# Simplification of boolean expression

I am asking you for a help with following boolean expression that i can't solve by myself. I can minimize expression in Karnaugh map but i have to use boolean algebra too.

Expression:

!(C + !D) * !(B + D) + !!(C + !D) * !!(B + D)

1. (!C * D) * (!B * !D) + (C + !D) * (B+D) - De Morgan law
2. (!C * D) * (!B * !D) equals 0 because of A * 0 = 0
3. (C + !D) * (B + D) - multiplying everything
4. BC + B!D + CD + !DD
5. !DD equals 0
6. BC + B!D + CD - that what remains

So i'm stucked and i don't know how should i continue.

Expression after simplification should be: CD + B!D - (SoP - Sum of Products)

Thank you very much for your time.

• Draw the Karnaugh Map and you'll see it.
– user16324
Commented Sep 30, 2017 at 15:30

You got most of the way. Follow what's below:

\begin{align*} F &= \overline{(C + \overline{D})}\: \overline{(B + D)} + \overline{\overline{(C + \overline{D})}}\: \overline{\overline{(B + D)}}\tag{0}\\\\ &=\overline{C} \:D\: \overline{B}\: \overline{D} + (C + \overline{D})\: (B + D)\tag{1}\\\\ &=(C + \overline{D})\: (B + D)\tag{2}\\\\ &=B\:C + C\: D + B\:\overline{D} + D\:\overline{D}\tag{3}\\\\ &=B\:C + C\: D + B\:\overline{D}\tag{4}\\\\ &=B\:C\:D + B\:C\:\overline{D} + C\: D + B\:\overline{D}\tag{5}\\\\ &=(B\:C + C)\: D+(B\:C + B)\:\overline{D}\tag{6}\\\\ &=(C\:[B + 1])\: D+(B\:[C + 1])\:\overline{D}\tag{7}\\\\ &=(C\cdot1)\: D+(B\cdot1)\:\overline{D}\tag{8}\\\\ &=C\: D+B\:\overline{D}\tag{9} \end{align*}

Here, you can see that I've expanded the BC term in step 5. This is just turning a single case into two cases, which does NOT change the result. I think you can see that it doesn't, by inspection.

Then, in step 6 I organize the summed terms so that I can factor out D and Not-D, to create two somewhat more complex terms. But now, simple inspection tells you that in the first term (based on D) that if C is true that it doesn't matter whether or not B C is true and that if C is false then so is B C. So that can be reduced down to just C. The same idea also applies to the second term (based on Not-D.) I've added steps 7 and 8 to show this transition.

The final result is in 9, now.

• @maro if you observe you implemented de Morgan's transform wrong by not inverting the output, you will be able to fix this on your own now. N.B. Commented Sep 30, 2017 at 16:40
• @TonyStewart.EEsince'75 i don't see a problem with my application of de Morgan's law. My 1. step in question is the same as jonk's 1. step. I got stucked because i didn't know about absorption property X + XY = X. Thanks to jonk's answer and Lorenzo's answer i was able to figure this out. Am i still missing something? Why would i change !(C + !D) * !(B + D) to (!C * D) + (!B * !D) when i am applying de Morgan's law only to variables in parantheses? Sorry to bother you. It's completely new to me so bear with me. Commented Sep 30, 2017 at 17:58
• @maro_vargovcik I added steps to better formalize that transition from step 6 onward. I know you understand, already. But it doesn't hurt to add it.
– jonk
Commented Sep 30, 2017 at 18:21

BC + B!D + CD = BC*(D+!D) + B!D + CD = BCD + BC!D + B!D + CD

Because of the absorption property: X + XY = X

BCD + CD = CD and BC!D + B!D = B!D

hence:

BCD + BC!D + B!D + CD = CD + B!D

which is what you expected.

• Also great answer. Thank you very much for your explanation. Commented Sep 30, 2017 at 18:20