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.