1. Verify whether the statement $\sim(\sim q \to p) \to \sim q$ is a tautology. - Recall the implication equivalence: $a \to b \equiv \sim a \lor b$. - Rewrite $\sim q \to p$ as $\sim (\sim q) \lor p = q \lor p$. - Then $\sim(\sim q \to p) = \sim(q \lor p) = \sim q \land \sim p$. - The whole statement becomes $(\sim q \land \sim p) \to \sim q$. - Using implication equivalence again: $\sim(\sim q \land \sim p) \lor \sim q = (q \lor p) \lor \sim q$. - Simplify: $q \lor p \lor \sim q$. - Since $q \lor \sim q$ is always true, the entire expression is always true. **Conclusion:** The statement is a tautology. 2. Verify the validity of the argument: - Premises: 1. If a number is even, then it is divisible by 2: $E \to D$. 2. The number is not divisible by 2: $\sim D$. - Conclusion: The number is not even: $\sim E$. - This is a classic modus tollens argument: From $E \to D$ and $\sim D$, infer $\sim E$. **Conclusion:** The argument is valid. 3. College software usage problem: - Total students: 120 - None used: 25 - Only P: 14 - Only Q: 22 - Only R: 11 - Both P and Q: 18 - All three: 7 - Twice as many used Q as P. Let $x$ = number of students who used P, $y$ = number who used Q. (i) Write simultaneous equations: - From "twice as many used Q as P": $$y = 2x$$ - Total students using packages: $$120 - 25 = 95$$ - Sum of users: Only P + Only Q + Only R + Both P and Q + All three + others involving P, Q, R = 95 - Using given data: $$14 + 22 + 11 + 18 + 7 + \text{others} = 95$$ - Calculate others: $$14 + 22 + 11 + 18 + 7 = 72$$ - So others = $95 - 72 = 23$ - Since others involve P and Q usage, set up equations for $x$ and $y$ including these values. (ii) Solve for $x$ and $y$: - Using $y = 2x$ and total users, solve accordingly. 4. Verify by membership table that: $$B - A = A^c \cap B$$ - Recall: - $B - A$ means elements in $B$ but not in $A$. - $A^c$ is the complement of $A$. - Intersection $A^c \cap B$ means elements in both $A^c$ and $B$. - Construct membership table for $A$ and $B$ with all combinations. - Show that membership values for $B - A$ and $A^c \cap B$ match for all cases. **Conclusion:** The equality holds.