Let $F_{1}(A, B, C)=(A \wedge \sim B) \vee[\sim C \wedge(A \vee B)] \vee \sim A$ and $F_{2}(A, B)=(A \vee B) \vee(B \rightarrow \sim A)$ be two logical expressions. Then :

(B) $F_{1}$ is not a tautology but $F_{2}$ is a tautology

(A) Both $F_{1}$ and $F_{2}$ are not tautologies

(C) $F_{1}$ and $F_{2}$ both are tautologies

(D) $F_{1}$ is a tautology but $F_{2}$ is not a tautology

Mathematical Reasoning * — JEE Main · 2021 · 26 Feb 2021 · Shift 2 Question with Solution

Let

F_{1} (A, B, C) = (A \land \sim B) \lor [\sim C \land (A \lor B)] \lor \sim A

and

F_{2} (A, B) = (A \lor B) \lor (B \to\sim A)

be two logical expressions. Then :

Answer: (B)

The implication

B \to\sim A

is equivalent to

\sim B \lor \sim A

, so

F_{2} = (A \lor B) \lor (\sim B \lor \sim A) = (A \lor \sim A) \lor (B \lor \sim B)

which is true under every assignment, so

F_{2}

is a tautology.
For

F_{1}

take

A

B

and

C

all true. Then

A \land \sim B

is false,

\sim C \land (A \lor B)

is false because

\sim C

is false, and

\sim A

is false, so

F_{1}

is false for this assignment and fails to be a tautology.
Thus

F_{1}

is not a tautology but

F_{2}

is a tautology.

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option