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Proof of the Week

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.
$\blacksquare$
Sources

