Equivalence Classes are Disjoint/Proof 2

Proof
Suppose that for $x, y \in S$:
 * $\eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O$

Let:
 * $z \in \eqclass x {\mathcal R}$
 * $z \in \eqclass y {\mathcal R}$

Then by definition of equivalence class:
 * $\tuple {x, z} \in \mathcal R$
 * $\tuple {y, z} \in \mathcal R$

Let $c \in \eqclass x {\mathcal R}$.

That is:
 * $\tuple {x, c} \in \mathcal R$

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

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

and
 * $\tuple {y, z} \in \mathcal R \land \tuple {z, c} \in \mathcal R \implies \tuple {y, c} \in \mathcal R$

So we have $c \in \eqclass y {\mathcal R}$.

By definition of subset:


 * $\eqclass x {\mathcal R} \subseteq \eqclass y {\mathcal R}$

Similarly, let $c \in \eqclass y {\mathcal R}$.

That is:
 * $\tuple {y, c} \in \mathcal R$

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

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

and
 * $\tuple {x, z} \in \mathcal R \land \tuple {z, c} \in \mathcal R \implies \tuple {x, c} \in \mathcal R$

So we have $c \in \eqclass x {\mathcal R}$.

By definition of subset:


 * $\eqclass y {\mathcal R} \subseteq \eqclass x {\mathcal R}$

That is:
 * $\eqclass x {\mathcal R} \subseteq \eqclass y {\mathcal R}$

and
 * $\eqclass y {\mathcal R} \subseteq \eqclass x {\mathcal R}$

By definition of set equality:


 * $\eqclass x {\mathcal R} = \eqclass y {\mathcal R}$

Thus:
 * $\eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O \implies \eqclass x {\mathcal R} = \eqclass y {\mathcal R}$

and the result follows.