GO Classes DPP | GATE CS | Digital Logic | Boolean Algebra Laws

Answer 1

Given $F=(A+B+\overline{C})(A+B+C)(A+\overline{B})$.

Let $X=A+B$.

Then the first two factors become:

$(X+\overline{C})(X+C)$.

Use the law $(X+Y)(X+\overline{Y})=X$.

So, $(A+B+\overline{C})(A+B+C)=A+B$.

Now,

$F=(A+B)(A+\overline{B})$.

Again use $(X+Y)(X+Z)=X+YZ$.

Here, $X=A$, $Y=B$, and $Z=\overline{B}$.

So,

$F=A+B\overline{B}$.

Since $B\overline{B}=0$,

$F=A+0=A$.

Answer: A.

Answer 2

Here is the Explanation