Match the following : \(\begin{array}{clcl} & \textbf{List-I} &&…
2015
Match the following :
\(\begin{array}{clcl} & \textbf{List-I} && \textbf{List-II} \\ \text{(a)}&(p \rightarrow q) \Leftrightarrow (\neg q \rightarrow \neg p) & \text{(i)} & \text{Contrapositive} \\ \text{(b)}& [(p \wedge q) \rightarrow r] \Leftrightarrow [p \rightarrow (q \rightarrow r)]& \text{(ii)} & \text{Exportation law} \\ \text{(c)} & (p \rightarrow q) \Leftrightarrow [(p \wedge \neg q) \rightarrow o] & \text{(iii)} & \text{Reduction as absurdum} \\ \text{(d)} & (p \leftrightarrow q) \Leftrightarrow [(p \rightarrow q) \wedge (q \rightarrow p)]& \text{(iv)} & \text{Equivalence} \\ \end{array}\)
Codes :
- A.
(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
- B.
(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
- C.
(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
- D.
(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
Attempted by 71 students.
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Correct answer: A
Answer: (a) Contrapositive; (b) Exportation law; (c) Reduction as absurdum; (d) Equivalence.
(a) (p → q) ⇔ (¬q → ¬p): This is the contrapositive equivalence. Replacing an implication by its contrapositive preserves truth because p → q is false exactly when p is true and q is false, which is equivalent to ¬q being true and ¬p being false.
(b) [(p ∧ q) → r] ⇔ [p → (q → r)]: This is the exportation law. It moves a conjunctive antecedent into a nested implication: if p and q together imply r, then p implies that q implies r, and vice versa.
(c) (p → q) ⇔ [(p ∧ ¬q) → o]: This expresses reduction as absurdum (proof by contradiction). Here o denotes a contradiction or falsehood. If p implies q, then assuming p and ¬q leads to a contradiction; conversely, if p together with ¬q leads to contradiction, p must imply q.
(d) (p ↔ q) ⇔ [(p → q) ∧ (q → p)]: This is the definition of logical equivalence (biconditional): p if and only if q holds exactly when both directions of implication hold.