If $a\equiv b \ \mathrm{mod}(n)$ and $m|n$, then $a\equiv b \mathrm{mod}(m)$.

If $a\equiv b \ \mathrm{mod}(n)$ and $m|n$, then $a\equiv b \
\mathrm{mod}(m)$.

Is this correct:
If $a\equiv b \ \mathrm{mod}(n)$ and $m|n$, then $a\equiv b \
\mathrm{mod}(m)$.
Let $a=q_{1}n+r$, $b=q_{2}n+r$ and $n=mc$. Then we have \begin{align*}
\frac{q_{1}mc+r -(q_{2}mc+r)}{mc}=\frac{q_{1}m+r -(q_{2}m+r)}{m},
\end{align*} which imply that $a\equiv b \ \mathrm{mod}(m)$.
Thank you in advance.