 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/discrete-mathematics-continuous-mathematics-and-probability/1/b/i/solution

x \in A ∖ (A ∖ B)

. Then

x \in A

but

x \in / A ∖ B

. Since membership in

A

is already known, failure to lie in that relative complement forces

x \in B

. Hence

x \in A \cap B

Conversely, if

x \in A \cap B

, then

x \in A

and its membership in

B

excludes it from

A ∖ B

. Therefore

x \in A ∖ (A ∖ B)

. The two inclusions prove the stated equality with the set intersection.

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