Is it possible to show that given $a,b,d,n \in \mathbb{N}$, if $a \equiv b \pmod{m}$ and $d\mid n$, then $a \equiv b \pmod{d}$?
I'm able to see it computationally by checking numbers but can't seem to formulate a proof so I'm not sure if I'm just picking lucky numbers that work out?
I'm trying to use this as a intermediate step in a larger proof but am unsure if it's valid.
A divisor of a divisor is a divisor, so $$a\equiv b\mod n\iff n\mid a-b\implies d\mid a-b\iff a\equiv b\mod d.$$