$\color{Green}{Background:}$
$\textbf{Theorem:}$ Let $R$ be an integral domain. Show that every associate of an irreducible element is irreducible
Proof 1: Let $a\in R$ be an irreducible element and let $b$ be an associate of $a.$ Then, there exists a unit $u\in R$ such that $a=bu.$ Also as $a$ is non-zero, non-unit, we have that $b$ is non-zero, non-unit. We now show that $b$ is irreducible element. Let $b=xy$ for some $x,y\in R.$ Then, $a=bu=xyu.$ Since $a$ is irreducible, either $x$ or $yu$ is a unit. If $yu$ is a u nit, then there exists some $v\in R$ such that $(yu)v=1.$ This gives that $y(uv)=1$ and so $y$ is a unit. Thus, either $x$ or $y$ is a unit and hence $b$ is an irreducible element.
Proof 2: Suppose $a=bu$ where $b$ is irreducible and $u$ is a unit. Suppose also $a=xy,$ then $xy=bu$ and hence $b=(bu)u^{-1}=x(yu^{-1}).$ Since $b$ is irreducible, $x$ is a unit or $yu^{-1}$ is a unit. If $yu^{-1}$ is a unit, say $yu^{-1}v=1,$ then $y$ is a unit. Therefore $a$ is irreducible.
$\color{Red}{Questions:}$
In both proofs above, don't I have to say or show that $x$ is a unit? Or in the conclusion of the definition of irreducible element involving a disjunction where it would usually have either "$x$ is a unit or $y$ is a unit." I only need to show one of the cases to be true and that will be enough?
Thank you in advance
You have to show that always at least one of $x$ or $y$ is a unit. From the irreducibility of $y$, we deduce that $x$ or $yu^{−1}$ is a unit. Now we make a case distinction: if $x$ is a unit, then clearly it's true that $x$ or $y$ is a unit. If $yu^{−1}$ is a unit, we proceed differently to show that $y$ is a unit, as the proof does.