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MHT CET · Maths · Mathematical Reasoning

Which of the following statement pattern is a contradiction?
\(\mathrm{S}_{1} \equiv(\mathrm{p} \rightarrow \mathrm{q}) \wedge(\mathrm{p} \wedge \sim \mathrm{q}) \ \mathrm{S}_{2} \equiv[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})] \rightarrow \mathrm{q} \ \mathrm{S}_{3} \equiv\) \((\mathrm{p} \vee \mathrm{q}) \rightarrow \sim \mathrm{p} \ \mathrm{S}_{4} \equiv[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})] \leftrightarrow \mathrm{q}\)

  1. A \(\mathrm{S}_{4}\)
  2. B \(\mathrm{~S}_{1}\)
  3. C \(\mathrm{~S}_{2}\)
  4. D \(\mathrm{~S}_{3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\mathrm{~S}_{1}\)

Step-by-step Solution

Detailed explanation

Image
All entries in last column are \(\mathrm{F}\). So \(\mathrm{S}_{1}\) is a contradiction.
This problem can be alternatively solved as follows :
\(\begin{aligned}
S_{1} & \equiv(p \rightarrow q) \wedge(p \wedge \sim q) \\
& \equiv[(\sim p \vee q)] \wedge[\sim(\sim p \vee q)] \\
& \equiv F
\end{aligned}\)