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