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This question already has an answer here:

how can i prove Demorgan's law (X+Y)'=X'. Y'?I already tried studying book but doesn't understand the first step so plz answer in detail.(beginner here)

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marked as duplicate by David Richerby, Evil, Rick Decker, Kyle Jones, Yuval Filmus Feb 1 '18 at 0:07

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  • $\begingroup$ Use a truth table. $\endgroup$ – Yuval Filmus Jan 24 '18 at 13:30
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In order to prove $A = B$, It is sufficient to prove that $A'B = 0$ and $A' + B = 1$. Try to think of why this should be the case intuitively. In case you are unable to understand, then think of $A$ and $B$ as sets, Boolean $+$ operation as set union operation and Boolean $.$ operation as set intersection operation.

Therefore, take $A = (X+Y)'$ and $B = X'Y'$.

So, $A'B = (X+Y)X'Y' = XX'Y' + YX'Y' = 0$.

Also, $A' + B = X + Y + X'Y' = 1$.

(Use the property $A + BC = (A + B )(A + C)$ for simplifying above expression.

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