Equivalence Classes are Disjoint/Proof 1

Proof
First we show that:
 * $\tuple {x, y} \notin \mathcal R \implies \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} = \O$

Suppose two $\mathcal R$-classes are not disjoint:

Thus we have shown that $\eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O \implies \tuple {x, y} \in \mathcal R$.

Therefore, by the Rule of Transposition:


 * $\tuple {x, y} \notin \mathcal R \implies \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} = \O$

Now we show that:
 * $\eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} = \O \implies \tuple {x, y} \notin \mathcal R$

Suppose $\tuple {x, y} \in \mathcal R$.

Thus we have shown that:
 * $\tuple {x, y} \in \mathcal R \implies \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O$

Therefore, by the Rule of Transposition:
 * $\eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} = \O \implies \paren {x, y} \notin \mathcal R$

Using the rule of Biconditional Introduction on these results:
 * $\eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} = \O \iff \paren {x, y} \notin \mathcal R$

and the proof is finished.