Simplify the Boolean expression A'BC + A'BD' + AB' + AC + ABC + ACD

Question

I'm trying to simplify the expression: A'BC + A'BD' + AB' + AC + ABC + ACD

I got to the expression: BC + AC + A'BD' + AB'

I know this expression is equvalant to the one above because I verified using truth tables.

The simplified expression, according to online calculators is: BC + A'BD' + AB'

How do I continue with my simplification?

Answer 1

This is an application of the Consensus Theorem:

$xy + y'z = xy + y'z + xz$

Applied to your case, we thus have that:

$AB' + BC + AC = AB' +BC$

So, if you have the Consensus Theorem in your arsenal, you can just do in 1 step:

$BC + AC + A'BD' + AB' \overset{Consensus}{=}$

$BC + A'BD'+AB'$

If you can't use Consensus, but you do have Adjacency ($xy +xy'=x$) and Absorption ($p + pq =p$), then you can do:

$BC + AC + A'BD' + AB' \overset{Adjacency}{=}$

$BC + ABC + AB'C + A'BD' + AB' \overset{Absorption \ \times \ 2}{=}$

$BC+ A'BD' + AB'$

And if you don't have those either, do what tranceloction does in the other answer.

Answer 2

You can finish as follows writing $1$ for "True":

\begin{eqnarray*}BC+AC+A'BD' + AB' & = & BC + A(B+B')C +A'BD' + AB' \\ & = & (1+A)BC + AB'(C+1) + A'BD' \\ & = & BC + AB' + A'BD' \end{eqnarray*}