Boolean algebra: why is $a\overline bc + ab = ac+ab$?

Question

Why is $a\overline bc + ab = ab + ac$? I think it has something to do with the rule $a + \overline a = 1$, right?

Answer 1

We have the following (do have a look here, especially example $T10(a)$):- $$\begin{align}a\overline{b}c+ab&=a(\overline{b}c+b)\\&=a(b+\overline{b}c)\\&=a(b(1+c)+\overline{b}c)\\&=a(b+bc+\overline{b}c)\\&=a(b+c)\\&=ab+ac\end{align}$$

Answer 2

Yes, start from the right to show it:

$ab + ac = ab+a(b+\overline{b})c = ab + abc + a\overline{b}c = ab + a\overline{b}c$

(the first step used your suspicion, the next is the distributive law and the last step used the absorbtion law)