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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!!

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1 Answer 1

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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.

As far as I can tell, you will be left with the same six terms on each side.

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  • $\begingroup$ Indeed so!! Thank you for your help! I think with my approach I would have to use the same technique. $\endgroup$
    – Nen
    Commented Sep 21, 2014 at 19:17

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