Question

Let $F_1\left(A,B,C\right)=\left(A\wedge ~B\right)\vee \left[~C\wedge \left(A\vee B\right)\right]\vee ~A$ and $F_2\left(A,B\right)=\left(A\vee B\right)\vee \left(B\to ~A\right)$be two logical expressions. Then:

• A. $F_1$ is not a tautology but $F_2$ is a tautology
• B. $F_1$ is a tautology but $F_2$ is not a tautology
• C. $F_1$ and $F_2$ both area tautologies
• D. Both $F_1$ and $F_2$ are not tautologies

Answer 1

The correct option is A

$F_1$ is not a tautology but $F_2$ is a tautology

Simplifying the given logical expressions.

Given data

$F_1\left(A,B,C\right)=\left(A\wedge ~B\right)\vee \left[~C\wedge \left(A\vee B\right)\right]\vee ~A$

$F_2\left(A,B\right)=\left(A\vee B\right)\vee \left(B\to ~A\right)$

Consider the given equation as:

$$\begin{gathered} F_1\left(A,B,C\right)=\left(A\wedge ~B\right)\vee \left[~C\wedge \left(A\vee B\right)\right]\vee ~A \\ {F}_1\left(A,B,C\right)=\left\{\left(A\wedge ~B\right)\vee ~A\right\}\vee \left[\left(A\vee B\right)\wedge ~C\right] \\ {F}_1\left(A,B,C\right)=\left\{\left(A\vee ~A\right)\wedge \left(~A\vee ~B\right)\right\}\vee \left[\left(A\vee B\right)\wedge ~C\right] \\ {F}_1\left(A,B,C\right)=\left\{t\wedge \left(~A\vee ~B\right)\right\}\vee \left[\left(A\vee B\right)\wedge ~C\right] \\ {F}_1\left(A,B,C\right)=\left\{\left(~A\vee ~B\right)\right\}\vee \left[\left(A\vee B\right)\wedge ~C\right] \\ {F}_1\left(A,B,C\right)=\left[\left(~A\vee ~B\right)\vee \left(A\vee B\right)\right]\left[\left(~A\vee ~B\right)\wedge ~C\right] \end{gathered}$$

$\left[\left(~A\vee ~B\right)\vee \left(A\vee B\right)\right]=t$

$F_1\left(A,B,C\right)=\left[\left(~A\vee ~B\right)\wedge ~C\right]\ne t\left(\text{tautology}\right)$

$F_2\left(A,B\right)=\left(A\vee B\right)\vee \left(B\to ~A\right)=t\left(\text{tautology}\right)$

Hence, the correct answer is Option A.