GATE CSE 2021 Set 2 | Question: 15

Question

Choose the correct choice(s) regarding the following proportional logic assertion $S$:

$$S: (( P \wedge Q) \rightarrow R) \rightarrow (( P \wedge Q) \rightarrow (Q \rightarrow R))$$

• A. $S$ is neither a tautology nor a contradiction
• B. $S$ is a tautology
• C. $S$ is a contradiction
• D. The antecedent of $S$ is logically equivalent to the consequent of $S$

Answer 1

option b and are correct ..see the solution 

Answer 2

$S : ((P \wedge Q) → R) → ((P \wedge Q) → (Q → R))$ 

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)

Answer 3

S : ((P $\wedge$ Q) → R) → ((P $\wedge$ Q) → (Q → R))

Consequent of S :-  (P $\wedge$ Q)’ V (Q → R) $\equiv$ (P’ V Q’) V (Q’ V R) $\equiv$ (P’ V Q’ V R) $\equiv$ ((P $\wedge$ Q)’ V R) $\equiv$ ((P $\wedge$ Q) → R)

=> S : ((P $\wedge$ Q) → R) → ((P $\wedge$ Q) → R) $\equiv$ True.

Answer 4

First part

Second part. Hope this helps!