Equivalence Classes are Disjoint

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

Then all $\mathcal{R}$-classes are mutually 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.