2
$\begingroup$

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?

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

1 Answer 1

1
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ (Also noting that getting $B+C$ from $B'+C$ is a problem...) $\endgroup$
    – abiessu
    Commented Apr 8, 2021 at 13:11
  • 1
    $\begingroup$ @abiessu Fortunately, the OP goes the other way around in the last step. Two wrongs make a right? :P $\endgroup$
    – Bram28
    Commented Apr 8, 2021 at 23:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .