Match List-I with List-II: \(\begin{array}{cccc} {} & \text{List-I} & {} &…
2019
Match List-I with List-II:
\(\begin{array}{cccc} {} & \text{List-I} & {} & \text{List-II} \\ (a) & p \rightarrow q & (i) & \rceil ( q \rightarrow \rceil p) \\ (b) & p \vee q & (ii) & p \wedge \rceil q \\ (c) & p \wedge q & (iii) & \rceil p \rightarrow q \\ (d) & \rceil ( p \rightarrow q) & (iv) & \rceil p \vee q \\ \end{array}\)
Choose the correct option from those given below:
- A.
\((a) – (ii); (b) – (iii); (c) – (i); (d) – (iv) \) - B.
\((a) – (ii); (b) – (i); (c) – (iii); (d) – (iv) \) - C.
\((a) – (iv); (b) – (i); (c) – (iii); (d) – (ii) \) - D.
\((a) – (iv); (b) – (iii); (c) – (i); (d) – (ii)\)
Attempted by 184 students.
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Correct answer: D
Key equivalences: use implication and De Morgan identities to transform each expression.
For p → q: p → q ≡ ¬p ∨ q, so it matches ¬p ∨ q.
For p ∨ q: ¬p → q ≡ p ∨ q because ¬(¬p) ∨ q = p ∨ q, so p ∨ q matches ¬p → q.
For p ∧ q: ¬(q → ¬p) ≡ p ∧ q because q → ¬p ≡ ¬q ∨ ¬p, and negating gives q ∧ p.
For ¬(p → q): ¬(p → q) ≡ p ∧ ¬q because p → q ≡ ¬p ∨ q and negating gives p ∧ ¬q.
Therefore the correct matching is: (a) p → q matches ¬p ∨ q; (b) p ∨ q matches ¬p → q; (c) p ∧ q matches ¬(q → ¬p); (d) ¬(p → q) matches p ∧ ¬q.