# Fully Simplifying a Boolean Expression with DeMorgan's Theorem

The expression I have is $$\\overline{(\bar{A}B+\bar{A}C})+(BC + \bar{B}\bar{C}) \$$, and I've been asked to simplify it using DeMorgan's Theorems. I've already made a start, which was to discount the double-negation, thus giving $$\(AB + A\bar{C})+(BC +\bar{B}\bar{C})\$$, but find myself unsure of what further steps I could take to simplify the circuit. Can it be further simplified, or have I done all that can be done to implement the circiut more efficiently?

• Are you absolutely sure that you handled the "double-negation" correctly?
– jonk
Oct 30, 2019 at 23:59
• I've edited your question (waiting for approval), please make sure I have edited it properly. I did make sure that the overbar above AC or BC were separate because $\overline{xy} \neq \bar{x}\bar{y}$, for an example.
– user103380
Oct 31, 2019 at 0:03
• Are we talking about DeMorgan's or Captain Morgan here? A Few people need to seriously review the theorem rules: electronics-tutorials.ws/boolean/demorgan.html
– Nedd
Oct 31, 2019 at 7:22

$$\\overline{(\bar{A}B+\bar{A}C})+(BC + \bar{B}\bar{C})\Leftrightarrow \$$

You made a mistake with the first $$\\bar{B}\$$ and the operator should also be inverted:

$$\\\(\overline{\bar{A}B}\cdot\overline{\bar{A}C})+(BC + \bar{B}\bar{C}) \Leftrightarrow\$$

Now you can apply De Morgan's law again

$$\\\((\bar{\bar{A}}+\bar{B})\cdot(\bar{\bar{A}}+\bar{C}))+(BC + \bar{B}\bar{C}) \Leftrightarrow\$$

$$\\\((A+\bar{B})\cdot(A+\bar{C}))+(BC + \bar{B}\bar{C}) \Leftrightarrow\$$

$$\\\(AA+\bar{B}A+A\bar{C}+\bar{B}\bar{C})+(BC + \bar{B}\bar{C}) \Leftrightarrow\$$

$$\\\(A+\bar{B}\bar{C})+(BC + \bar{B}\bar{C}) \Leftrightarrow\$$

$$\\\A+\bar{B}\bar{C}+BC + \bar{B}\bar{C} \$$

To not finish your homework (probably) , you can continue from here.

• @Downvoter: why the downvote? Oct 31, 2019 at 9:28
• Firstly I've not downvoted but, I suspect that someone seeing your answer might be thinking you should use proper latex script. They might think this because the original question was edited some time after you left your answer and it makes your answer (retrospectively) look a little untidy. You have my sympathy because you just copied and pasted the original formula from the "untidy" original question. Oct 31, 2019 at 10:29
• Thanks for that comment, indeed, he changed the layout, so I also updated my answer accordingly (long time ago I used Latex). Oct 31, 2019 at 11:12
• @HarrySvensson You are right (well I saw the incorrect B negation, but the operator should also be inverted) … I will try again (it's also too long ago for me I think). Oct 31, 2019 at 11:33
• I down voted. I noticed the error just before having to run to leave my house. It was all I had time for. Oct 31, 2019 at 16:19

Same answer as Michel Keijzers, just a simpler process.

$$\\overline{(\bar{A}B+\bar{A}C})+(BC + \bar{B}\bar{C}) \$$

$$\\overline{(\bar{A} (B+C}))+(BC + \bar{B}\bar{C}) \$$

$$\(\bar {\bar{A}} + \overline{(B+C}))+(BC + \bar{B}\bar{C}) \$$

$$\A + \bar B \bar C + BC + \bar{B}\bar{C}\$$