I have to prove, using the concept of divisibility (and no division), that if $c\ne 0$ then $a\mid b \leftrightarrow ac\mid bc$. I first proved $a \mid b \to ac\mid bc$, then got stuck on the opposite process. Here's what I did on the first part (please assume all numbers to be integers):
Is this right?
As for $ac\mid bc \implies a\mid b$, I can't figure out a way to remove $c$ from $b$ without messing everything on $a$'s side.
Googling for an answer, I only found solutions that used division itself, which was of no help. Thanks in advance for your help!