Boolean logic simplification

Question

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!!

Answer 1

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