Inverse of Non-Symmetric Relation is Non-Symmetric

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

If $\mathcal R$ is non-symmetric, then so is $\mathcal R^{-1}$.

Proof
Let $\mathcal R$ be non-symmetric.

Then:
 * $\exists \left({x_1, y_1}\right) \in \mathcal R \implies \left({y_1, x_1}\right) \in \mathcal R$

and also:
 * $\exists \left({x_2, y_2}\right) \in \mathcal R \implies \left({y_2, x_2}\right) \notin \mathcal R$

Thus:
 * $\exists \left({y_1, x_1}\right) \in \mathcal R^{-1} \implies \left({x_1, y_1}\right) \in \mathcal R^{-1}$

and also:
 * $\exists \left({y_2, x_2}\right) \in \mathcal R^{-1} \implies \left({x_2, y_2}\right) \notin \mathcal R^{-1}$

and so $\mathcal R^{-1}$ is non-symmetric.