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The propositions $(p \Rightarrow \sim p) \wedge(\sim p \Rightarrow p)$ is a
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2546 Upvotes
Verified Answer
The correct answer is:
contradiction
\begin{array}{c|c|c|c|c}
\hlinep & \sim p & p \Rightarrow \sim p & \sim p \Rightarrow p & (p \Rightarrow \sim p) \wedge(\sim p \Rightarrow p) \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}
Clearly,$(p \Rightarrow \sim p) \wedge(\sim p \Rightarrow p)$ is a
contradiction.
\hlinep & \sim p & p \Rightarrow \sim p & \sim p \Rightarrow p & (p \Rightarrow \sim p) \wedge(\sim p \Rightarrow p) \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}
Clearly,$(p \Rightarrow \sim p) \wedge(\sim p \Rightarrow p)$ is a
contradiction.
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