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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:

  1. $E_a \cap E_b = \varnothing$ (Premise)
  2. $\exists x (x \in (E_a \cap E_b))$ (intersection not empty)
  3. $\exists (x \in E_a \wedge x \in E_b)$ (dfn. intersection)
  4. $a \mathcal{R} x \wedge b \mathcal{R} x$ (dfn equivalence class)
  5. $x \mathcal{R} b$ ($\mathcal{R}$ is symmetric)
  6. $a \mathcal{R} b$ (by 4, 5, $\mathcal{R}$ symmetric) --- Why?
  7. $E_a = E_b$

Can someone explain how I get to step 6?

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    I think you are using the transitivity of $\mathcal{R}$ in point 6.2012-10-11

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From (4) you have $a\mathcal{R}x$ and $b\mathcal{R}x$. Applying symmetry to $b\mathcal{R}x$ gives you $x\mathcal{R}b$; that was (5). Now use transitivity, not symmetry: from $a\mathcal{R}x$ and $x\mathcal{R}b$ you get $a\mathcal{R}b$ by transitivity of $\mathcal{R}$.