 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/b/i/solution

A \oplus B = (A ∖ B) \cup (B ∖ A)

A \oplus \emptyset = (A ∖ \emptyset) \cup (\emptyset ∖ A) = A \cup \emptyset = A,

and

A \oplus A = (A ∖ A) \cup (A ∖ A) = \emptyset \cup \emptyset = \emptyset .

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