# COT 3100 - Quiz 1
*Michael Akinyemi*
*Tue/Thu - 9 AM*
### Question 1: Inference (10)
f = fish can fly
b = birds can swim
s = squirrels can talk
c = capybaras can write
Accepting the following premises:
1. $f \implies b$
- “If fish can fly, birds can swim”
2. $b \implies s$
- “If birds can swim, squirrels can talk”
3. $(b \land s) \implies c$
- “If birds can swim and squirrels can talk, capybaras can write”
4. $f$
- “Fish can fly”
Show, using the rules of inference, that capybaras can write
$ \begin{aligned}
\\ \text{5. } & b && \text{Modus Ponens 1, 4}
\\ \text{6. } & s && \text{Modus Ponens 2, 5}
\\ \text{7. } & s \land b && \text{Conjunction 5, 6}
\\ \text{8. } & c && \text{Modus Ponens 3, 7}
\end{aligned}$
### Question 2: Universal Generalization (10)
Show by universal generalization that:
$((A \cap C) \cup B) \cap D \subseteq (A \cap C) \cup (B \cap D)$
**Proof:**
By definition of distributivity $((A \cap C) \cap D) \cup (B \cap D) \iff ((A \cap C) \cup B) \cap D$.
Let some arbitrary $x \in (A \cap C) \cap D$, by definition of $\cap$, $x \in (A \cap C)$ AND $x \in D$. By definition of $\subseteq$, we have $(A \cap C) \cap D \subseteq A \cap C$. Thus, $x \in (A \cap C)$.
By definition of $\cup$, if $(A \cap C) \cup (B \cap D)$ and $(A \cap C) \subseteq (A \cap C) \cup (B \cap D)$
$\therefore ((A \cap C) \cup B) \cap D \subseteq (A \cap C) \cup (B \cap D)$