 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/1/d/i/solution

For every set

S

S \in P (A \cap B) ⟺ S \subseteq A \cap B ⟺ (S \subseteq A and S \subseteq B) ⟺ S \in P (A) \cap P (B) .

Equality by identical membership therefore proves

P (A \cap B) = P (A) \cap P (B) .

The essential observation is that being a subset of a set intersection means being a subset of both factors.

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