Proof using modulo

Question

Suppose $n|m$, then $n\cdot d=m,~d\in\Bbb Z$.

If $a\equiv b\mod m$, then $a\equiv b\mod{n\cdot d}$.
Additionally, $d\equiv d\mod{n\cdot d}$.

So we know $ad\equiv bd\mod{n\cdot d}$.
$\therefore a\equiv b\mod n$

Would this be a valid proof?