Which of the following pairs of propositions are not logically equivalent?
2020
Which of the following pairs of propositions are not logically equivalent?
Answer: D. \(((p \wedge q) \rightarrow r)\) and \(((p \rightarrow r) \wedge (q \rightarrow r))\) — Summary: The pair "(p ∧ q) → r" and "(p → r) ∧ (q → r)" are not logically equivalent; the other three pairs are equivalent. Pair: (p → r) ∧ (q → r) and (p ∨…
- A.
\(((p \rightarrow r) \wedge (q \rightarrow r))\)and\(((p \vee q) \rightarrow r)\) - B.
\(p \leftrightarrow q\)and\((\neg p \leftrightarrow \neg q)\) - C.
\(((p \wedge q) \vee (\neg p \wedge \neg q))\)and\(p \leftrightarrow q\) - D.
\(((p \wedge q) \rightarrow r)\)and\(((p \rightarrow r) \wedge (q \rightarrow r))\)
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Correct answer: D
Summary: The pair "(p ∧ q) → r" and "(p → r) ∧ (q → r)" are not logically equivalent; the other three pairs are equivalent.
Pair: (p → r) ∧ (q → r) and (p ∨ q) → r: They are equivalent because (p → r) ∧ (q → r) = (¬p ∨ r) ∧ (¬q ∨ r) = (¬p ∧ ¬q) ∨ r = ¬(p ∨ q) ∨ r = (p ∨ q) → r.
Pair: p ↔ q and ¬p ↔ ¬q: They are equivalent because negating both operands does not change whether the two operands have the same truth value; both biconditionals have identical truth tables.
Pair: (p ∧ q) ∨ (¬p ∧ ¬q) and p ↔ q: They are equivalent because the left expression is the disjunctive normal form of the biconditional; it is true exactly when p and q have the same truth value.
Pair: (p ∧ q) → r and (p → r) ∧ (q → r): These are not equivalent.
Counterexample: Let p = true, q = false, r = false. Then p ∧ q is false, so (p ∧ q) → r is true. But p → r is false (true → false is false) and q → r is true (false → false is true), so (p → r) ∧ (q → r) is false. The formulas differ under this assignment.
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