MHT CET · Maths · Mathematical Reasoning
Given p: A man is a judge, q: A man is honest If \(\mathrm{S} 1\) : If a man is a judge, then he is honest S2: If a man is a judge, then he is not honest S3: A man is not a judge or he is honest S4: A man is a judge and he is honest Then
- A \(\mathrm{S}_2 \equiv \mathrm{S}_3\)
- B \(\mathrm{S}_1 \equiv \mathrm{S}_2\)
- C \(\mathrm{S}_2 \equiv \mathrm{S}_4\)
- D \(\mathrm{S}_1 \equiv \mathrm{S}_3\)
Answer & Solution
Correct Answer
(D) \(\mathrm{S}_1 \equiv \mathrm{S}_3\)
Step-by-step Solution
Detailed explanation
We will write logical form of given statements
\(
\begin{array}{ll}
\mathrm{S}_1=\mathrm{p} \rightarrow \mathrm{q} & \mathrm{S}_2=\mathrm{p} \rightarrow-\mathrm{q} \\
\mathrm{S}_3=\sim \mathrm{p} \vee \mathrm{q} & \mathrm{S}_4=\mathrm{p} \wedge \mathrm{q}
\end{array}
\)
We know that \(\mathrm{p} \rightarrow \mathrm{q} \equiv \sim \mathrm{q} \rightarrow \sim \mathrm{p} \equiv \sim(\sim \mathrm{q}) \vee \sim \mathrm{p} \equiv \mathrm{q} \vee \sim \mathrm{p}\)
\(
\begin{array}{ll}
\mathrm{S}_1=\mathrm{p} \rightarrow \mathrm{q} & \mathrm{S}_2=\mathrm{p} \rightarrow-\mathrm{q} \\
\mathrm{S}_3=\sim \mathrm{p} \vee \mathrm{q} & \mathrm{S}_4=\mathrm{p} \wedge \mathrm{q}
\end{array}
\)
We know that \(\mathrm{p} \rightarrow \mathrm{q} \equiv \sim \mathrm{q} \rightarrow \sim \mathrm{p} \equiv \sim(\sim \mathrm{q}) \vee \sim \mathrm{p} \equiv \mathrm{q} \vee \sim \mathrm{p}\)
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