How is the distribution law working in this Boolean Algebraic equation?

Question

I am working on a sample boolean algebraic question as study for my upcoming exam, my full expression is $(A\overline B(C + BD) + \overline{AB})C$

I have simplified my expression to

$A\overline BC + \overline {AB}C$

My example papers solution book says the next step uses the distributive law to reduce the expression to

$\overline BC ( A + A)$

But I do not understand this distribution, shouldn't $\overline {AB}$ become $\overline A + \overline B$ , how are we getting two positive A variables?

Answer 1

Yes, $\overline {AB}$ becomes $\overline A + \overline B$, so you get:

$A \overline BC + (\overline A + \overline B) C=$

$A \overline B C + \overline A C + \overline B C=$

$(A \overline B + \overline A + \overline B)C=$ (Absorption)

$(\overline A + \overline B)C =$

$\overline {AB}C$

...So no, I have no idea where the book gets their expression from ...

Answer 2

First, $$(A\overline{B}(C + BD) + \overline{AB})C$$ is likely to be $$(A\overline{B}(C + BD) + \overline{A}\overline{B})C \enspace.$$

In case you can't see the difference (whether you can depends on how MathJax works) in the second case it's $\overline{A} \cdot \overline{B}$.

Second, and regardless of the previous observation, you should get

$$ A\overline{B}C + \overline{A}\overline{B}C \enspace. $$

You cannot drop the $C$ from the first term. Finally, assuming $\overline{B} \cdot \overline{C}$, distributivity yields $(A + \overline{A})\overline{B}C = \overline{B}C$. In all likelihood they forgot an overline in the solutions.