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

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??


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

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

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  • $\begingroup$ Yes! I see the mistake now. I just started learning proof writing and I confused the order of quantifiers. That's a mistake I shouldn't have made. Nevertheless, thank you for your answer. I don't Know if it's Worth leaving the question online but maybe it helps someone. $\endgroup$ Oct 8 '19 at 20:34
  • $\begingroup$ @ToTheSpace2 You're welcome! And yes, I think it might be helpful to someone ... your title is helpful in that regard as well. $\endgroup$
    – Bram28
    Oct 8 '19 at 20:40
  • $\begingroup$ True, I could've written a more informative title. $\endgroup$ Oct 8 '19 at 21:01
  • $\begingroup$ @ToTheSpace2 I think your title is just fine, actually! $\endgroup$
    – Bram28
    Oct 8 '19 at 21:11
  • $\begingroup$ Now that it's corrected, It's perfect! $\endgroup$ Oct 8 '19 at 21:42

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