 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/1/b/ii/solution

x \in \overline{A \oplus B}

. If

x \in A

, then

x \in B

, since otherwise

x \in A ∖ B \subseteq A \oplus B

. Hence in this case

x \in A \cap B

. If

x \in / A

, then

x \in / B

, since otherwise

x \in B ∖ A \subseteq A \oplus B

. Hence in this case

x \in \overline{A \cup B}

Conversely, an element of

A \cap B

belongs to neither

A ∖ B

nor

B ∖ A

. An element of

\overline{A \cup B}

also belongs to neither set difference. Thus every element of

(A \cap B) \cup \overline{A \cup B}

lies outside

A \oplus B

, proving the reverse set inclusion. Therefore

\overline{A \oplus B} = (A \cap B) \cup \overline{A \cup B} .

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