# Boolean logic simplification

To simplify $$A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + AABC'D + ABCD' + AB'C'D' + AB'CD$$

I have no idea how to start the first step.

Thanks in advance!!

## 1 Answer

Use distribution, a number of times! We're given:

$$\color{red}{\bf A'B'C'D + A'B'CD'} + \color{purple}{A'BC'D' + A'BCD} + \color{blue}{\bf AABC'D + ABCD'} + \color{purple}{AB'C'D' + AB'CD}$$

We can use the distributive law, for example, on the terms $$\color{red}{\bf A'B'C'D + A'B'CD'} = (A'B')\color{green}{\bf (C'D + CD')}$$

and on the terms $$\color{blue}{\bf AABC'D + ABCD'} = ABC'D + ABCD' = AB\color{green}{\bf (C'D + CD')}$$

Now, the righthand sides of the above have a common "factor": $$A'B'\color{green}{\bf (C'D + CD')} + AB\color{green}{\bf (C'D + CD')} = (A'B'+ AB)(C'D + CD')$$

That leaves us with \begin{align} &\quad (A'B' + AB)(C'D + CD') + \color{purple}{{\bf A'B} C'D' + {\bf A'B} CD+ {\bf AB'}C'D' + {\bf AB'}CD} \\ \\ & = (A'B' + AB)(C'D + CD') + A'B(C'D' + CD) + AB'(C'D'+ CD) \\ \\ & = (A'B' + AB)(C'D + CD') + (A'B + AB')(C'D' + CD)\end{align}

• the last step cannot be simply further? – rMath Oct 1 '13 at 12:52
• Work on it...I'll follow up with comments, corrections, and/or further simplification. I've been answering your posts, and trying to be helpful, but you aren't upvoting anyone's answers, nor accepting my answers. That makes me wonder if I'm helping or not... – Namaste Oct 1 '13 at 22:19
• rMath - I took another look at this, and what I end with is as "simplified" as it gets. It can be expressed in various forms, (conjunctive normal form, disjunctive normal form, SOP...etc. But all are just expansions of the above, and much longer. – Namaste Oct 2 '13 at 12:01
• @amWhy: You just own these! +1 – Amzoti Oct 3 '13 at 2:03