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This question already has an answer here:

So I'm trying to prove the validity of this formula and I am a bit lost, not sure how to start. I know generally speaking a valid formula is one where if all the premises are true, then the conclusion can not be false, but I don't know how to prove this really. any help is appreciated

http://imgur.com/jop9Gq3

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marked as duplicate by Rick Decker, Claude Leibovici, user91500, hardmath, Henning Makholm logic Jun 29 '16 at 13:04

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    $\begingroup$ See the comments on this question. $\endgroup$ – Git Gud Jun 28 '16 at 21:55
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Convert to the contrapositive:$$\exists y((\exists x.S(x))\implies S(y))$$with $S=\lnot R$, and maybe it's a little easier to analyse. Simply put, if $\exists x.S(x)$, then there exists a $y$ such that the conclusion, $S(y)$ is true, and the formula is valid. If there is no such $x$, then the formula is vacuously true.

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