Prove $(R^{-1})^{-1} = R$

Question

The problem is actually pretty simple, but because of the double inverse, I got confused on how to properly write the proof. I just want to make sure that What I wrote is actually valid.

proof:

(i)Suppose (x,y)$\in (R^{-1})^{-1}.$ Then by the defn of inverse, (y,x)$\in R^{-1}$. Then by the defn of inverse again, (x,y)$\in R$. Thus, $(R^{-1})^{-1} \subseteq R$.

(ii) Suppose (x,y)$\in R$. Then by the defn of inverse, (y,x) $\in R^{-1}$. Then by the defn of inverse again, (x,y) $\in (R^{-1})^{-1}$. Thus, R $\subseteq (R^{-1})^{-1}$.

Therefore, from (i) and (ii), $(R^{-1})^{-1} = R$.

Answer 1

This is fine. It could be simpler :

$$ (x,y)\in (R^{-1})^{-1} \quad \Leftrightarrow \quad (y,x)\in R^{-1} \quad \Leftrightarrow \quad (x,y)\in R $$ which means $\forall (x,y) : (x,y)\in R \ \Leftrightarrow \ (x,y)\in (R^{-1})^{-1}$, and, which means $R=(R^{-1})^{-1}$.