Q. The proposition $(p \rightarrow \sim p) \wedge(\sim p \rightarrow p)$ is
ManipalManipal 2015
Solution:
The truth table of $(p \rightarrow \sim p) \wedge(\sim p \rightarrow p)$ is as follows:
$p$
$\sim p$
$p \to \sim p$
$\sim p \to p$
$(p \to \sim p) \wedge (\sim p \to p)$
T
F
F
T
F
F
T
T
F
F
Clearly, last column of the above truth table contains $F$ only. So, the given statement is a contradiction.
| $p$ | $\sim p$ | $p \to \sim p$ | $\sim p \to p$ | $(p \to \sim p) \wedge (\sim p \to p)$ |
|---|---|---|---|---|
| T | F | F | T | F |
| F | T | T | F | F |