Integer Coprime to all Factors is Coprime to Whole/Proof 2

Theorem
Let $a, b, c \in \Z$.

Let $c$ be coprime to each of $a$ and $b$.

Then $c$ is coprime to $a b$.

Proof
This proof follows the structure of 's proof, if not its detail.

Let $a, b, c \in \Z$ such that $c$ is coprime to each of $a$ and $b$.

Let $d = a b$.

$c$ and $d$ are not coprime.

Then:
 * $\exists e \in \Z_{>1}: e \divides c, e \divides d$

We have that $c \perp a$ and $e \divides c$

From :
 * $e \perp a$

As $e \divides d$ it follows by definition of divisibility that:
 * $\exists f \in \Z: e f = d$

We also have that:
 * $a b = d$

So:
 * $a b = e f$

But from :
 * $e : a = b : f$

or in more contemporary language:
 * $\dfrac a e = \dfrac b f$

But $a \perp e$.

From, $\dfrac a e$ is in canonical form.

From :
 * $e \divides b$

But we already have that:
 * $e \divides c$

So $e$ is a divisor of both $b$ and $c$.

But this is a contradiction of the assertion that $c$ and $b$ are coprime.

Hence the result.