Integer Divisor is Equivalent to Subset of Ideal

Theorem
Let $\Z$ be the set of all integers.

Let $\Z_{>0}$ be the set of strictly positive integers.

Let $m \in \Z_{>0}$ and let $n \in \Z$.

Let $\left({m}\right)$ be the principal ideal of $\Z$ generated by $m$.

Then:
 * $m \backslash n \iff \left({n}\right) \subseteq \left({m}\right)$

Proof
The Ring of Integers is a Principal Ideal Domain.

The result follows directly from Principal Ideals in Integral Domain.