Positive Integer Greater than 1 has Prime Divisor

Lemma
Every positive integer greater than $$1$$ has at least one divisor which is prime.

Proof
By the Fundamental Theorem of Arithmetic, every natural number greater than one can be factored into a unique set of prime numbers.

Therefore, every positive integer greater than one has at least one prime factor.

Alternative Proof (not using FTA)
Suppose the contrary, and that there are some positive integers which are not divisible by some prime.

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

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 Minimal Element.

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

Now $$n$$ can not be prime itself:
 * $$\left({\left({n \in \mathbb{P}}\right) \and \left({n \backslash n}\right) \implies n \notin S}\right) \implies n \notin \mathbb{P}$$

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:
 * Neither $$r$$ nor $$s$$ has a prime divisor;
 * At least one of $$r$$ and $$s$$ has a prime divisor.

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

This contradicts our claim that $$n$$ is 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.