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> **Proposition.** If $\text{Ran}(R) \subseteq \text{Dom}(S)$, then $\text{Dom}(S \circ R) = \text{Dom}(R)$
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My attempt:

Suppose $\text{Ran}(R) \subseteq \text{Dom}(S)$

We need to show:

$(\rightarrow) $ $\text{Dom}(S \circ R) \subseteq \text{Dom}(R)$

$(\leftarrow)$ $\text{Dom}(R) \subseteq \text{Dom}(S \circ R) $

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$(\rightarrow)$

Consider arbitrary element $a$, where $a \in \text{Dom}(S \circ R)$. Then there must be some element $p$, where $p = (a,x)$ and $p \in R$. It implies that $a \in Dom(R)$. Since $a$ was arbitrary, we have $\text{Dom}(S \circ R) \subseteq \text{Dom}(R)$

$(\leftarrow$)

Consider element $(x,y)$ where $(x,y) \in R$. Since $y \in Ran(R)$ and $\text{Ran}(R) \subseteq \text{Dom}(S)$, it follows that $y \in Dom(S)$, which means that there must be some element $(y,p)$ such that $(y,p) \in S$. 

Since $(y,p) \in S$ and $(x,y) \in R$, we have $(x,p) \in S \circ R$, and it means that $x \in Dom(S \circ R)$.

Since we've considered arbitrary element, we have $\text{Dom}(R) \subseteq \text{Dom}(S \circ R)$

We've shown both sides, hence $\text{Dom}(S \circ R) = \text{Dom}(R)$. $\Box$

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Is it correct?

If it is, are there better ways to prove $(\rightarrow)$?