- Home
- Standard 11
- Mathematics
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 ...... .
$F _{1}$ and $F _{2}$ both are tautologies
$F _{1}$ is a tautology but $F _{2}$ is not a tautology
$F _{1}$ is not tautology but $F _{2}$ is a tautology
Both $F _{1}$ and $F _{2}$ are not tautologies
Solution
$F_{1}:(A \wedge \sim B) \vee[\sim C \wedge(A \vee B)] \vee \sim A$
$F_{2}:(A \vee B) \vee(B \rightarrow \sim A)$
$F_{1}:\{(A \wedge \sim B) \vee \sim A\} \vee[(A \vee B) \wedge \sim C]$
$:\{( A \vee \sim A ) \wedge(\sim A \vee \sim B )\} \vee[( A \vee B ) \wedge \sim C ]$
$:\{ t \wedge(\sim A \vee \sim B )\} \vee[( A \vee B ) \wedge \sim C ]$
$:(\sim A \vee \sim B ) \vee[( A \vee B ) \wedge \sim C ]$
$: \underbrace{[(\sim A \vee \sim B ) \vee( A \vee B )]}_{ t } \wedge[(\sim A \vee \sim B ) \wedge \sim C ]$
$F_{1}:(\sim A \vee \sim B) \wedge \sim C \neq t($ tautology $)$
$F_{2}:(A \vee B) \vee(\sim B \vee \sim A)=t($ tautology $)$
