Inverse of Diagonal Relation
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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:
- $\tuple {x, x} \in \Delta_S \iff \tuple {x, x} \in {\Delta_S}^{-1}$
$\blacksquare$