# Diagonal Relation is Many-to-One

## Theorem

That is:

$\forall x \in \Dom {\Delta_S}: \tuple {x, y_1} \in \Delta_S \land \tuple {x, y_2} \in \Delta_S \implies y_1 = y_2$

where $\Delta_S$ is the diagonal relation on a set $S$.

## Proof

Let $S$ be a set and let $\Delta_S$ be the diagonal relation on $S$.

Let $\tuple {x, y_1} \in \Delta_S \land \tuple {x, y_2} \in \Delta_S$.

From the definition of the diagonal relation:

$\tuple {x, y_1} = \tuple {x, x}$
$\tuple {x, y_2} = \tuple {x, x}$

and so $y_1 = y_2$.

$\blacksquare$