Statement -1: The statement A to (B to A) is | vedclass

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(C) For Statement $-1$: $A 	o (B 	o A) \equiv \sim A \vee (\sim B \vee A) \equiv (\sim A \vee A) \vee \sim B \equiv T \vee \sim B \equiv T$.$A 	o (

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Statement $-1$ is true; Statement $-2$ is false.

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(C) For Statement $-1$: $A 	o (B 	o A) \equiv \sim A \vee (\sim B \vee A) \equiv (\sim A \vee A) \vee \sim B \equiv T \vee \sim B \equiv T$.$A 	o (A \vee B) \equiv \sim A \vee (A \vee B) \equiv (\sim A \vee A) \vee B \equiv T \vee B \equiv T$.Since both are equivalent to $T$ (Tautology),Statement $-1$ is true.For Statement $-2$: Let $P = (A \wedge B) 	o (\sim A \vee B)$. If $A=T, B=T$,then $P = (T \wedge T) 	o (F \vee T) = T 	o T = T$.Then $\sim P = \sim T = F$. Since the statement is not true for all truth values,it is not a tautology. Thus,Statement $-2$ is false.

Statement $-1$: The statement $A \to (B \to A)$ is equivalent to $A \to (A \vee B)$.
Statement $-2$: The statement $\sim [(A \wedge B) \to (\sim A \vee B)]$ is a tautology.

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Statement $-1$: The statement $A \to (B \to A)$ is equivalent to $A \to (A \vee B)$.Statement $-2$: The statement $\sim [(A \wedge B) \to (\sim A \vee B)]$ is a tautology.