Boolean Simplification (1.2)

Question

I'm going to attempt to ask a question again. I'm new to this site and I think that this site could be useful to me through out the 4 years in University for me. Can you tell me whether I'm going the right direction with this simplification

I came across a Boolean Simplifcation question and I would like your help.

A'BC'D' + AB'C'D' + AB'CD' + ABC'D + ABCD' = A'BC'D' + AB'D' (C'+ C) + ABC'D + ABCD' = A'BC'D' + AB'D' + ABC'D + ABCD'

A'BC'D' + AB'D' + ABC'D + ABCD' AB(C'D + CD') AB(1.D + 1.D') AB(D + D') AB(1) AB

A'BC'D + AB'D' + AB D(A'BC' + AB') = D (A⊕B + C) = D + C' + A⊕B + AB

Answer 1

$A'BC'D'+AB'C'D'+AB'CD'+ABC'D+ABCD' =$ (use Adjacency to combine 2nd and 3rd as well as 3rd and 5th)

$A'BC'D'+ AB'D' + ACD' + ABC'D =$ (use Distribution to combine 1st and 4th as wel as 2nd and third)

$BC'(A'D'+AD) + AD'(B'+C) =$ (use XOR)

$BC'(A \bigoplus D') + AD'(B'+C)$

That's about as simple as you can get.

Answer 2

If we want a minimal sum of products (without using XOR), then a Karnaugh map shows that the best we can do is simplify from $5$ to $3$: \begin{align*} &A'BC'D' + AB'C'D' + AB'CD' + ABC'D + ABCD \\ &= A'BC'D' + AB'D'(C' + C) + ABD(C' + C) \\ &= A'BC'D' + AB'D' + ABD \\ \end{align*}