Simplifying Boolean expression $B'D' + CD' + ABC'D$ I'm struggling with simplifying $B'D' + CD' + ABC'D$.
Isn't it that it's already simplified? I tried doing $(B' + C)(D'+D) + ABC'D$ to get $B'+C+ABC'D$, but I am getting different truth tables.
What should the steps be?
EDIT:
I now have the following steps:
So, basically, I have
$D'(B'+C)+ABC'D$ to
$D'(B+C)+D(ABC')$
$(D'+D)((B+C)+ABC')$
$B'+C+ABC'$
Can it be simplified any further?

The original expression is as simplified as it can be.
In your EDIT, the step from:
$D'(B +C) + D(ABC')$
to:
$(D'+D)((B+C)+ABC')$
is incorrect.  That is like saying that $ab+cd=(a+c)(b+d)$, which from basic algebra you should know is not correct
