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I have two parts to my question, the first being how do I use Boolean algebra to reduce A'C'+ABC+AC' to have only 3 literals. I sort of understood the concept where you can factor things out but for some reason it won't come to mind with this one. My next question is how do I use DeMorgan's law with (a+c)(a+b')(a'+b+c').

I don't necessarily need the answers, I need to know the process to get the solution.

Edit: For DeMorgan's law, I'm not really sure if this is right but I multiplied out all 3 and got aa'+aa'b'+aa'c+a'b'c+ab+abb"+abc+bb'c+ac'+ab'c'+acc'+b'cc'.

If that is right then the next step would be to simplify correct?

15quince15
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  • Although it is reasonable to use "multipication" to mean logical and with addition to mean logical or, one should state that this is the intended interpretation. Less ambiguous would be to use actual mathematical notation $\land$ for logical and (resp. $\lor$ for logical or). See this introduction to using mathematical expressions. – hardmath Sep 22 '17 at 02:04

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HINT

For the first problem focus on the terms $A'C'$ and $AC'$

You can do: $A'C' + AC' = (A' + A)C' = (1)C' = C'$

So as the first few steps you get:

$A'C' + ABC + AC' = $

$A'C' + AC' + ABC = $

$(A' + A)C' + ABC = $

$(1)C' + ABC = $

$C' + ABC$

Now, do you see how you can simplify this just a little bit more?

Bram28
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  • Is it right if I use the complement for A' and A and for C' and C from ABC so I end up with (1)(1) + AB + (1)C' and ends up with AB + C' – 15quince15 Sep 22 '17 at 10:10
  • @15quince15 That actually ends up being correct ... but I am unsure if you are really thinking about that the right way ... In fact, you're using a pretty advanced version of a Reduction rule there which goes a good bit beyond the 'standard fare' of Boolean Algebra. I'll add a little bit to my post to help you think about this with more elementary principles. – Bram28 Sep 22 '17 at 13:53
  • thank you very much for your help, I understood what you are doing by arranging it and separating it then using the complement you are able to get to that point. So to continue from here you would do (C' + AB)(C'+C) then (C'+AB)(1). – 15quince15 Sep 22 '17 at 16:31
  • @15quince15 Exactly! Step by simple step typically does the job :) – Bram28 Sep 22 '17 at 17:22