Equivalence Classes are Disjoint/Proof 2

Theorem
Let $\mathcal R$ be an equivalence relation on a set $S$.

Then all $\mathcal R$-classes are pairwise disjoint:


 * $\left({x, y}\right) \notin \mathcal R \iff \left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R = \varnothing$

Proof
Suppose that for $x, y \in S$:
 * $\left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R \ne \varnothing$

Let:
 * $z \in \left[\!\left[{x}\right]\!\right]_\mathcal R$
 * $z \in \left[\!\left[{y}\right]\!\right]_\mathcal R$

Then by definition of equivalence class:
 * $\left({x, z}\right) \in \mathcal R$
 * $\left({y, z}\right) \in \mathcal R$

Let $c \in \left[\!\left[{x}\right]\!\right]_\mathcal R$.

That is:
 * $\left({x, c}\right) \in \mathcal R$

By definition of equivalence relation, $\mathcal R$ is symmetric so:
 * $\left({z, x}\right) \in \mathcal R$

By definition of equivalence relation, $\mathcal R$ is transitive so:
 * $\left({z, x}\right) \in \mathcal R \land \left({x, c}\right) \in \mathcal R \implies \left({z, c}\right) \in \mathcal R$

and
 * $\left({y, z}\right) \in \mathcal R \land \left({z, c}\right) \in \mathcal R \implies \left({y, c}\right) \in \mathcal R$

So we have $c \in \left[\!\left[{y}\right]\!\right]_\mathcal R$.

By definition of subset:


 * $\left[\!\left[{x}\right]\!\right]_\mathcal R \subseteq \left[\!\left[{y}\right]\!\right]_\mathcal R$

Similarly, let $c \in \left[\!\left[{y}\right]\!\right]_\mathcal R$.

That is:
 * $\left({y, c}\right) \in \mathcal R$

By definition of equivalence relation, $\mathcal R$ is symmetric so:
 * $\left({z, y}\right) \in \mathcal R$

By definition of equivalence relation, $\mathcal R$ is transitive so:
 * $\left({z, y}\right) \in \mathcal R \land \left({y, c}\right) \in \mathcal R \implies \left({z, c}\right) \in \mathcal R$

and
 * $\left({x, z}\right) \in \mathcal R \land \left({z, c}\right) \in \mathcal R \implies \left({x, c}\right) \in \mathcal R$

So we have $c \in \left[\!\left[{x}\right]\!\right]_\mathcal R$.

By definition of subset:


 * $\left[\!\left[{y}\right]\!\right]_\mathcal R \subseteq \left[\!\left[{x}\right]\!\right]_\mathcal R$

That is:
 * $\left[\!\left[{x}\right]\!\right]_\mathcal R \subseteq \left[\!\left[{y}\right]\!\right]_\mathcal R$

and
 * $\left[\!\left[{y}\right]\!\right]_\mathcal R \subseteq \left[\!\left[{x}\right]\!\right]_\mathcal R$

By definition of set equality:


 * $\left[\!\left[{x}\right]\!\right]_\mathcal R = \left[\!\left[{y}\right]\!\right]_\mathcal R$

Thus: $\left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R \ne \varnothing \implies \left[\!\left[{x}\right]\!\right]_\mathcal R = \left[\!\left[{y}\right]\!\right]_\mathcal R$

and the result follows.