Solve the Following Questions

Question:

Let $\mathrm{F}_{1}(\mathrm{~A}, \mathrm{~B}, \mathrm{C})=(\mathrm{A} \wedge \sim \mathrm{B}) \vee[\sim \mathrm{C} \wedge(\mathrm{A} \vee \mathrm{B})] \vee \sim \mathrm{A}$ and $\mathrm{F}_{2}(\mathrm{~A}, \mathrm{~B})=(\mathrm{A} \vee \mathrm{B}) \vee(\mathrm{B} \rightarrow \sim \mathrm{A})$ be two logical expressions. Then :

  1. $\mathrm{F}_{1}$ and $\mathrm{F}_{2}$ both are tautologies

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

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

  4. Both $F_{1}$ and $F_{2}$ are not tautologies


Correct Option: , 3

Solution:

$\mathrm{F}_{1}:(\mathrm{A} \wedge \sim \mathrm{B}) \vee[\sim \mathrm{C} \wedge(\mathrm{A} \vee \mathrm{B})] \vee \sim \mathrm{A}$

$\mathrm{F}_{2}:(\mathrm{A} \vee \mathrm{B}) \vee(\mathrm{B} \rightarrow \sim \mathrm{A})$

$\begin{aligned} \mathrm{F}_{1} &:\{(\mathrm{A} \wedge \sim \mathrm{B}) \vee \sim \mathrm{A}\} \vee[(\mathrm{A} \vee \mathrm{B}) \wedge \sim \mathrm{C}] \\ &:\{(\mathrm{A} \vee \sim \mathrm{A}) \wedge(\sim \mathrm{A} \vee \sim \mathrm{B})\} \vee[(\mathrm{A} \vee \mathrm{B}) \wedge \sim \mathrm{C}] \\ &:\{\mathrm{t} \wedge(\sim \mathrm{A} \vee \sim \mathrm{B})\} \vee[(\mathrm{A} \vee \mathrm{B}) \wedge \sim \mathrm{C}] \\ &:(\sim \mathrm{A} \vee \sim \mathrm{B}) \vee[(\mathrm{A} \vee \mathrm{B}) \wedge \sim \mathrm{C}] \end{aligned}$

$: \underbrace{[(\sim A \vee \sim B) \vee(A \vee B)]}_{t} \wedge[(\sim A \vee \sim B) \wedge \sim C]$

$\mathrm{F}_{1}:(\sim \mathrm{A} \vee \sim \mathrm{B}) \wedge \sim \mathrm{C} \neq \mathrm{t}($ tautology $)$

$\mathrm{F}_{2}:(\mathrm{A} \vee \mathrm{B}) \vee(\sim \mathrm{B} \vee \sim \mathrm{A})=\mathrm{t}($ tautolog $\mathrm{y})$

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