elementary number theory - Modular arithmetic r = c (mod a), s = c (mod b), find (mod a)

Assume that $a,b\in \mathbb{N}$ and that $a | b$. Let $r \equiv c \pmod a$ and $s \equiv c
\pmod b$. What is $s \pmod a$?

So I know the following:
$r \equiv c \mod a$
$s \equiv c\mod b$

Which implies by rule $a \equiv b \pmod m$ is the same as $b \equiv a \pmod m$,
$c \equiv r \pmod a$
$c \equiv s \pmod b$

Which implies that
$r \pmod a = s \pmod b$

However after that I do not know how to prove $s \pmod a$ from here. Am I overlooking a modular arithmetic rule or should I solve it differently?

Answer

Well, this may help you. Since $a|b$ and $b|s-c$ we get $a|s-c$, i.e., $s\equiv c \pmod a$, but $c\equiv r \pmod a$. So by transitivity we deduce that $s\equiv r\pmod a$.