Question

The proposition $(p\Rightarrow \sim p)\wedge (\sim p\Rightarrow p)$ is

• A. Tautology and contradiction
• B. Neither tautology nor contradiction
• C. Contradiction
• D. Tautology

Answer 1

The correct option is C Contradiction

$\begin{array}{ccccc}p& \sim p& p\Rightarrow \sim p& \sim p\Rightarrow p& (p\Rightarrow \sim p)\wedge (\sim p\Rightarrow p)\\ T& F& F& T& F\\ F& T& T& F& F\end{array}$
Clearly, $(p\Rightarrow \sim p)\wedge (\sim p\Rightarrow q)$ is a contradiction.