Contraposition of an implication with quantifiers

Question

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?

Answer 1

You are correct. $\forall$ turns into $\exists$ and each logical expression gets negated.

Answer 2

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.