Steps to Simplify Boolean Expression

Question

Could someone please example how

$(\bar A . \bar B . \bar C) + (\bar A . \bar B . C) + (A . B . \bar C) + (A . B . C)$

can be simplified to

$(\bar A . \bar B) + (A . B)$ ?

Here's what I've tried, but I don't trust all the steps:

$(\bar A . \bar B . \bar C) + (\bar A . \bar B . C) + (A . B . \bar C) + (A . B . C)$

$(\bar A . \bar B . \bar C) + (A . B . \bar C) + (\bar A . \bar B . C) + (A . B . C)$

$\bar C . ((\bar A . \bar B ) + (A . B)) + C . ((\bar A . \bar B) + (A . B))$

$(\bar C + C) . ((\bar A . \bar B ) + (A . B) + (\bar A . \bar B) + (A . B))$

$(\bar A . \bar B ) + (A . B) + (\bar A . \bar B) + (A . B)$

$2 . (\bar A . \bar B ) + 2 . (A . B)$

Now I'm stuck. Am I right so far? If so, why can I cancel those $2$s?

Answer 1

@Michael Rozenberg has already explained how to reduce the expression. Your method is also correct until this point:

$(\bar A . \bar B ) + (A . B) + (\bar A . \bar B) + (A . B)$

At this point, remember the laws of Boolean Algebra, which state that A + A = A. That allows you to simplify the expression further to $(\bar A . \bar B ) + (A . B) $, which is the desired result

Answer 2

Because $$\bar A \bar B \bar C + \bar A \bar BC + AB\bar C + ABC=$$ $$=\bar{A}\bar{B}(\bar{C}+C)+AB(\bar{C}+C)=\bar{A}\bar{B}+AB.$$