# Negation of nested quantifiers

The problem is:

$$\exists x \forall y (x \ge y)$$

With a domain of all real positive integers.

The negation is:

$$\forall x \exists y (x < y)$$

so, if $y = x + 1$, the negation is true.

That means the negation of the negation (i.e. the original problem) is false.

My question is, that if the original problem is $\exists x \forall y (x \ge y)$, why can't I take $x = y$ and prove the problem true?

• I don't see a paradox and I am also unable to understand your question. Can you kindly rephrase your question? – Mohammad Alaggan Sep 8 '12 at 2:53
• Sorry. I'm new at this. The problem can not be true because it's negation is true. However, it seems to me that x = y would make the problem true. Why is doesn't x = y satisfy the initial problem? – david.keck Sep 8 '12 at 2:58
• It's fine, now worries. I am just asking for a clarification of your question so I can help you :) – Mohammad Alaggan Sep 8 '12 at 2:59
• I edited my initial comment. I didn't know that enter submitted at first. Sorry. – david.keck Sep 8 '12 at 3:01