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Let F_1(A,B,C)=(A\wedge 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 respective logical expressions. Then:
Option: 1 F_1 is not a tautology but F_2 is a tautology
Option: 2 F_1 is a tautology but F_2 is not a tautology
Option: 3 Both F_1 \text{and}F_2 are not tautologies
Option: 4 F_1 \text{and}F_2 both are tautologies

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\begin{aligned} F_{1} &:(A \wedge \sim B) \vee[\sim C \wedge(A \vee B)] \vee \sim A \\ F_{2} &:(A \vee B) \vee(B \rightarrow \sim A) \\ F_{1} &:\{(A \wedge \sim B) \vee \sim A\} \vee[(A \vee B) \wedge \sim C] \\ &:\{(A \vee \sim A) \wedge(\sim A \vee \sim B)\} \vee[(A \vee B) \wedge \sim C] \\ &:\{t \wedge(\sim A \vee \sim B)\} \vee[(A \vee B) \wedge \sim C] \\ &:(\sim A \vee \sim B) \vee[(A \vee B) \wedge \sim C]\\ &:[(\sim A \vee \sim B) \vee(A \vee B)] \wedge[(\sim A \vee \sim B) \wedge \sim C] \end{aligned}

\\F_{1}:(\sim A \vee \sim B) \wedge \sim C \neq t(\text { tautology }) \\ F_{2}:(A \vee B) \vee(\sim B \vee \sim A)=t(\text { tautology })

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himanshu.meshram

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