MHT CET · Maths · Mathematical Reasoning
The negation of the statement pattern \(\sim \mathrm{S} \vee(\sim \mathrm{r} \wedge \mathrm{s})\) is equivalent to
- A \(\mathrm{s} \wedge \mathrm{r}\)
- B \(\mathrm{s} \wedge(\mathrm{r} \wedge \sim \mathrm{s})\)
- C \(\mathrm{s} \wedge \sim \mathrm{r}\)
- D \(\mathrm{S} \vee(\mathrm{r} \vee \sim \mathrm{s})\)
Answer & Solution
Correct Answer
(A) \(\mathrm{s} \wedge \mathrm{r}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \sim(\sim s \vee(\sim r \wedge s)) \\
& \equiv s \wedge \sim(\sim r \wedge s)...[De Morgan's law] \\
& \equiv s \wedge(r \vee \sim s)......[De Morgan's law] \\
& \equiv(s \wedge r) \vee(s \wedge \sim s)...[Distributive law] \\
& \equiv(s \wedge r) \vee F...[Complement law] \\
& \equiv s \wedge r...[Identity law]
\end{aligned}\)
& \sim(\sim s \vee(\sim r \wedge s)) \\
& \equiv s \wedge \sim(\sim r \wedge s)...[De Morgan's law] \\
& \equiv s \wedge(r \vee \sim s)......[De Morgan's law] \\
& \equiv(s \wedge r) \vee(s \wedge \sim s)...[Distributive law] \\
& \equiv(s \wedge r) \vee F...[Complement law] \\
& \equiv s \wedge r...[Identity law]
\end{aligned}\)
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