Inverse Relation Properties

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

If $\mathcal R$ has any of the properties:


 * Reflexive
 * Antireflexive
 * Non-reflexive
 * Symmetric
 * Asymmetric
 * Antisymmetric
 * Non-symmetric
 * Transitive
 * Antitransitive
 * Non-transitive

... then its inverse $\mathcal R^{-1}$ has the same properties.

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

Thus:


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

So if $\mathcal R$ is reflexive then so is $\mathcal R^{-1}$.


 * $\forall x \in \mathcal R: \left({x, x}\right) \notin \mathcal R \implies \left({x, x}\right) \notin \mathcal R^{-1}$.

So if $\mathcal R$ is antireflexive then so is $\mathcal R^{-1}$.


 * If $\mathcal R$ is non-reflexive, it is neither reflexive nor antiflexive and therefore $\exists x \in S: \left({x, x}\right) \in \mathcal R$ and $\exists y \in S: \left({y, y}\right) \notin \mathcal R$.

Thus, the same applies to $\mathcal R^{-1}$.

Symmetry

 * Suppose $\mathcal R$ is symmetric.

Then from Relation equals Inverse iff Symmetric it follows that $\mathcal R^{-1}$ is also symmetric.


 * Suppose $\mathcal R$ is asymmetric.

Then $\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \notin \mathcal R$.

Thus if $\left({x, y}\right) \in \mathcal R$ then $\left({y, x}\right) \in \mathcal R^{-1}$ and $\left({x, y}\right) \notin \mathcal R^{-1}$.

Thus it follows that $\mathcal R^{-1}$ is also asymmetric.


 * Suppose $\mathcal R$ is antisymmetric.

Then $\left({x, y}\right), \left({y, x}\right)\in \mathcal R \implies x = y$.

It follows that $\left({y, x}\right), \left({x, y}\right)\in \mathcal R^{-1} \implies x = y$.

Thus it follows that $\mathcal R^{-1}$ is also antisymmetric.


 * Suppose $\mathcal R$ is 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.

Transitivity

 * Suppose $\mathcal R$ is transitive.

Then $\left({x, y}\right), \left({y, z}\right) \in \mathcal R \implies \left({x, z}\right) \in \mathcal R$.

Thus $\left({y, x}\right), \left({z, y}\right) \in \mathcal R^{-1} \implies \left({z, x}\right) \in \mathcal R^{-1}$ and so $\mathcal R^{-1}$ is transitive.


 * Suppose $\mathcal R$ is antitransitive.

Then $\left({x, y}\right), \left({y, z}\right) \in \mathcal R \implies \left({x, z}\right) \notin \mathcal R$.

Thus $\left({y, x}\right), \left({z, y}\right) \in \mathcal R^{-1} \implies \left({z, x}\right) \notin \mathcal R^{-1}$ and so $\mathcal R^{-1}$ is antitransitive.


 * Suppose $\mathcal R$ is non-transitive.

Then:
 * $\exists x_1, y_1, z_1 \in S: \left({x_1, y_1}\right), \left({y_1, z_1}\right) \in \mathcal R, \left({x_1, z_1}\right) \in \mathcal R$.
 * $\exists x_2, y_2, z_2 \in S: \left({x_2, y_2}\right), \left({y_2, z_2}\right) \in \mathcal R, \left({x_2, z_2}\right) \notin \mathcal R$.

So:
 * $\exists x_1, y_1, z_1 \in S: \left({y_1, x_1}\right), \left({z_1, y_1}\right) \in \mathcal R^{-1}, \left({z_1, x_1}\right) \in \mathcal R^{-1}$.
 * $\exists x_2, y_2, z_2 \in S: \left({y_2, x_2}\right), \left({z_2, y_2}\right) \in \mathcal R^{-1}, \left({z_2, x_2}\right) \notin \mathcal R^{-1}$.

So $\mathcal R^{-1}$ is non-transitive.