Equivalence Classes are Disjoint

Theorem
Let $$\mathcal R$$ be an equivalence relation on a set $$S$$.

Then all $\mathcal R$-classes are pairwise disjoint:


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

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 Material Equivalence 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.