Divisibility and Principal Ideal Domain Proof Let R be a ring. Show that a|b iff $b \in (a)$ iff $(b) \subseteq $ (a). 
I first just want to write out what I know about this statement:
a|b means that a divides b or a is divisible by b and there exists $x \in R$ with ax=b
(a) is a principal ideal domain which is a domain in which every proper ideal can be generated by a single element
(a) can also be written as (a)=Ra
I'm just not sure how to go about this proof. Any help would be appreciated!

If $a|b$, then $b=ra$ for some $r\in R$, which means $b\in (a)$.
If $b\in (a)$, then I claim $(b)\subset (a)$. Indeed, an element of $(b)$ has the form $sb$ for some $s\in R$. Since $b\in (a)$, we must have $sb\in (a)$ also since $(a)$ is an ideal and is closed under multiplication by elements of $R$. 
Finally, if $(b)\subset (a)$, then $b\in (b)\subset (a)$, and so $b= ra$ for some $r\in R$. This is the same as saying $a|b$.
This shows the first statement implies the second, which implies the third, which finally implies the first again, hence all three are equivalent.
