "1", and the sum (in Boolean algebra) of 1 + 1 is itself equal to 1. So you have 'C'D(1), which equals 'C'D.

Your answer looks correct to me.


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