boolean simplification with k map

Question

A'B'CD + A'BCD + ABCD + AB'CD + ABC'D' +ABC'D + ABCD'

So I am trying to simplify this Boolean expression I need to use a K map but I also was trying to simplify it algebraically. This is what I got so far

CD + ABC' + ABC'D

So can this be simplified even more and how

Answer 1

I think you made a typo when you said you got to:

$CD+ABC'+ABC'D$

Because that should really be:

$CD+ABC'+ABCD'$

This can be simplified more using:

Nested Reduction

$PQ+PQ'R=PQ+PR$

Applied to your statement, given $CD$, you can reduce $ABCD'$ to just $ABC$. And using Adjacency, that combines with $ABC'$ to just $AB$

Final result: $CD+AB$

Of course, you could have gotten this immediately from the original expression as well. Indeed, you made the 'mistake' of thinking that you can use terms only once when combining with other terms. This is why you combined the first four terms to $CD$, the fifth and sixth term to $ABC'$, and were left with the last one. However, please realize that you can use terms more than once, because of:

Idempotence

$P+P=P$

Applied to your original expression, we can use Idempotence to duplicate the third term $ABCD$, and now you have:

$A'B'CD + A'BCD + ABCD + AB'CD + ABC'D' +ABC'D + ABCD'+ ABCD $

$=A'CD+ACD+ABC'+ABC$

$=CD+AB$