Prove that:
$$(a): [(a|b)\land(c|d)]\Longrightarrow ac|bd;$$
$$(b): a|b\Longrightarrow ac|bc;$$
$$(c): ac|bc\Longrightarrow a|b.$$
Proof:
(a) By definition of divisibility, there exists two integers $n_{1}$ and $n_{2}$ such that: $$a|b\Longrightarrow b = a\cdot n_{1} \,\text{and}\,c|d\Longrightarrow d = c\cdot n_{2}.$$ Multiplying the two statements, we have $$b\cdot d= (a\cdot n_{1})\cdot(c\cdot n_{2})=ac(n_{1}n_{2}).$$ Let $n'=n_{1}n_{2}$. Since the product of $n_{1}$ and $n_{2}$ are integers, it follows that $n'$ is an integer. Thus, $$bd=ac(n_{1}n_{2})=ac\cdot n'$$ and by definition of divisibility, $ac|bd$.
(b) $a|b \Longrightarrow b= a\cdot n\,$ for some integer $n$. Multiplying both sides by $c$, we have $bc = ac \cdot n$. Thus, by definition of divisibility, $ac|bc$.
(c) $ac|bc \Longrightarrow bc = ac\cdot n\,$ for some integer $n$. Dividing both sides by $c$, we have $b = a\cdot n$. Thus, by definition of divisibility, $a|b$.
Are these proofs sufficient?
I also have some question about (c). Isn't it assumed that $c\neq 0$? Since dividing by $0$ is undefined? Also, after proving $(b)$ and $(c)$, we can simply say $a|b\Longleftrightarrow ac|bc$? Is this correct?
Thanks for your time.
Those proofs look good to me. As for (c), yes you have to assume $c\neq 0$ (otherwise the statement is false, for example $2\cdot 0 | 1 \cdot 0$ but $2$ doesn't divide $1$.
You are correct, for $c\neq0$ $a|b \Leftrightarrow ac|bc$.