Relation both Symmetric and Asymmetric is Null

Theorem

Let $\RR$ be a relation in $S$ which is both symmetric and asymmetric.

Then:

$\RR = \O$

Let $\RR \ne \O$.

Then:

$\exists \tuple {x, y} \in \RR$

Now, either $\tuple {y, x} \in \RR$ or $\tuple {y, x} \notin \RR$.

If $\tuple {y, x} \in \RR$, then $\RR$ is not asymmetric.

If $\tuple {y, x} \notin \RR$, then $\RR$ is not symmetric.

Therefore, if $\RR \ne \O$, $\RR$ can not be both symmetric and asymmetric.

Let $\RR = \O$.

Then:

$\forall x, y \in S: \tuple {x, y} \notin \RR \land \tuple {y, x} \notin \RR$

$\tuple {p \land q} \lor \tuple {\neg p \land \neg q} \vdash p \iff q$

Thus:

$\forall x, y \in S: \tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$

So the condition for symmetry is fulfilled, and $\RR = \O$ is symmetric.

As $\RR = \O$:

$\forall x, y \in S: \tuple {x, y} \notin \RR$

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

$\forall x, y \in S: \tuple {x, y} \notin \RR$

therefore:

$\forall x, y \in S: \tuple {x, y} \in \RR \implies \forall x, y \in S: \tuple {y, x} \notin \RR$

This is the definition of asymmetric.

Thus the only relation $\RR$ to be both symmetric and asymmetric is $\RR = \O$.

$\blacksquare$

1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations