Prime not Divisor implies Coprime/Proof 2

Theorem
Let $p, a \in \Z$.

If $p$ is a prime number then:
 * $p \nmid a \implies p \perp a$

where:
 * $p \nmid a$ denotes that $p$ does not divide $a$
 * $p \perp a$ denotes that $p$ and $a$ are coprime.

Proof
Let $p$ be a prime number.

Let $a \in \Z$ be such that $p$ is not a divisor of $a$.

Suppose $p$ and $a$ are not coprime.

Then:
 * $\exists c \in \Z_{>1}: c \mathop \backslash p, c \mathop \backslash a$

But then by definition of prime:
 * $c = p$

Thus:
 * $p \mathop \backslash a$

The result follows by Proof by Contradiction.