Equivalence Classes are Disjoint/Proof 2

Proof
Suppose that for $x, y \in S$:
 * $\eqclass x \RR \cap \eqclass y \RR \ne \O$

Let:
 * $z \in \eqclass x \RR$
 * $z \in \eqclass y \RR$

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

Let $c \in \eqclass x \RR$.

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

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

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

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

So we have $c \in \eqclass y \RR$.

By definition of subset:


 * $\eqclass x \RR \subseteq \eqclass y \RR$

Similarly, let $c \in \eqclass y \RR$.

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

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

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

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

So we have $c \in \eqclass x \RR$.

By definition of subset:


 * $\eqclass y \RR \subseteq \eqclass x \RR$

That is:
 * $\eqclass x \RR \subseteq \eqclass y \RR$

and
 * $\eqclass y \RR \subseteq \eqclass x \RR$

By definition of set equality:


 * $\eqclass x \RR = \eqclass y \RR$

Thus:
 * $\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \eqclass x \RR = \eqclass y \RR$

and the result follows.