How can $(\exists x\in Z)(\forall y\in Z)(x>y)$ and its negation, $(\forall x\in Z)(\exists y\in Z)(x\le y)$, both be true?

Question

$$(\exists x\in Z)(\forall y\in Z)(x>y)$$

This statement is true since we can take any $y\in Z$, add $1$ to it which would yield $x\in Z$ always greater than $y$.

If we now negate this statement we get: $$(\forall x\in Z)(\exists y\in Z)(x\le y)$$ This statement should be false, but if we take any $x\in Z$, add $1$ to it, we get $y\in Z$ such that $x\le y$ which makes the negation of a true statement a true statement??

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Now there's probably something really wrong in my reasoning so can someone clarify this a little bit?

Answer 1

$(\exists x\in Z)(\forall y\in Z)(x>y)$ is False, not True ...

There is an integer greater than all integers (including itself)?! No.

Indeed, when you say:

$(\forall x\in Z)(\exists y\in Z)(x\le y)$

This statement is true since we can take any $y\in Z$, add $1$ to it which would yield $x\in Z$ always greater than $y$.

what you really show is that:

$(\forall y\in Z)(\exists x\in Z)(x>y)$ is True (which indeed it is)

So, in case you had not yet realized this: the order of the quantifiers matters!