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

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Question

$F_{1}:(A \wedge \sim B) \vee[\sim C \wedge(A \vee B)] \vee \sim A$

$F_{2}:(A \vee B) \vee(B \rightarrow \sim A)$

$F_{1}:\{(A \wedge \sim B) \ve

Vedclass · Accepted Answer

$F _{1}$ is not tautology but $F _{2}$ is a tautology

Vedclass · Answer

$F_{1}:(A \wedge \sim B) \vee[\sim C \wedge(A \vee B)] \vee \sim A$

$F_{2}:(A \vee B) \vee(B ightarrow \sim A)$

$F_{1}:\{(A \wedge \sim B) \vee \sim A\} \vee[(A \vee B) \wedge \sim C]$

$:\{( A \vee \sim A ) \wedge(\sim A \vee \sim B )\} \vee[( A \vee B ) \wedge \sim C ]$

$:\{ t \wedge(\sim A \vee \sim B )\} \vee[( A \vee B ) \wedge \sim C ]$

$:(\sim A \vee \sim B ) \vee[( A \vee B ) \wedge \sim C ]$

$: \underbrace{[(\sim A \vee \sim B ) \vee( A \vee B )]}_{ t } \wedge[(\sim A \vee \sim B ) \wedge \sim C ]$

$F_{1}:(\sim A \vee \sim B) \wedge \sim C 
eq t($ tautology $)$

$F_{2}:(A \vee B) \vee(\sim B \vee \sim A)=t($ 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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