 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-resit/1/a/ii/solution

Using set difference as intersection with a complement of a set, then De Morgan's laws and distribution, gives

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

For the third equality, expand the two set unions: the terms

A \cap C^{c}

and

A \cap B

are absorbed by

A

, and

A \cap A = A

. This proves the required identity entirely by set algebra.

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