Relation both Symmetric and Asymmetric is Null

Theorem
Let $\mathcal R$ be a relation in $S$ which is both symmetric and asymmetric.

Then $\mathcal R = \varnothing$.

Proof

 * If $\mathcal R \ne \varnothing$, then $\exists \left({x, y}\right) \in \mathcal R$.

Now, either $\left({y, x}\right) \in \mathcal R$ or $\left({y, x}\right) \notin \mathcal R$.

If $\left({y, x}\right) \in \mathcal R$, then $\mathcal R$ is not asymmetric.

If $\left({y, x}\right) \notin \mathcal R$, then $\mathcal R$ is not symmetric.

Therefore, if $\mathcal R \ne \varnothing$, $\mathcal R$ can not be both symmetric and asymmetric.


 * Now suppose $\mathcal R = \varnothing$.

Then $\forall x, y \in S: \left({x, y}\right) \notin \mathcal R \land \left({y, x}\right) \notin \mathcal R$.

From Biconditional Equivalences, we have $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q$ and thus $\forall x, y \in S: \left({x, y}\right) \in \mathcal R \iff \left({y, x}\right) \in \mathcal R$.

So the condition for symmetry is fulfilled, and $\mathcal R = \varnothing$ is symmetric.

As $\mathcal R = \varnothing$, we have $\forall x, y \in S: \left({x, y}\right) \notin \mathcal R$.

From False Statement Implies Every Statement: $\neg p \vdash p \implies q$, we can deduce:

$\forall x, y \in S: \left({x, y}\right) \notin \mathcal R$ therefore $\forall x, y \in S: \left({x, y}\right) \in \mathcal R \implies \forall x, y \in S: \left({y, x}\right) \notin \mathcal R$.

This is the definition of asymmetric.

Thus the only relation $\mathcal R$ to be both symmetric and asymmetric is $\mathcal R = \varnothing$.