# Simplification of Boolean Expression

Thread Starter

#### mathsfailure

Joined Nov 28, 2009
4
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.

Thread Starter

#### mathsfailure

Joined Nov 28, 2009
4
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.

#### mik3

Joined Feb 4, 2008
4,843
I cant see bar letters in the equation. If you want to write bar letters write them like this: A'.

Thread Starter

#### mathsfailure

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

yup here it is.

#### mik3

Joined Feb 4, 2008
4,843
I will give you a hint:

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

Try to simply it more.

#### Fraser_Integration

Joined Nov 28, 2009
142
de morgans theorem:

A' + B' = (AB)'

and

A'B' = (A+B)'

Thread Starter

#### mathsfailure

Joined Nov 28, 2009
4
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

#### mik3

Joined Feb 4, 2008
4,843
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.

#### Ratch

Joined Mar 20, 2007
1,070
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

http://forum.allaboutcircuits.com/showthread.php?p=135582#post135582

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