Validity of equation

[anacondaonline]

I'm trying to solve this question from my book.

Book has given this solution :

I don't understand that red marked part in the given solution.

Do you think book solution is correct? Could you please explain that red marked part. how did the book arrive to that from the earlier step? I'm stuck right at that part.

Need help.

 

[anacondaonline]

I have understood the above part. ..... book is using "*" definition there.

But I'm stuck in the next step. Is it a law ? which law ? Please see below

 

[Laplace]

But I'm stuck in the next step. Is it a law ? which law ?

Have you tried multiple application of DeMorgan's law?

 

[Ratch]

The book solution is correct, but the solution method is in error. They made a mistake at the place you cannot understand. The whole expression should be have a bar over it. They must have made another mistake later to get the correct answer. Two wrongs make it right.

Since the "*" means Exclusive OR, why not rewrite the equation as B(AB'+A'B)'+B'(AB'+A'B) and solve that instead?

I am enclosing a program I wrote some years ago to give the minterms of a logical expression.

Ratch

 

[Ratch]

I have understood the above part. ..... book is using "*" definition there.

But I'm stuck in the next step. Is it a law ? which law ? Please see below

The book is correct here. It is a simple application of DeMorgan's Theorem.

Ratch

 

[Harald Kapp]

the solution method is in error.

I don't think so. The "*" in the task description is not the usual AND function. "*" is defined as A*B = AB + /A/B instead.
Assuming that AB means logical AND and "+" means logical OR, then "*" represents an Exclusive NOR.

 

[Ratch]

I don't think so. The "*" in the task description is not the usual AND function. "*" is defined as A*B = AB + /A/B instead.
Assuming that AB means logical AND and "+" means logical OR, then "*" represents an Exclusive NOR.

Did I not say in post #4 that the asterisk in that problem means XOR? I stand by my statement that the solution is correct, but the method contains an error.

Ratch

 

[Harald Kapp]

Did I not say in post #4 that the asterisk in that problem means XOR?

I'm sorry, I missed that part. It's still XNOR, however.

 

[Laplace]

The original question was how did the author arrive at the circled expression? It was done by multiple application of DeMorgan's Theorem, {When breaking an overbar, the operation enclosed by the overbar changes between AND / OR}.

 

[Ratch]

Too much time has been spent on this problem already. DeMorgan's theorem is unnecessary.

B*A*B
B*(A*B)
B(A*B)'+B'(A*B)
B(AB+A'B')+B'(AB'+A'B)
ABB+A'B'B+AB'B'+A'BB'
AB +0 +AB' +0
A(B+B')
A

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[Ratch]

The original question was how did the author arrive at the circled expression? It was done by multiple application of DeMorgan's Theorem, {When breaking an overbar, the operation enclosed by the overbar changes between AND / OR}.
View attachment 41099

DeMorgan's theorem is not necessary.

(A'B'+AB)'
1-(A'B'+AB) ; Complement of (A'B'+AB)
(A'B'+A'B+AB'+AB) - (A'B'+AB) ; 1 = (A'B'+A'B+AB'+AB)
A'B+AB'

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[Laplace]

DeMorgan's theorem is not necessary.

True; however, they do teach DeMorgan's Theorem for a reason. But one thing I was never taught is the use of the Boolean minus "-" operation. There is the AND gate, the OR gate, the NOT gate. Where does one get a MINUS gate?

 

[Ratch]

True; however, they do teach DeMorgan's Theorem for a reason. But one thing I was never taught is the use of the Boolean minus "-" operation. There is the AND gate, the OR gate, the NOT gate. Where does one get a MINUS gate?

It is not a gate, it is an operation, like add or multiply. To complement any Boolean term or expression, just subtract it from "1". Just like you would complement a decimal number by subtracting it from "10". It corresponds to a K-map with the unmarked squares taken as the complement.

Ratch

 

[Laplace]

It corresponds to a K-map with the unmarked squares taken as the complement.

So you are offering a K-map as the solution to a Boolean equation. When has that ever been considered rigorous?

 

[Ratch]

So you are offering a K-map as the solution to a Boolean equation. When has that ever been considered rigorous?

I am using the K-map as an explanation of why the subtraction of a Boolean term from "1" gives the complement of the term. The K-map is as rigorous as a slide rule is for what it does.

Ratch