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The statement pattern $[(p \vee q) \wedge \sim p] \wedge(\sim q)$ is
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Verified Answer
The correct answer is:
a contradiction
\begin{array}{|c|c|c|c|c|c|c|}
\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline \mathrm{p} & \mathrm{q} & \sim \mathrm{p} & \sim \mathrm{q} & \mathrm{p} \vee \mathrm{q} & (\mathrm{p} \vee \mathrm{q}) \wedge \sim \mathrm{p} & {[(\mathrm{p} \vee \mathrm{q}) \wedge \sim \mathrm{p}] \wedge \sim \mathrm{q}} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}
All entries in last column are F.
$\therefore$ It is contradiction.
\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline \mathrm{p} & \mathrm{q} & \sim \mathrm{p} & \sim \mathrm{q} & \mathrm{p} \vee \mathrm{q} & (\mathrm{p} \vee \mathrm{q}) \wedge \sim \mathrm{p} & {[(\mathrm{p} \vee \mathrm{q}) \wedge \sim \mathrm{p}] \wedge \sim \mathrm{q}} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}
All entries in last column are F.
$\therefore$ It is contradiction.
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