GATE CS 2021 Set 2 Q.25 (CS Q.15): Propositional and First-Order Logic

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

S: ((P \land Q) \to R) \to ((P \land Q) \to (Q \to R))

$S$ is neither a tautology nor a contradiction.

$S$ is a tautology.

$S$ is a contradiction.

The antecedent of $S$ is logically equivalent to the consequent of $S$ .

Official answer and solution

Official answer: B and D
From: Official GATE 2021 answer key

Solution

Use logical equivalences to simplify both sides of the main implication in statement $S$ .

Assertion $S$ is given by:

S: ((P \land Q) \to R) \to ((P \land Q) \to (Q \to R))

Consider the antecedent of $S$ , denoted $A$ :

A = (P \land Q) \to R \equiv \neg(P \land Q) \lor R \equiv \neg P \lor \neg Q \lor R

Now consider the consequent of $S$ , denoted $C$ :

C = (P \land Q) \to (Q \to R)

Using the identity $X \to Y \equiv \neg X \lor Y$ :

C \equiv \neg(P \land Q) \lor (\neg Q \lor R) \equiv \neg P \lor \neg Q \lor \neg Q \lor R \equiv \neg P \lor \neg Q \lor R

Since both $A$ and $C$ simplify to $\neg P \lor \neg Q \lor R$ , the antecedent of $S$ is logically equivalent to the consequent of $S$ (Statement D is correct).

Furthermore, since $A \equiv C$ , the formula $S$ is of the form $A \to A$ , which is always true regardless of the truth values of $P, Q,$ and $R$ . Hence, $S$ is a tautology (Statement B is correct).

Final answer

B, D

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2021 Computer Science and Information Technology - Session 2 (CS2)
Question: 25 of 65, printed as CS Q.15
Section: Computer Science and Information Technology

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