Let's analyze each part of the problem step by step, using truth tables to determine if the propositions are logically equivalent.

(a) ¬(p → ¬q) ⇔ p ∧ q

1. Construct truth table for each part:
   • p → ¬q: This is the conditional statement "if p, then not q."
   • ¬(p → ¬q): This is the negation of "p → ¬q."
   • p ∧ q: This is the conjunction of "p" and "q."

The truth table for $p, q, p \to \neg q, \neg (p \to \neg q), p \land q$:

p	q	p → ¬q	¬(p → ¬q)	p ∧ q
T	T	F	T	T
T	F	T	F	F
F	T	T	F	F
F	F	T	F	F

Analysis:

• For each truth value assignment to $p$ and $q$, the columns for $¬(p → ¬q)$ and $p ∧ q$ match exactly.
• Therefore, $\neg(p \to \neg q)$ is logically equivalent to $p \land q$.

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(b) ¬p ∧ q ⇔ ¬(p → ¬q)

1. Construct truth table for each part:
   • ¬p ∧ q: This is the conjunction of "not p" and "q."
   • ¬(p → ¬q): This is the negation of "p → ¬q" (from the previous part).

The truth table for $p, q, \neg p, \neg p \land q, p \to \neg q, \neg (p \to \neg q)$:

p	q	¬p	¬p ∧ q	p → ¬q	¬(p → ¬q)
T	T	F	F	F	T
T	F	F	F	T	F
F	T	T	T	T	F
F	F	T	F	T	F

Analysis:

• For each truth value assignment to $p$ and $q$, the columns for $¬p ∧ q$ and $¬(p → ¬q)$ match exactly.
• Therefore, $\neg p \land q$ is logically equivalent to $\neg(p \to \neg q)$.

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Final Conclusion:

• (a) $¬(p → ¬q) ⇔ p ∧ q$ is true.
• (b) $¬p ∧ q ⇔ ¬(p → ¬q)$ is true.

Both propositions are logically equivalent as shown by the truth tables.

Would you like further details or clarification?

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Here are 5 related questions:

1. How can logical equivalence be proven without a truth table?
2. What are common logical equivalences involving conditionals and negations?
3. Can De Morgan’s laws help in simplifying logical propositions?
4. How does logical implication ($\to$) relate to conjunction ($\land$) and disjunction ($\lor$)?
5. Why is negating a conditional statement tricky in logical reasoning?

Tip: When working with conditional statements, translating them into equivalent expressions can simplify complex logical propositions.