# Preimage of Singleton

## Theorem

Let $\mathcal R$ be a relation.

Let $\mathcal R^{-1} \left({t}\right)$ denote the preimage of $t$ under $\mathcal R$.

Let $\mathcal R^{-1} \left({ \left\{{t}\right\} }\right)$ denote the preimage of $\left\{{t}\right\}$ under $\mathcal R$.

Then:

$\mathcal R^{-1} \left({ \left\{{t}\right\} }\right) = \mathcal R^{-1} \left({t}\right)$

## Proof

 $\displaystyle \mathcal R^{-1} \left({ t }\right)$ $=$ $\displaystyle \left\{ {s : \left({ s, t }\right) \in \mathcal R}\right\}$ by definition of preimage of element $\displaystyle$ $=$ $\displaystyle \left\{ {s : \exists y: \left({ y = t \land \left({ s, y }\right) \in \mathcal R }\right)}\right\}$ by Equality implies Substitution $\displaystyle$ $=$ $\displaystyle \left\{ {s : \exists y \in \left\{ t \right\}: \left({ s, y }\right) \in \mathcal R}\right\}$ by definition of singleton $\displaystyle$ $=$ $\displaystyle \mathcal R^{-1} \left({ \left\{ {t}\right\} }\right)$ by definition of preimage of subset

$\blacksquare$