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
Let $\mathcal R \ne \varnothing$.

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.

Let $\mathcal R = \varnothing$.

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

From Biconditional as Disjunction of Conjunctions:
 * $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right) \vdash p \iff q$

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

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 $\mathcal R = \varnothing$.