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I need to prove that: $$ !(!(X.W) + !(X.Z))) = X.W.Z $$ I have tried multiple approaches but cannot figure this out.

Using DeMorgan's theorem, I break the negative sign binding $XW$, and $XZ$, and change their sign: $$ !((!X + !W + !X + !Z)) $$ By Indempotent Law, I can keep only one $!X$: $$ !(!X + !W + !Z) $$ What are the next steps?

Edit, adding angryavian's solution: Using DeMorgan's Theorem again I can negate !X,!Y,!Z as a group instead of individually, and change their sign (. instead of +): $$ !(!(X.W.Z)) $$

Double not (!) negates each other, leaving with the solution: $$ X.W.Z $$

zyked
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1 Answers1

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CW answer (from comments to push it from unanswered queue:

Hint: Use DeMorgan's law again

Jesse P Francis
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