Inverse of Diagonal Relation

Theorem

Let $S$ be a set.

Let $\Delta_S$ denote the diagonal relation on $S$.

Let ${\Delta_S}^{-1}$ denote the inverse of $\Delta_S$.

Then:

${\Delta_S}^{-1} = \Delta_S$

Proof

By definition of diagonal relation:

$\Delta_S = \set {\tuple {x, x} \in S \times S: x \in S}$

By definition of inverse relation:

${\Delta_S}^{-1} = \set {\tuple {x, x} \in S \times S: x \in S}$

Hence it follows that:

$\tuple {x, x} \in \Delta_S \iff \tuple {x, x} \in {\Delta_S}^{-1}$

$\blacksquare$