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(D) To check if the statement $[A \wedge (A$ $\rightarrow B)]$ $\rightarrow B$ is a tautology,we simplify it using logical laws:$[A \wedge (A$ $\right

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$[A \wedge (A$ $\rightarrow B)]$ $\rightarrow B$

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(D) To check if the statement $[A \wedge (A$ $\rightarrow B)]$ $\rightarrow B$ is a tautology,we simplify it using logical laws:$[A \wedge (A$ $\rightarrow B)]$ $\rightarrow B$$= [A \wedge (\sim A \vee B)] \rightarrow B$$= [(A \wedge \sim A) \vee (A \wedge B)] \rightarrow B$$= [F \vee (A \wedge B)] \rightarrow B$$= (A \wedge B) \rightarrow B$$= \sim (A \wedge B) \vee B$$= (\sim A \vee \sim B) \vee B$$= \sim A \vee (\sim B \vee B)$$= \sim A \vee T$$= T$Since the final result is $T$ (a tautology),the statement $[A \wedge (A$ $\rightarrow B)]$ $\rightarrow B$ is a tautology.

The statement among the following that is a tautology is

Similar Questions

Which of the following statements is/are False?
$S_1: \exists n \in N$, such that $n^2 + n + 2$ is divisible by $4$.
$S_2: \exists x \in N$, such that $x - 17 < 20$.
$S_3: \forall n \in N, x^2 + 3x - 10 = 0$.
$S_4: \forall n \in N, n^2 \ge 1$.

Statement-$1$: $\sim (p \Leftrightarrow \sim q)$ is equivalent to $p \Leftrightarrow q$.
Statement-$2$: $\sim (p \Leftrightarrow \sim q)$ is a tautology.

The dual of the statement $[p \vee (\sim q)] \wedge (\sim p)$ is

The expression $((p \wedge q) \vee (p \vee \sim q)) \wedge (\sim p \wedge \sim q)$ is equivalent to

Statement-$I$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a contradiction.
Statement-$II$: $(p$ $\rightarrow q) \Leftrightarrow (\sim q$ $\rightarrow \sim p)$ is a tautology.

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