[SOLVED] [Moved]How can a find the midterm and maxterm?

[u24c02]

Hi.
Now I'm trying to sove this problem this.
ABC +AB'C'+AB'C+ABC'+A'B'C*

And
I have just found some discussion like this.
F(A,B,C) = m7+m4+m5+m6+m1
= A'B'C' + AB'C' + AB'C + ABC' + A'B'C
F'(A,B,C)= m0 + m2 + m3

If we complement F', we get F;
F = (m0 + m2 + m3)'
By DeMorgan's Theorem;

F = (A'B'C')'(A'BC')'(A'BC)'*
F = (A''+B''+C'')(A''+B'+C'')(A''+B'+C')
F = (A+B+C)(A+B'+C)(A+B'+C')


I want to know how can him did like this to solve?

p.s. How can found the midterm and maxterm?

 

[u24c02]

Can I bind the not at boolean algebra?

Hi.

I wonder that is this right?

1. (A'B')' = A+B
2. (AB)'=A'+B'
3. A'B' = (AB)'
4. (A+B)'=A'B'
5.(A'B')=(A+B)'
6.A'B'C' = A'+B'+C'
I especially do not understand regard 6.
I think it should be (A+B+C)'.

Is this same between A'+B'+C' and (A+B+C)'?

 

[donniebhai]

Re: Can I bind the not at boolean algebra?

it should be (A+B+C)'
(A+B+C)' and A'+B'+C' are certainly different

 

[u24c02]

Re: Can I bind the not at boolean algebra?

it should be (A+B+C)'
(A+B+C)' and A'+B'+C' are certainly different

Thanks but the following answer is so different with you.
https://www.indiabix.com/digital-electronics/boolean-algebra-and-logic-simplification/discussion-138

What is the truth? Who is right?

 

[harpv]

Re: Can I bind the not at boolean algebra?

Thanks but the following answer is so different with you.
https://www.indiabix.com/digital-electronics/boolean-algebra-and-logic-simplification/discussion-138

What is the truth? Who is right?

You are not paying attention here. The expression given in the link is (ABC)' = A'+B'+C' , but the one in your question is A'B'C' which is equal to (A+B+C)' and not equal to A'+B'+C'.