Euclid's Lemma for Prime Divisors/Proof 3

Proof
Let $p \mathrel \backslash a b$.

Suppose $p \nmid a$.

Then by :
 * $p \perp a$

As $p \mathrel \backslash a b$, it follows by definition of divisor:
 * $\exists e \in \Z: e p = a b$

So by :
 * $p : a = b : e$

But as $p \perp a$, by:

and:

it follows that:
 * $p \mathrel \backslash b$

Similarly, if $p \perp b$ then $p \mathrel \backslash a$.

Hence the result.