 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/1/a/solution

x \in (A \cap B) ∖ C

x \in A

x \in B

and

x \in / C

. Consequently

x \in A ∖ C

and

x \in B ∖ C

, so

x \in (A ∖ C) \cap (B ∖ C)

For the reverse set inclusion, take

x \in (A ∖ C) \cap (B ∖ C)

. Membership in the first set difference gives

x \in A

and

x \in / C

, while membership in the second gives

x \in B

and

x \in / C

. Thus

x \in A \cap B

and

x \in / C

, proving

x \in (A \cap B) ∖ C

. The two inclusions establish the required equality.

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