1. Problem (a): Find the power set of $A = \{a,b,c,d,e\}$. The power set is the set of all subsets of $A$. Since $|A|=5$, the power set has $2^5=32$ subsets. These include the empty set, all singletons, pairs, triples, quadruples, and the set itself. 2. Problem (b): Test validity of the argument: Premises: - A student is hardworking or lucky: $H \lor L$. - If hardworking, then passes: $H \to P$. - If lucky, then passes: $L \to P$. Conclusion: Every student passes: $P$. To check validity, use logical inference: From $H \lor L$ and $H \to P$, $L \to P$, we conclude $P$ must hold regardless, so the argument is valid. 3. Problem (c): Soda preference data from 800 consumers: - Coke ($C$): 230 - Fanta ($F$): 245 - Sprite ($S$): 325 - All three ($C \cap F \cap S$): 30 - Coke and Sprite ($C \cap S$): 70 - Coke only: 110 - Sprite only: 185 (i) Venn diagram: three overlapping circles labeled $C, F, S$ with intersections: - Center (all three): 30 - $C$ only: 110 - $S$ only: 185 - $C \cap S$ total is 70, so $C \cap S$ only is $70 - 30 = 40$ (ii) Find Fanta only: From $|F|=245$, subtract those who took $F$ with others. Let $x$ be $F$ only. Other Fanta intersections: - All three: 30 - $C \cap F$ only: to find - $F \cap S$ only: to find (iii) & (iv) We'll find $C \cap F$ only and $F \cap S$ only. Start by determining known quantities: Total $C=230$ consists of $C$ only (110) + $C \cap S$ only (40) + $C \cap F$ only + all three (30). Thus: $$230 = 110 + 40 + (C \cap F \text{ only}) + 30$$ $$C \cap F \text{ only} = 230 - 110 - 40 - 30 = 50$$ Similarly, total $S=325$ is $S$ only (185) + $C \cap S$ only (40) + $F \cap S$ only + all three (30). So: $$325 = 185 + 40 + (F \cap S \text{ only}) + 30$$ $$F \cap S \text{ only} = 325 - 185 - 40 - 30 = 70$$ Now find $F$ only: Total $F=245 = F$ only + $C \cap F$ only (50) + $F \cap S$ only (70) + all three (30) $$F \text{ only} = 245 - 50 - 70 - 30 = 95$$ (v) None of the brands: Add all only and overlapping counts: $$110 (C) + 95 (F) + 185 (S) + 50 (C \cap F) + 40 (C \cap S) + 70 (F \cap S) + 30 (all three) = 580$$ Consumers who took none: $$800 - 580 = 220$$ 4. Problem (d) Tautologies: (i) Show $(p \land q) \to (p \lor q)$ is a tautology. Since if $p$ and $q$ are true, then certainly $p$ or $q$ is true. Logical implication holds always true. (ii) Show $\neg p \land (p \lor q) \to q$ is tautology. If $\neg p$ is true, then $p$ is false. So $p \lor q$ simplifies to $q$, so the implication is $q \to q$ which is always true. (iii) Show $[(p \to q) \land (q \to r)] \to (p \to r)$ is tautology. This is transitivity of implication. If $p$ implies $q$ and $q$ implies $r$, then $p$ implies $r$. 5. Problem (e): Cartesian product $A \times B$ with $A=\{z,a\}$ and $B=\{0,5,4\}$: $$A \times B = \{(z,0),(z,5),(z,4),(a,0),(a,5),(a,4)\}$$ 6. Problem (f): Use truth table to check equivalence of $p \to q$ and $\neg p \lor q$. Both have same truth values in all cases, so they are logically equivalent. 7. Problem (g): Show $p \lor (q \land r)$ is equivalent to $(p \lor q) \land (p \lor r)$. This is distributive law in logic. They have same truth table values, so they are equivalent. Final answers summarized above.