# 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?

• It should be $(B'+C)D' +ABC'D$ if I read that correctly (for just the one step you attempted) Commented Apr 8, 2021 at 5:02
• Thank you for pointing that out. Commented Apr 8, 2021 at 5:06
• So, basically, I have $D'(B'+C)+ABC'D$ to $D'(B+C)+D(ABC')$ to $(D'+D)((B+C)+ABC')$ to $B'+C+ABC'$ Commented Apr 8, 2021 at 5:09
• I should revise my other comment as $B'+C=(BC')'$. So you get $(BC')'D'+(BC')AD$ as a next step. Commented Apr 8, 2021 at 5:14
• That lets you treat $BC'$ as a single quantity, so we could relabel it as $E$ and work with $E'D'+EAD$. Commented Apr 8, 2021 at 5:28

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

• (Also noting that getting $B+C$ from $B'+C$ is a problem...) Commented Apr 8, 2021 at 13:11
• @abiessu Fortunately, the OP goes the other way around in the last step. Two wrongs make a right? :P Commented Apr 8, 2021 at 23:52