Proof of $\text{Dom}(R^{-1}) = \text{Ran}(R)$ where $R$ is a relation from $A$ to $B$
This exercise is taken from Velleman's "How to Prove It". Velleman starts his proof off by noting that $\text{Dom}(R^{-1})$ and $\text{ Ran}(R)$ are both subsets of $B$ and thus lets $b$ be an arbitrary element of $B$. The proof is then begun with assuming $b \in \text{Dom}(R^{-1})$.
My question: Why must we note that either side is a subset of $B$? Would it not be sufficient to introduce an arbitrary variable $x$ such that $x \in \text{Dom}(R^{-1})$ and show it to be an element of $\text{Ran}(R)$? It seems superfluous to say the variable is an arbitrary element of $B$ as stating it is an element of either side of the identity implies that it is.
The proof most likely has two parts: (i) show that $\text{Dom}(R^{-1})\subseteq \text{Ran}(R)$, and (ii) show that $\text{Ran}(R)\subseteq \text {Dom}(R^{-1})$. At the beginning the author then makes some definitions which hold for both parts. When he then says "assume $b\in \text{Dom}(R^{-1})$, this most likely also means "begin of part (i)", while the statement $b\in B$ probably means "this holds for both parts, so I additionally state it before the first part where I list everything which holds for either part".