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

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Here is the problem statement in my book

Printed textbook problem: define the connective * for two-valued variables A, B, C as A * B = AB + A'B', then let C = A * B and verify the validity of (i) A = B * C.
Here is the solution as per my book
Handwritten Boolean algebra derivation of A = B * C, substituting B * A * B and expanding to B-bar(overline(A*B)) + B(A*B), then simplifying step by step to A-B-bar + AB = A(B-bar + B) = A.


I don't understand that marked part. Is that correct? Can we write that? How?
Need help to understand that part.
 
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  • Printed textbook problem: define the connective * for two-valued variables A, B, C as A * B = AB + A'B', then let C = A * B and verify the validity of (i) A = B * C.
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  • Handwritten Boolean algebra derivation of A = B * C, substituting B * A * B and expanding to B-bar(overline(A*B)) + B(A*B), then simplifying step by step to A-B-bar + AB = A(B-bar + B) = A.
    TmHEB8y.jpg
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Discussion

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*}
 
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?

Handwritten Boolean algebra derivation showing A = B*C, then A = B*A*B, then substitutions with complemented terms, ending with A̅B + AB = A(B̅ + B) = A. A red dotted ellipse circles the step B̅(ĀB + A B̅), with an arrow and red note reading How?
 
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  • Handwritten-style printed derivation of a Boolean algebra simplification, with a red annotation asking how the circled step follows and a dotted ellipse around the term (A-bar B + A B-bar).
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  • Handwritten Boolean algebra derivation showing A = B*C, then A = B*A*B, then substitutions with complemented terms, ending with A̅B + AB = A(B̅ + B) = A. A red dotted ellipse circles the step B̅(ĀB + A B̅), with an arrow and red note reading How?
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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##
 
Mark44 said:
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 ,

Handwritten-style question asking whether the Boolean expression B̄AB = B̄B̄A can be written this way and whether ordering matters, with overlines on the letters.


I just rearranged those values ...I am not sure if this is allowed ...could you please confirm ?
 
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  • Handwritten-style question asking whether the Boolean expression B̄AB = B̄B̄A can be written this way and whether ordering matters, with overlines on the letters.
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  • Handwritten-style question asking whether the Boolean expression B̅AB̅ = B̅B̅A̅ can be written this way and whether ordering matters, with overlines on the variables in red and the expression in blue.
    2cDlv1Y.png
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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.
 
Last edited:
Mark44 said:
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.
 
momentum said:
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.