Intersection of Relation with Inverse is Symmetric Relation

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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.

$\blacksquare$


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