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Prove/disprove: $$\forall a>0:a\in\mathbb R: \exists N\in\mathbb R:\forall x\in \mathbb R:\exists z\in\mathbb R:\forall n\in \mathbb N:|n-99|<N\Rightarrow n>10 \vee \frac {n^2} 4 \le 25$$

So as you can see, $a,x,z$ aren't used, so can I just ignore them and continue?

Also, $|n-99|<N$ is false, since $n\to\infty$ but $N$ is a constant, so if what we suppose is false then the whole statement will always be true.

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  • $\begingroup$ The only variables that matter are $N$ and $n$. $\endgroup$ Dec 22 '14 at 6:24
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The only variables that matter are $N$ and $n$. And the assertion is true. Pick for example $N=0$, or $N=-18.3$. Then there are no $n$ satisfiying the inequality $|n-99|\lt N$, so the implication is trivially true.

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Yes. $\;\langle \exists x :: E \rangle\;$ and $\;\langle \forall x :: E \rangle\;$ are completely equivalent to $\;E\;$, if the variable $\;x\;$ does not occur (or at least does not occur as a free variable) in $\;E\;$.

Note however that one needs to be careful that the variables don't range over empty sets: for an $\;E\;$ not containing $\;x\;$, we have that $\;\langle \exists x : x \in V : E \rangle\;$ is equivalent to $\;V \not= \emptyset \;\land\; E\;$, and $\;\langle \forall x : x \in V : E \rangle\;$ to $\;V = \emptyset \;\lor\; E\;$.

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