Equivalence Classes are Disjoint/Proof 1

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

Suppose two $\RR$-classes are not disjoint:

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

Therefore, by the Rule of Transposition:


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

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

Suppose $\tuple {x, y} \in \RR$.

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

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

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

and the proof is complete.