Q60 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
The statement pattern $(p \vee q) \rightarrow \sim r$ is logically equivalent to
- A$(\sim p \vee \sim q) \vee \sim r$
- B$(\sim p \wedge \sim q) \wedge \sim r$
- Correct answerC$(\sim p \wedge \sim q) \vee \sim r$
- D$(\sim p \vee \sim q) \wedge \sim r$
Solution
The given statement pattern is $(p \vee q) \rightarrow \sim r$.
Using the logical equivalence $A \rightarrow B \equiv \sim A \vee B$, the expression can be rewritten by taking $A = p \vee q$ and $B = \sim r$.
$(p \vee q) \rightarrow \sim r \equiv \sim(p \vee q) \vee \sim r$
Applying De Morgan's law, $\sim(p \vee q) \equiv \sim p \wedge \sim q$.
Substituting this into the expression yields:
$(\sim p \wedge \sim q) \vee \sim r$
Answer: $(\sim p \wedge \sim q) \vee \sim r$
Using the logical equivalence $A \rightarrow B \equiv \sim A \vee B$, the expression can be rewritten by taking $A = p \vee q$ and $B = \sim r$.
$(p \vee q) \rightarrow \sim r \equiv \sim(p \vee q) \vee \sim r$
Applying De Morgan's law, $\sim(p \vee q) \equiv \sim p \wedge \sim q$.
Substituting this into the expression yields:
$(\sim p \wedge \sim q) \vee \sim r$
Answer: $(\sim p \wedge \sim q) \vee \sim r$
Hence, the correct answer is option: 'C'.