 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-resit/1/c/i/solution

Take every complement of a set relative to

U

A ∖ (B \cap A \cap C) = A \cap (B^{c} \cup A^{c} \cup C^{c}) = (A \cap B^{c}) \cup (A \cap A^{c}) \cup (A \cap C^{c}) = A \cap (B^{c} \cup C^{c}) = A ∖ (B \cap C) .

Here the middle term is the empty set, since a set and its complement of a set are disjoint.

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