I need to reduce this expression:
$$F(A,B,C,D) = A'B'C'D + A'B'CD + A'BC'D + A'BCD' + AB'C'D + ABC'D' + ABCD'$$
I also have the following solution:
\begin{align*} &= \bar A \bar B D + \bar A \bar B \bar D + A \bar B \bar C D + AB\bar D\\ &= \bar B D(\bar A + A\bar C) + B \bar D\\ &= \bar B D(\bar A + \bar C) + B \bar D \end{align*}
However, I don't understand it very well, could someone explain me in detail what happened here?
$(A'B'C'D + A'B'CD) + (A'BC'D'+A'BCD') + AB'C'D + (ABC'D' + ABCD')$
$= A'B'D + A'BD' + AB'C'D + ABD'$ (1)
$= (A'B'D + AB'C'D) + (A'BD' + ABD')$ (2)
$= B'D(A'+AC') + BD'$ (3)
$= B'D(A'+C') + BD'$ (4)
(1): You can undistribute in three expressions. The first two terms, for example, can be undistributed as follows: $A'B'D(C'+C) = A'B'D$ since $(C'+C = 1)$. Similarly for the other two pairs.
(2): Just rearranged the terms.
(3): Undistributed both pairs.
(4): $A' +AC' = A'+C'$ because you know that either $A'$ or $A$ is asserted a priori. If $A'$ is asserted, you get $1$. If $A$ is asserted instead, well, $AC' = C'$.