$\bar{A}\bar{B}+A\bar{B}\bar{C} \equiv \bar{A}\bar{B}+\bar{B}\bar{C}$

Question

When simplifying an expression I managed to get as far as the left hand side of the below. $$\bar{A}\bar{B}+A\bar{B}\bar{C} \equiv \bar{A}\bar{B}+\bar{B}\bar{C}$$ The answer was the right hand side. Without a truth table, I don't see how I could have gotten to the solution.

Could someone please explain how the solution is simplified to remove $A$?

Answer 1

Factorizing the $\bar B$ out from both sides, it is enough to show $\bar A + A \bar C = \bar A + \bar C$. But then , start from the RHS: \begin{align} \bar A + \bar C &= \bar A + (\bar A + A)\bar C \\&= \bar A + \bar A \bar C + A \bar C \\&= \bar A(1 + \bar C) + A\bar C \\&= \bar A + A\bar C \end{align}