Simplification of Boolean Expression

Discussion in 'Homework Help' started by mathsfailure, Nov 28, 2009.

1. mathsfailure Thread Starter New Member

Nov 28, 2009
4
0
Hi guys,
I need help with my homework regarding simplification of boolean algebra.

X = A.B + C . (A.C +B)

I'm not sure how to do the compliment as in the bar above the letter.

2. mathsfailure Thread Starter New Member

Nov 28, 2009
4
0
is it correct to do it this way

(A+B+C) * (A+C+B)
AA+AC+AB+BA+BC+BB+CA+CC+CB

I'm kinda lost.

3. mik3 Senior Member

Feb 4, 2008
4,846
70
I cant see bar letters in the equation. If you want to write bar letters write them like this: A'.

4. mathsfailure Thread Starter New Member

Nov 28, 2009
4
0
X = ]A.B' + C]' . (A'.C' +B')

yup here it is.

5. mik3 Senior Member

Feb 4, 2008
4,846
70
I will give you a hint:

[A.B'+C]'=(A'+B).C'

Try to simply it more.

6. Fraser_Integration Member

Nov 28, 2009
142
6
de morgans theorem:

A' + B' = (AB)'

and

A'B' = (A+B)'

7. mathsfailure Thread Starter New Member

Nov 28, 2009
4
0
ive simplified the equation to somewhat where i got confused

(A'+B+C')(A'+C'+B')
A'A' + A'C' + A'B' + BA' +BC'+BB'+C'A'+C'C'+C'B'
A'+A'C'+A'B'+BA'+BC'+C'A'+C'+C'B'

Confused

8. mik3 Senior Member

Feb 4, 2008
4,846
70
You can continue by taking common factors as to make some terms zero. However, I think it is easier to proceed like this:

(AB'+C')'(A'C'+B')=(A'+B)C(A'C'+B')=(A'C+BC)(A'C'+B')=A'CA'C'+A'B'C+BCA'C'+BB'C

Then apply the rule:

AA'=0

to simplify the final expression.

9. Ratch New Member

Mar 20, 2007
1,068
4
mathsfailure,

You will find the program in the link below to be useful. It gives the minterms of any Boolean expression. If the minterms of the problem match your solution or your intermediate simplifications, then you know you did not make a mistake with the algebra. Once you know the minterms, you can simplify the Boolean expression with a Karnaugh map. Also search for threads in this forum for "logic".

Ratch