MHT CET · Maths · Mathematical Reasoning
The negation of the statement pattern \(\sim \mathrm{p} \vee(\mathrm{q} \rightarrow \sim \mathrm{r})\) is
- A \(\mathrm{p} \rightarrow(\mathrm{q} \wedge \sim \mathrm{r})\)
- B \(\mathrm{p} \vee(\mathrm{q} \wedge \mathrm{r})\)
- C \(\mathrm{p} \wedge(\mathrm{q} \wedge \mathrm{r})\)
- D \(\sim \mathrm{p} \wedge(\mathrm{q} \wedge \mathrm{r})\)
Answer & Solution
Correct Answer
(C) \(\mathrm{p} \wedge(\mathrm{q} \wedge \mathrm{r})\)
Step-by-step Solution
Detailed explanation
(A)
\(\begin{array}{l}
\sim[\sim p \vee(q \rightarrow \sim r)] \\
\equiv p \wedge \sim(q \rightarrow \sim r) \\
\equiv p \wedge \sim(\sim q \vee \sim r) \\
\equiv p \wedge \sim[\sim(q \wedge r)] \\
\equiv p \wedge(q \wedge r)
\end{array}\)
\(\begin{array}{l}
\sim[\sim p \vee(q \rightarrow \sim r)] \\
\equiv p \wedge \sim(q \rightarrow \sim r) \\
\equiv p \wedge \sim(\sim q \vee \sim r) \\
\equiv p \wedge \sim[\sim(q \wedge r)] \\
\equiv p \wedge(q \wedge r)
\end{array}\)
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