Intersection of Equivalences

Theorem
The intersection of two equivalence relations is itself an equivalence relation.

Proof
Let $$\mathcal{R}_1$$ and $$\mathcal{R}_2$$ be equivalences on $$S$$, and let $$\mathcal{R}_3 = \mathcal{R}_1 \cap \mathcal{R}_2$$.

1. $$\mathcal{R}_3$$ is reflexive:

2. $$\mathcal{R}_3$$ is symmetric:

3. $$\mathcal{R}_3$$ is transitive:

Thus $$\mathcal{R}_3$$ is an equivalence.