Contraposition of an implication with quantifiers I am trying to prove a theorem, and a method is by using contraposition. What is the contraposition of the phrase:

$\exists x$ satisfying P $\Rightarrow$ $\forall y$ satisfy P

I thought it as

$\exists y$ that does not satisfy P $\Rightarrow$ $\forall x$ do not satisfy P.

Is this correct?
You are correct. $\forall$ turns into $\exists$ and each logical expression gets negated.
The contraposition of: $\exists x P(x) \Rightarrow \forall y P(y) $ is $ \exists x \lnot P(x)\Rightarrow \forall y \lnot P(y) $

(the two other answers are wrong) 

Proof:


$\exists x P(x) \Rightarrow \forall y P(y) $
$ \lnot \forall y P(y) \Rightarrow \lnot \exists x P(x) $  Modus tolens
$ \lnot \lnot \exists y \lnot P(y) \Rightarrow \lnot \exists x P(x) $ Replace  $\forall y $ with $ \lnot \exists y  \lnot $
$ \exists y \lnot P(y) \Rightarrow \lnot \exists x P(x) $ Double negation elimination 
$ \exists y \lnot P(y) \Rightarrow \lnot \lnot \forall x  \lnot P(x) $ Replace  $\exists x $ with $ \lnot \forall y  \lnot $
$ \exists y \lnot  P(y) \Rightarrow \forall x  \lnot P(x) $ Double negation elimination 
$ \exists x \lnot  P(x) \Rightarrow \forall y  \lnot P(y) $ Relettering.