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I have been stuck on a homework problem which asks first to simplify a boolean equation using a k-map and second to reach the same result using boolean algerbra. The k-map was no issue and I have verified the k-map result, but no matter what I do, I am unable to achieve the same result using algerbra techniques. Any pointers would be much appreciated.

Original Equation: A'C (A'BD)' + A'BC'D' + AB'C

Kmap Solution: A'BD' + B'C

Written Work on Whiteboard Image

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  • $\begingroup$ Are you familiar with the consensus theorem? You may know it under the name of resolution. It says that $ab+b'c+ac = ab+b'c$. $\endgroup$
    – Fabio Somenzi
    Commented Feb 19, 2017 at 15:32
  • $\begingroup$ I am uncertain how to apply it. Every way I attempt it, I always end up with A'BC' + A'BD' + B'C $\endgroup$
    – Seth Duke
    Commented Feb 19, 2017 at 16:23
  • $\begingroup$ Your first term is incorrect. Can you show your work? $\endgroup$
    – Fabio Somenzi
    Commented Feb 19, 2017 at 16:29
  • $\begingroup$ I added a link to my written out work on the main question. $\endgroup$
    – Seth Duke
    Commented Feb 19, 2017 at 18:10
  • $\begingroup$ Your board work is correct; you just copied one term incorrectly. From A'CD' + A'BD' + B'C observe that the first term is the consensus term of the other two; hence it can be dropped. To apply consensus, you match A'D' to $a$, B to $b$ and C to $c$. $\endgroup$
    – Fabio Somenzi
    Commented Feb 19, 2017 at 19:41

1 Answer 1

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You have done everything correctly, but to get to the desired answer, let's pick it up a few lones earlier, where you have:

$B'C + A'CD' + A'BC'D'$

From here:

$= B'C + A'BCD' + A'B'CD' + A'BC'D'$ (By expansion of A'CD')

$= B'C + A'BCD' + A'BC'D'$ (Third term got Absorbed by first)

$= B'C + A'BD'$

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  • $\begingroup$ I just finished looking through the instructors manual for this questions and this is exactly as it is listed. I just don't understand why I am allowed to throw the B term back in? $\endgroup$
    – Seth Duke
    Commented Feb 19, 2017 at 19:24
  • $\begingroup$ You mean in the first step I do? It's because in general $A = AB + AB'$. So I expand A'CD' to A'BCD' + AB'CD' $\endgroup$
    – Bram28
    Commented Feb 19, 2017 at 21:20

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