Proof if $\mathcal{R}$ is an equivalence relation, either $E_a \cap E_b = \varnothing$ or $E_a= E_b$
The proof given looks like:
- $E_a \cap E_b = \varnothing$ (Premise)
- $\exists x (x \in (E_a \cap E_b))$ (intersection not empty)
- $\exists (x \in E_a \wedge x \in E_b)$ (dfn. intersection)
- $a \mathcal{R} x \wedge b \mathcal{R} x$ (dfn equivalence class)
- $x \mathcal{R} b$ ($\mathcal{R}$ is symmetric)
- $a \mathcal{R} b$ (by 4, 5, $\mathcal{R}$ symmetric) --- Why?
- $E_a = E_b$
Can someone explain how I get to step 6?