---
# System prepended metadata

title: CSC-208 Practice Problems 2 solution

---

# CSC-208 Practice Problems 2 solution

1. 
    _Proof_: Let $x \in (A \cap C) \cup (B \cap C)$

    Case 1: Let $x \in A \cap C$, 
    + $x \in A \cap C$ [assumption]
    + $x \in A \wedge x \in  C$ [def. of $\cap$]
    + $x \in A$ [def. of $\wedge$]
    + $x \in A \cup B$ [def. of $\cup$]


    Case 2: Let $x \in B - A$, 
    + $x \in B - A$ [assumption]
    + $x \in B \wedge x \notin A$ [def. of set complement]
    + $x \in B$ [def. of $\wedge$]
    + $x \in A \cup B$ [def. of $\cup$]

    Thus we complete the proof.

2. $R = \{(Amy, Kate), (Amy, Tom), (Tom, Kate)\}$
3. For the sake of contradiction, we assume that if m and n are integers and mn is odd, m is even or n is even. If both m and n are even, then their product must be even as well; if m is even and n is odd, their product would still be even; lastly, if m is odd and n is even, their product is also even. However, we have a contradiction because we claim that mn is odd. Thus, it must be the case that m is odd and n is odd. 

4. Surjective function



