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

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii
- iii c
  - Solution iii

i

Solution

Take every complement of a set relative to

U

. By set difference, De Morgan's laws, and distribution of set intersection over set union,

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.

ii

Solution

Use set difference, associativity of set intersection, and De Morgan's laws:

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

iii

Solution

The commutativity and associativity of set intersection give

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

Equivalently, part (ii) reduces both sides to removal of the same set union

B \cup C = C \cup B

.

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