I'm very new to elementary number theory proofs and have been trying to figure out how to prove these seemingly straightforward identities with divisibility with no success.
For some integers $a,b,c \in \mathbb{Z}$
1) If ${a\mid bc}$ and ${a \not\mid b}$ then ${a\mid c}$
2) If ${a\mid c}$, ${b\mid c}$ and ${\gcd(a,b) = 1}$ then ${ab\mid c}$
For 1), example, if 2 divides 3a, then 2 clearly divides a because 2 does not divide 3... not sure how to formalize
For 2), I think it is somewhat related to the divisibility rule (https://en.wikipedia.org/wiki/Divisibility_rule). Let's say $6\mid 12$ and $3\mid 12$ is true but ${6\cdot 3\mid 12}$ is not and that relates to the fact that 6 is a multiple of 3. However, suppose ${2\mid a}$ and ${3\mid a}$. 2 and 3 are not multiples of each other and so the smallest number that is divisible by 2 and 3 must be a multiple of 2 and 3 (6 being the smallest), hence ${a}$ is divisible by 6. Is there a way to formalize this a better way?
The first claim is false: $4$ divides $(2\cdot 6)$ and $4$ does not divide $2$, do not imply that $4$ divides $6$.
For the second one, recall that if $\gcd(a,b)=1$ then, by the Bezout's identity, there are integers $m$ and $n$ such that $am+bn=1$. Moreover, $c=ra=sb$ for some integers $r$ and $s$. Hence $$c=c(am+bn)=cam+cbn=sbam+rabn=(ab)(sm+rn).$$