Question

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

• A. a tautology
• B. a fallacy
• C. neither a tautology nor a fallacy
• D. logically equivalent to $p\vee (\sim p)$

Answer 1

The correct option is B a fallacy

$\begin{array}{ccccc}p& \sim p& a=(p\Rightarrow \sim p)& b=(\sim p\Rightarrow p)& a\wedge b\\ T& F& F& T& F\\ F& T& T& F& F\end{array}$

Since, $(p\Rightarrow \sim p)\wedge (\sim p\Rightarrow p)$ is always false, so its a fallacy.

Alternate solution:
We know that $x\Rightarrow y\equiv \sim x\vee y$
$\therefore (p\Rightarrow \sim p)\wedge (\sim p\Rightarrow p)$
$\equiv (\sim p\vee \sim p)\wedge (p\vee p)$
$\equiv \sim p\wedge p$
$\equiv F$