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