Given:
$F_1: P → \sim P$
$F_2: (P → \sim P) \vee (\sim P → P) $
Lets find out by case method for each of the above equation.
$F_1: P → \sim P$
Case 1: $P=True$ |
Case 2: $P=False$ |
$F_1: True → \sim True$
$F_1: True → False$
$F_1: False$
|
$F_1: False→ \sim False$
$F_1: False → True$
$F_1: True$
|
We observe that $F_1$ is both $True$ and $False$ depending upon the input value. This is the property of Satisfiability.
Therefore $F_1$ is Satisfiable.
$F_2: (P →\sim P) \vee (\sim P → P) $
Case 1: $P=True$ |
Case 2: $P=False$ |
$F_2: (True → \sim True) \vee (\sim True → True)$
$F_2: (True → False) \vee (False → True)$
$F_2: False \vee True$
$F_2: True$
|
$F_2: (False → \sim False) \vee (\sim False → False)$
$F_2: (False → True) \vee (True → False)$
$F_2: True \vee False$
$F_2: True$
|
We observe that $F_2$ is $True$ for both the cases. It doesn’t depends upon the input. This is the property of Tautology.
Therefore $F_2$ is Valid $(Valid \equiv Tautology)$.
Answer is (A).