# Validity of equation

Discussion in 'Electronics Homework Help' started by anacondaonline, May 13, 2018.

1. ### anacondaonline

6
0
May 5, 2018
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.

2. ### anacondaonline

6
0
May 5, 2018
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 3. ### Laplace

1,252
184
Apr 4, 2010
Have you tried multiple application of DeMorgan's law?

4. ### Ratch

1,094
334
Mar 10, 2013
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

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5. ### Ratch

1,094
334
Mar 10, 2013
The book is correct here. It is a simple application of DeMorgan's Theorem.

Ratch

6. ### Harald KappModeratorModerator

11,806
2,749
Nov 17, 2011
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.

7. ### Ratch

1,094
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Mar 10, 2013
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

8. ### Harald KappModeratorModerator

11,806
2,749
Nov 17, 2011
I'm sorry, I missed that part. It's still XNOR, however.

1,252
184
Apr 4, 2010
10. ### Ratch

1,094
334
Mar 10, 2013
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

Ratch

11. ### Ratch

1,094
334
Mar 10, 2013
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'

Ratch

12. ### Laplace

1,252
184
Apr 4, 2010
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?

13. ### Ratch

1,094
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Mar 10, 2013
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

14. ### Laplace

1,252
184
Apr 4, 2010
So you are offering a K-map as the solution to a Boolean equation. When has that ever been considered rigorous?

15. ### Ratch

1,094
334
Mar 10, 2013
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  