 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability/1/a/ii/solution

x \in (A \cap B) ∖ C

. Membership in the intersection gives

x \in A

and

x \in B

, while the set difference gives

x \in / C

. Consequently

x \in (A ∖ C) \cap (B ∖ C)

For the reverse set inclusion, an element of

(A ∖ C) \cap (B ∖ C)

belongs to both

A

and

B

and does not belong to

C

. It therefore belongs to

(A \cap B) ∖ C

(A \cap B) ∖ C = (A ∖ C) \cap (B ∖ C) .

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