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

From Material Equivalence, we have $$\left({p \and q}\right) \or \left({\neg p \and \neg q}\right) \vdash p \iff q$$ 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 "A false proposition implies anything": $$\neg p \vdash p \implies q$$, 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 $$\mathcal{R} = \varnothing$$.