Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 1 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

The set difference consists of elements belonging to the first set but not the second:

A ∖ B = {x \in U : x \in A and x \in / B} .

The symmetric difference consists of elements belonging to exactly one of the two sets:

A \oplus B = (A ∖ B) \cup (B ∖ A) .

In particular, the two sets in this union are disjoint sets.

ii

Solution

Every complement of a set is taken in the same universal set

U

. Apply De Morgan's laws for propositional logic to the membership propositions. For every

x \in U

,

x \in \overline{A \cap B} x \in \overline{A \cup B} ⟺ \neg (x \in A \land x \in B) ⟺ (x \in / A \lor x \in / B) ⟺ x \in \overline{A} \cup \overline{B}, ⟺ \neg (x \in A \lor x \in B) ⟺ (x \in / A \land x \in / B) ⟺ x \in \overline{A} \cap \overline{B} .

Equality of membership for every element proves the set identities

\overline{A \cap B} = \overline{A} \cup \overline{B}, \overline{A \cup B} = \overline{A} \cap \overline{B} .

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