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 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.

Also see

 * Fundamental Theorem on Equivalence Relations


 * Union of Equivalence Classes is Whole Set
 * Equivalence Class is Not Empty