Let $$F_1(A, B, C) = (A \wedge \sim B) \vee [\sim C \wedge (A \vee B)] \vee \sim A$$ and $$F_2(A, B) = (A \vee B) \vee (B \to \sim A)$$ be two logical expressions. Then:

We are given $$F_1(A, B, C) = (A \wedge \sim B) \vee [\sim C \wedge (A \vee B)] \vee \sim A$$ and $$F_2(A, B) = (A \vee B) \vee (B \to \sim A)$$.

First, we check whether $$F_1$$ is a tautology. Consider $$A = T$$, $$B = T$$, $$C = T$$: $$(T \wedge F) \vee [F \wedge (T \vee T)] \vee F = F \vee F \vee F = F$$. Since $$F_1$$ evaluates to $$F$$ for this assignment, $$F_1$$ is not a tautology.

Next, we check $$F_2$$. Recall that $$B \to \sim A \equiv \sim B \vee \sim A$$. So $$F_2 = (A \vee B) \vee (\sim B \vee \sim A) = (A \vee \sim A) \vee (B \vee \sim B) = T \vee T = T$$.

More explicitly, rearranging: $$F_2 = A \vee B \vee \sim B \vee \sim A = (A \vee \sim A) \vee (B \vee \sim B) = T$$. So $$F_2$$ is always true, making it a tautology.

Therefore $$F_1$$ is not a tautology but $$F_2$$ is a tautology.