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