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:

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
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.

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

  • $\begingroup$ Indeed so!! Thank you for your help! I think with my approach I would have to use the same technique. $\endgroup$ – Nen Sep 21 '14 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.