Positive Integer Greater than 1 has Prime Divisor/Proof 2

Proof
Suppose the contrary, and that there are some positive integers which are not divisible by some prime.

Let $S = \left\{{n \in \Z: n > 1: \neg \exists p \in \Bbb P: p \mathop \backslash n}\right\}$.

That is:
 * $S = \left\{{\text{all integers not divisible by a prime}}\right\}$

Let $n \in S$ be the smallest of these.

As $S$ is bounded below by $1$, this is bound to exist, by Integers Bounded Below has Smallest Element.

So:
 * $\neg \exists x \in S: x < n$

Now $n$ cannot be prime itself:
 * $\left({\left({n \in \Bbb P}\right) \land \left({n \mathop \backslash n}\right) \implies n \notin S}\right) \implies n \notin \Bbb P$

So from Composite Number has Two Divisors Less Than It:
 * $\exists r, s \in \Z: n = r s, 1 < r < n, 1 < s< n$

There are two possibilities:
 * $(1):\quad$ Neither $r$ nor $s$ has a prime divisor
 * $(2):\quad$ At least one of $r$ and $s$ has a prime divisor.

If either $r$ or $s$ has a prime divisor, then:
 * $\exists p \in \Bbb P: \left({p \mathop \backslash r}\right) \lor \left({p \mathop \backslash s}\right) \implies p \mathop \backslash n$

This contradicts our claim that $n$ is not not divisible by some prime.

However, if neither $r$ nor $s$ has a prime divisor, it follows that $r, s \in S$.

But as $r, s < n$, this contradicts our choice of $n$ as the smallest element of $S$.

Therefore there can be no such $n$, therefore $S = \varnothing$, and all positive integers greater than one are divisible by some prime.