 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/i/solution

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

The complements turn the inner union into an intersection and vice versa. Thus

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

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