Simplifying Boolean expression $B'D' + CD' + ABC'D$

Question

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?

Answer 1

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