Prove the following, for positive integers $m$ and $n$. If $a \equiv b \pmod{n}$ and $m\mid n$, then $a \equiv b \pmod{m}$.
This seems to me to be simple transitivity with the Fundamental theorem of arithmetic. $a \equiv b \pmod{n}$ and $m\mid n$ means that $n \mid a - b$ so there is an integer k such that $kn = a - b$. Since $m|n$ there is an $l$ such that $lm = n$. Thus by substitution $klm = a - b$. Which means that $m \mid a - b$, this proves that $a \equiv b \pmod{m}$.
My question is firstly is this proof sound? And also if there is any more detail or a more concise way of writing this proof. I am new to proof writing in this format.
Any tips would be greatly appreciated. Thank you.
This is a perfect proof. This is sufficiently concise.