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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)

Can you help me, please?

Thank you very much for your time.

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

2 Answers 2

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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.

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  • \$\begingroup\$ @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. \$\endgroup\$ Commented Sep 30, 2017 at 16:40
  • \$\begingroup\$ @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. \$\endgroup\$ Commented Sep 30, 2017 at 17:58
  • \$\begingroup\$ @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. \$\endgroup\$
    – jonk
    Commented Sep 30, 2017 at 18:21
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Starting from your result:

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.

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

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