Let F_{1}(A, B, C) = (A wedge sim B) vee [sim | vedclass

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(C) For $F_{1}(A, B, C) = (A \wedge \sim B) \vee [\sim C \wedge (A \vee B)] \vee \sim A$: Using the distributive law: $F_{1} = [(A \wedge \sim B) \vee

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$F_{1}$ is not a tautology but $F_{2}$ is a tautology

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(C) For $F_{1}(A, B, C) = (A \wedge \sim B) \vee [\sim C \wedge (A \vee B)] \vee \sim A$: Using the distributive law: $F_{1} = [(A \wedge \sim B) \vee \sim A] \vee [\sim C \wedge (A \vee B)]$ $F_{1} = [(A \vee \sim A) \wedge (\sim B \vee \sim A)] \vee [\sim C \wedge (A \vee B)]$ Since $(A \vee \sim A) = t$ (tautology): $F_{1} = [t \wedge (\sim A \vee \sim B)] \vee [\sim C \wedge (A \vee B)]$ $F_{1} = (\sim A \vee \sim B) \vee [\sim C \wedge (A \vee B)]$. This expression depends on the values of $A, B, C$,so it is not a tautology. For $F_{2}(A, B) = (A \vee B) \vee (B \rightarrow \sim A)$: Using the implication rule $(P \rightarrow Q) = (\sim P \vee Q)$: $F_{2} = (A \vee B) \vee (\sim B \vee \sim A)$ By commutative and associative laws: $F_{2} = (A \vee \sim A) \vee (B \vee \sim B)$ $F_{2} = t \vee t = t$. Thus,$F_{2}$ is a tautology.

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 ...... .

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