1
$\begingroup$

I'm having some trouble getting a handle with this course. We are starting Boolean algebra and my professor wants us simplify the following:

(AB)'+(A'+B')'=

(AB)'+BC+A'B'C'=

I am assuming the "()" with "'" means the over-score above the variables.

Forgive my ignorance but my professor does not explain anything. He just says "Do!" in a Russian accent. I just want to understand.

1 Answers 1

1

By de Morgan’s law $(AB)'=A'+B\,'$, and it’s always true that $X+X'=1$, so $(AB)'+(A'+B\,')'=(A'+B\,')+(A'+B\,')'=1\;.$

Similarly, we can start simplifying $(AB)'+BC+A'B\,'C\,'$ by using de Morgan’s law to expand the first term, getting $A'+B\,'+BC+A'B\,'C'$. Now use one of the distributive laws to get $A'+A'B\,'C\,'=A'1+A'B\,'C\,'=A'(1+B\,'C\,')$ and then an absorption law to get $A'+A'B\,'C\,'=A'(1+B\,'C\,')=A'1=A'\;.$ Thus, $A'+B\,'+BC+A'B\,'C'=A'+B\,'+BC\;.$

Note that I could have reached the same final result by simplifying $B\,'+A'B\,'C\,'$ to $B\,'$, using exactly the same approach.

Added: As StainlessSteelRat notes in the comments, the simplification can be taken a step further. Specifically,

$B\,'+BC=(B\,'+B\,'C)+BC=B\,'+(B\,'+B)C=B\,'+C\;,$

so $A'+B\,'+BC=A'+B\,'+C$.

  • 0
    @Brian Since B' is there the B in BC is not required, so it reduces to A' + B' + C.2015-04-08