MHT CET · Maths · Mathematical Reasoning
The negation of contrapositive of the statement \(\mathrm{p} \rightarrow(\sim \mathrm{q} \wedge \mathrm{r})\) is
- A \((\sim q \vee \sim r) \wedge \sim p\)
- B \((q \vee \sim r) \wedge p\)
- C \((q \wedge \sim r) \vee p\)
- D \(\quad(\sim q \wedge \sim r) \vee \sim p\)
Answer & Solution
Correct Answer
(B) \((q \vee \sim r) \wedge p\)
Step-by-step Solution
Detailed explanation
Contrapositive of the statement \(\mathrm{p} \rightarrow(\sim \mathrm{q} \wedge \mathrm{r})\) is
\(\begin{aligned}
& \sim(\sim q \wedge r) \rightarrow \sim p \\
& \equiv(q \vee \sim r) \rightarrow \sim p
\end{aligned}\)
...[De Morgan's law]
Negation of contrapositive of \(p \rightarrow(\sim q \wedge r)\) is
\(\begin{aligned}
& \sim[(q \vee \sim r) \rightarrow \sim p] \\
& \equiv(q \vee \sim r) \wedge \sim(\sim p) \quad \ldots[\because \sim(p \rightarrow q) \equiv p \wedge \sim q] \\
& \equiv(q \vee \sim r) \wedge p
\end{aligned}\)
\(\begin{aligned}
& \sim(\sim q \wedge r) \rightarrow \sim p \\
& \equiv(q \vee \sim r) \rightarrow \sim p
\end{aligned}\)
...[De Morgan's law]
Negation of contrapositive of \(p \rightarrow(\sim q \wedge r)\) is
\(\begin{aligned}
& \sim[(q \vee \sim r) \rightarrow \sim p] \\
& \equiv(q \vee \sim r) \wedge \sim(\sim p) \quad \ldots[\because \sim(p \rightarrow q) \equiv p \wedge \sim q] \\
& \equiv(q \vee \sim r) \wedge p
\end{aligned}\)
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