Quotient Set Determined by Relation Induced by Partition is That Partition

Theorem
Let $S$ be a set.

Let $\mathcal P$ be a partition of $S$.

Let $\mathcal R$ be the relation induced by $\mathcal P$.

Then the quotient set $S / \mathcal R$ of $S$ is $\mathcal P$ itself.

Proof
Let $P \subseteq S$ such that $P \in \mathcal P$.

Let $x \in P$.

Then:

Therefore:
 * $P = \left[\!\left[{x}\right]\!\right]_\mathcal R$

and so:
 * $P \in S / \mathcal R$

and so:
 * $\mathcal P \subseteq S / \mathcal R$

Now let $x \in S$.

As $\mathcal P$ is a partition:
 * $\exists P \in \mathcal P: x \in P$

Then by definition of $\mathcal R$:
 * $\left({x, y}\right) \in \mathcal R \iff y \in \left[\!\left[{x}\right]\!\right]_\mathcal R$

Therefore:
 * $\left[\!\left[{x}\right]\!\right]_\mathcal R = P$

and so:
 * $\left[\!\left[{x}\right]\!\right]_\mathcal R \in \mathcal P$

That is:
 * $\mathcal S / \mathcal R \subseteq P$

It follows by definition of set equality that:
 * $\mathcal S / \mathcal R = P$

Hence the result.