How to minimise boolean algebra ~(a+~b)(a~b+c)(b+~c)

Question

I don't know exactly what should I do after using DeMorgans law. No matter what I do I can't get the simplest form which is ~abc

~(a+~b)(a~b+c)(b+~c) = ~ab(a~b+c)(b+~c)

Have you tried using a [K-map](https://en.wikipedia.org/wiki/Karnaugh_map) to minimize it using pen and paper? — Jules, Jun 05 '16 at 16:41
So what is your question exactly? Are you just looking for someone to do this homework problem for you, or a tutorial on Boolean algebra, or what? Ask a more specific question! — Eric Lippert, Jun 05 '16 at 16:42
You can use (twice) that conjuction distributes over disjunction — harold, Jun 05 '16 at 16:48
I don't know what should I do after using DeMorgans law. Is there some patern that I should follow. Any tutorial that isn't K-map. Maybe some software to help me minimise it. — user140345, Jun 05 '16 at 16:50

score 1 · Accepted Answer · answered Jun 05 '16 at 17:15

1

Do you see why this is true?

(a~b+c)(b+~c)
= a~bb + a~b~c + cb + c~c

?

First ensure that you thoroughly understand why that is true.

Now, what do you know about a~bb and c~c?

Can you complete the proof from here?

answered Jun 05 '16 at 17:15

Eric Lippert

I've got it, thanks! I just had to remember few parts. – user140345 Jun 05 '16 at 17:36

1 Answers1