Q62 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
The contrapositive of the statement pattern $[p \vee (p \to q)] \to (p \wedge \sim q)$ is
- A$(p \wedge \sim q) \to [p \wedge (p \to q)]$
- B$(\sim p \wedge \sim q) \to [\sim p \wedge (p \to \sim q)]$
- C$(\sim p \vee q) \wedge [\sim p \vee (p \wedge \sim q)]$
- Correct answerD$(\sim p \vee q) \to [\sim p \wedge (p \wedge \sim q)]$
Solution
Let the given statement be $X \to Y$, where $X = p \vee (p \to q)$ and $Y = p \wedge \sim q$.
The contrapositive of $X \to Y$ is $\sim Y \to \sim X$.
First, we find $\sim Y$:
$\sim Y = \sim (p \wedge \sim q) = \sim p \vee q$
Next, we find $\sim X$:
$\sim X = \sim [p \vee (p \to q)]$
Using De Morgan's Law, we get:
$\sim X = \sim p \wedge \sim (p \to q)$
Since $p \to q \equiv \sim p \vee q$, its negation is $\sim (p \to q) \equiv p \wedge \sim q$.
Substituting this back, we get:
$\sim X = \sim p \wedge (p \wedge \sim q)$
Therefore, the contrapositive $\sim Y \to \sim X$ is:
$(\sim p \vee q) \to [\sim p \wedge (p \wedge \sim q)]$
The contrapositive of $X \to Y$ is $\sim Y \to \sim X$.
First, we find $\sim Y$:
$\sim Y = \sim (p \wedge \sim q) = \sim p \vee q$
Next, we find $\sim X$:
$\sim X = \sim [p \vee (p \to q)]$
Using De Morgan's Law, we get:
$\sim X = \sim p \wedge \sim (p \to q)$
Since $p \to q \equiv \sim p \vee q$, its negation is $\sim (p \to q) \equiv p \wedge \sim q$.
Substituting this back, we get:
$\sim X = \sim p \wedge (p \wedge \sim q)$
Therefore, the contrapositive $\sim Y \to \sim X$ is:
$(\sim p \vee q) \to [\sim p \wedge (p \wedge \sim q)]$
Hence, the correct answer is option: 'D'.