 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/1/a/ii/solution

\overline{C}

(A \cap B) ∖ C (A \cup B) ∖ C = (A \cap B) \cap \overline{C} = (A \cap \overline{C}) \cap (B \cap \overline{C}), = (A \cup B) \cap \overline{C} = (A \cap \overline{C}) \cup (B \cap \overline{C}) .

In the first line the repeated factor

\overline{C}

changes nothing. Rewriting in terms of set differences yields

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

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