Intersection of Relation with Inverse is Symmetric Relation

Theorem
Let $\mathcal R$ be a relation on a set $S$.

Then $\mathcal R \cap \mathcal R^{-1}$, the intersection of $\mathcal R$ with its inverse, is symmetric.

Proof
Let $\left({x, y}\right) \in \mathcal R \cap \mathcal R^{-1}$

By definition of intersection:


 * $\left({x, y}\right) \in \mathcal R$


 * $\left({x, y}\right) \in \mathcal R^{-1}$

By definition of inverse relation:


 * $\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \in \mathcal R^{-1}$


 * $\displaystyle \left({x, y}\right) \in \mathcal R^{-1} \implies \left({y, x}\right) \in \left ({\mathcal R^{-1}} \right )^{-1}$

By Inverse of Inverse Relation the second statement may be rewritten:


 * $\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \in \mathcal R^{-1}$


 * $\left({x, y}\right) \in \mathcal R^{-1} \implies \left({y, x}\right) \in \mathcal R$

Then by definition of intersection:


 * $\left({y, x}\right) \in \mathcal R \cap \mathcal R^{-1}$

Hence $\mathcal R \cap \mathcal R^{-1}$ is symmetric.