Question

The statement $(p\to \sim p)\wedge (\sim p\to p)$ is

• A. a tautology.
• B. a contradiction.
• C. neither a tautology nor a contradiction.
• D. none of these.

Answer 1

Answer: (B)

a contradiction.

$\because (p\to \sim p)\wedge (\sim p\to p)$
$\equiv (\sim p\vee \sim p)\wedge (p\vee p)(\because p\to q\equiv \sim p\vee q)$
$\equiv \sim p\wedge p(\because p\vee p\equiv p)$
$\equiv c$
$c$ is a contradiction.