Solve Variables with Ease from My Book's Problem Statement to Solution

[momentum]

Here is the problem statement in my book

Here is the solution as per my book

I don't understand that marked part. Is that correct? Can we write that? How?
Need help to understand that part.

 

[andrewkirk]

It might be easier to follow if we put the implied parentheses in:

\begin{align*}
A&=B*C\\
&=B*(A*B)\\
&=\overline{B}\ \overline{(A*B)} + B (A*B)
\end{align*}
In the last step they are just applying the definition of the '$*$' operator to the first instance of '$*$' in the second line.

Another way to look at it is:
\begin{align*}
A&= B*C\\
&= BC + \overline B\ \overline C\\
&= \overline{B}\ \overline{(A*B)} + B (A*B)
\end{align*}

 

[momentum]

Thanks. I am clear on that part now...very nice.

But stuck with the next step :(

How the next step working. Is it any law? Could you please elaborate more on this red part below?

 

[Mark44]

To get to the circled expression, the author has used DeMorgan's Laws several times.

$\overline{\bar A\bar B + AB}= (\overline{\bar A \bar B}) (\overline{AB})$
Can you continue from there? This expression does simplify to $\bar A B + B\bar A$.

You will also need to use the fact that for a Boolean variable B, $B\bar B = 0$

 

[momentum]

Can you continue from there? This expression does simplify to

yes..applying DeMorgan's Laws two times I achieve that...not a problem.

one more query ,

I just rearranged those values ...I am not sure if this is allowed ...could you please confirm ?

 

[Mark44]

Yes, the Boolean AND and OR operations are commutative, so $\bar B \bar A B = \bar A \bar B B$, which can be further simplified to $\bar A$.

Edit: $\bar A \bar B B$ simplifies to 0.

 

[momentum]

Yes, the Boolean AND and OR operations are commutative, so $\bar B \bar A B = \bar A \bar B B$, which can be further simplified to $\bar A$.

I think you made a typo here.

since B!B =0 ... so the value of this expression should be 0

correct me if I'm wrong.

 

[Mark44]

I think you made a typo here.

since B!B =0 ... so the value of this expression should be 0

correct me if I'm wrong.

Right, I got ahead of myself.
I have edited my earlier reply.