Equivalence Classes are Disjoint/Proof 1

Proof
First we show that:
 * $\left({x, y}\right) \notin \mathcal R \implies \left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R = \varnothing$

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

Thus we have shown that $\left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R \ne \varnothing \implies \left({x, y}\right) \in \mathcal R$.

Therefore, by the Rule of Transposition:


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

Now we show that:
 * $\left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R = \varnothing \implies \left({x, y}\right) \notin \mathcal R$

Suppose $\left({x, y}\right) \in \mathcal R$.

Thus we have shown that:
 * $\left({x, y}\right) \in \mathcal R \implies \left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R \ne \varnothing$

Therefore, by the Rule of Transposition:
 * $\left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R = \varnothing \implies \left({x, y}\right) \notin \mathcal R$

Using the rule of Biconditional Introduction on these results:
 * $\left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R = \varnothing \iff \left({x, y}\right) \notin \mathcal R$

and the proof is finished.