 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/1/d/ii/solution

For every

x \in A

, membership in the relevant equivalence class satisfies

x \in [a]_{R \cap S} ⟺ a (R \cap S) x ⟺ (a R x and a S x) ⟺ x \in [a]_{R} \cap [a]_{S} .

Equality follows by axiom of extensionality. Thus $[a]_{R \cap S} = [a]_{R} \cap [a]_{S}$ .

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