MHT CET · Maths · Mathematical Reasoning
The statement pattern \(\mathrm{p} \rightarrow \sim(\mathrm{p} \wedge \sim \mathrm{q})\) is equivalent to
- A \(\mathrm{q}\)
- B \((\sim p) \vee q\)
- C \((\sim p) \wedge q\)
- D \((\sim p) \vee(\sim q)\)
Answer & Solution
Correct Answer
(B) \((\sim p) \vee q\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& p \rightarrow \sim(p \wedge \sim q) \\
& \equiv \sim p \vee \sim(p \wedge \sim q) \\
& \equiv \sim p \vee(\sim p \vee q) \\
& \equiv(\sim p \vee \sim p) \vee q \\
& \equiv \sim p \vee q
\end{aligned}
\)
\(\ldots[\because \mathrm{p} \rightarrow \mathrm{q} \equiv \sim \mathrm{p} \vee \mathrm{q}]\)
...[De Morgan's law]
...[Associative law]
...[Idempotent law]
\begin{aligned}
& p \rightarrow \sim(p \wedge \sim q) \\
& \equiv \sim p \vee \sim(p \wedge \sim q) \\
& \equiv \sim p \vee(\sim p \vee q) \\
& \equiv(\sim p \vee \sim p) \vee q \\
& \equiv \sim p \vee q
\end{aligned}
\)
\(\ldots[\because \mathrm{p} \rightarrow \mathrm{q} \equiv \sim \mathrm{p} \vee \mathrm{q}]\)
...[De Morgan's law]
...[Associative law]
...[Idempotent law]
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