# Equivalence of the two boolean expression

This is a question from a textbook on digital logic which I am having a difficult time with:
Prove that the following expression is valid:
$\bar{x_1}x_3+x_1x_2\bar{x_3}+\bar{x_1}x_2+x_1\bar{x_2}=\bar{x_2}x_3+x_1\bar{x_3}+x_2\bar{x_3}+\bar{x_1}x_2x_3$

My attempt at the proof:
Using Demorgan's Theorem I am able to simplify the LHS to
$x_1(\bar{x_2}+\bar{x_3})+\bar{x_1}(x_3+x_2)$ (Since $x+\bar{x}y=x+y$),
and applying the same procedure to the RHS I have
$x_3(\bar{x_2}+\bar{x_1})+\bar{x_3}(x_1+x_2)$,
which the two sides are clearly not identical to each other. So is there any other means of simplifying the expression further?

Any help is appreciated!!

You could try breaking these down. So $\bar{x_1}x_3+\cdots = \bar{x_1}x_2x_3+\bar{x_1}\bar{x_2}x_3 + \cdots$ and so on. Then remove duplicates.