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I have approximately no idea on how to solve the following problem, so any help would very much be appreciated:

$$x' y'+ x y = (x y' + x' y)'$$

I can't figure out how to prove the equalities using boolean algebra
Thank you for your help!

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  • $\begingroup$ Hint: Do you know DeMorgan's laws? Can you apply one of them the right hand side twice? First use one of them to get rid of that negation outside the parentheses, and then use the other law to get rid of the negations outside the newly created parentheses. $\endgroup$ – Dilip Sarwate Nov 18 '13 at 2:45
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Here's one way. First, use DeMorgan's law: $$ (xy' + x'y)' = (xy')'(x'y)' = (x' + y)(x + y') $$ From there, you can expand the product using the distributive property, noting that $x'x$, which means "$x$ and not $x$", is $0$ (i.e. false). $$ (x' + y)(x + y') = (x'x) + xy + x'y' + (yy') = xy + x'y' $$

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