$S : ((P \wedge Q) → R) → ((P \wedge Q) → (Q → R))$
Since Q is the most repetitive term here in this equation, let’s pick it to put True/False in the equation.
By-Case method
Case 1: $Q = True$ |
Case 1: $Q = False$ |
$S : ((P \wedge True) → R) → ((P \wedge True) → (True→ R))$
$S : (P → R) → (P → R)$
$S : True$
|
$S : ((P \wedge False) → R) → ((P \wedge False) → (False→ R))$
$S : ((False) → R) → ((False) → True)$
$S : (False → R) → True$
$S: True → True$
$S: True$
|
So, we can say that even if Q is True or False, the equation always simplifies to $True$.
Hence Tautology.
Also, we see that, even if Q is True or False, the Equation resolves to either:
$S: (P → R) → (P → R)$ ...(when $Q=True$) and
$S: (True) → (True)$ ...(when $Q=False$)
Therefore, we can say that antecedent (if) of $S$ is logically equivalent to the consequent (then) of $S$.
Therefore, answer is both (B) and (D)