# Boolean Simplification

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.

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

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Alright that makes sense. My question is now do we always follow those same steps? i.e. de Morgans, distribute laws, absorption laws. Or does each equation go through a different approach, depending how it is presented? – Leo Oct 10 '12 at 9:29
@Leo: In general it’ll be a different approach each time, just as it was back in eighth- or ninth-grade algebra when you were asked to simplify an expression. In fact it is just algebra, though the rules are a little different from the ones for the familiar algebra of real numbers. – Brian M. Scott Oct 10 '12 at 9:35
Thanks Brain for help. – Leo Oct 10 '12 at 9:48
@Leo: You’re welcome. – Brian M. Scott Oct 10 '12 at 10:03
@Brian Since B' is there the B in BC is not required, so it reduces to A' + B' + C. – StainlessSteelRat Apr 8 '15 at 15:56