# Simplifying Boolean Expression

Discussion in 'Homework Help' started by gino, Sep 17, 2005.

1. ### gino Thread Starter New Member

Sep 17, 2005
3
0
I am having trouble simplifying the following Boolean Expression in Sum of Products form:

AB' (C+D') + AB'C'D

= AB'C + AB'D' + AB'C'D
= ?

2. ### Brandon Senior Member

Dec 14, 2004
306
0
Go from step 1 and use DeMorgan's instead. Don't distribute the AB', that should put you on the right track.

3. ### gino Thread Starter New Member

Sep 17, 2005
3
0
I assume you apply DeMorgan's law to the first part of the equation?

e.g. [AB'(C+D')]'

A' + B + C'D

Not sure if its correct.

4. ### techduq New Member

Aug 31, 2005
7
0

Start again and factor out the AB': AB' (C + D' + C'D). Do you know the rule A + A'B = A + B? See how you can use it to simplify the expression in parentheses? C + C'D = C + D. Then you have AB'(C + D + D'). Another basic rule: A + A' =1. This leads us to AB'(C + 1). And of course C + 1 = 1. Hence: AB' (C + 1) = AB'. The original expression thus simplifies to AB'. You could also use a four-variable Karnaugh map to quickly obtain the same result (maybe you haven't done K-maps yet...).

You can utilize another rule: A + AB = A. Can you see how this rule allows you to get to the result immediately?

5. ### gino Thread Starter New Member

Sep 17, 2005
3
0
I didn't know the rule A + A'B = A + B. Is this derived from some Boolean Identities?

Not sure about using A + AB = A.

And we haven't covered Kmaps yet.

6. ### techduq New Member

Aug 31, 2005
7
0

To understand A + AB = A, factor as is regular algebra: A + AB = A(1 + B), where 1 + B = 1 (if an OR gate has a high on one input and logic level B on the other input, the output is high, regardless of what B is). Therefore, A + AB = A(1 + B) = A(1) = A. The last part follows from the fact that when you have a high on one input of an AND and logic level A on the other input, the output is A, regardless of what A is.

Now extend the rule: A + AB + AC + AD + .... = A!!! You can keep factoring terms as above, one after another, and you end up with just A. So consider your original expression: AB'C + AB'D' + AB'C'D. Let X = AB': So, you have XC + XD' + XC'D. Just as A is the only survivor of the above expression, so X is all that survives of this one, where X is your AB', of course.

The other rule I showed you, A + A'B = A + B, is a bit more involved to derive, but I'm sure you'll see it in class soon enough! Good luck.

Sep 30, 2007
1
0
Can anyone plz help me out to simplify boolean function
1) (A + B)' + (A' + B')'

2) y(wz' + wz) + xy

3) xy + x'z + yz

plzz do help me as my exams are supposed to start n plzz do tell me any site wherein i can refer n learn more bout system architecture.

8. ### agentofdarkness Active Member

Oct 9, 2007
42
0
You can you the consensus theorem to simplify this expression. YZ is a redundant term. In the terms XY + X'Z, ignore the Xs (since there is X + X') and look at the other variables. The redundant term for these two is YZ. Similarly, if you had AB + A'C, the redundant term would be BC. I have found that this theorem is quite helpful. You can also add redundant terms to any expression in order to simplify it. Sometimes adding a redundant term will allow you to cancel another term, and then you can also cancel the redundant term as well.

9. ### Stacey New Member

Jul 19, 2008
1
0
I'm trying to prove that the term AB is redundant in the function below.

F = BC + AB + AC'
F = A'BC + ABC + AB + AC' {multiplying (A' +A) by the first term}
F = A'BC + AB(C + 1) + AC'
F = A'BC + AB + AC'
F = B(A'C + A) + AC'
F = B(A + C) + AC'

I dont seem to be getting any closer, am I?

10. ### Ratch New Member

Mar 20, 2007
1,068
3
gino,

AB'C + AB'D' + AB'C'D = AB'C(D+D')+AB'D'(C+C')+AB'C'D

=AB'CD+AB'CD'+AB'CD'+AB'C'D'+AB'C'D

=AB'CD+AB'CD'+AB'C'D'+AB'C'D

11 10 8 9

So the SOP's are 8,9,10,11 . Now you can use a 4 variable K-map, or do as I did and use the Quine-McCluskey tabulation method.

The expression simplifies to AB'

You can tell that is the correct answer because no matter which of the 4 values you can assign to CD, the expression is always AB'. For instance C=0, D=0. The expression is 0+0+AB'+0 . Ratch

11. ### Ratch New Member

Mar 20, 2007
1,068
3

(A + B)' + (A' + B')' = A'B'+AB by DeMorgan's theorem. It cannot be simplified further.

y(wz' + wz) + xy = wyz'+wyz+xy

= wyz'(x+x')+wyz(x+x')+xy(w+w')

=wxyz'+wx'yz'+wxyz+wx'yz+wxy+w'xy

=wxyz'+wx'yz'+wxyz+wx'yz+wxyz+wxyz'+w'xyz+w'xyz'

=wxyz'+wx'yz'+wxyz+wx'yz+w'xyz+w'xyz'

SOP'S = 14 10 15 11 7 6

Using a 4 variable K-map or the Quine McClusky tabulation method we get xy+wy .

xy + x'z + yz = xy(z+z')+x'z(y+y')+yz(x+x')

=xyz+xyz'+x'yz+x'y'z+xyz+x'yz

=xyz+xyz'+x'yz+x'y'z

SOP = 7 6 3 1

Using a 3 variable K-map or the Quine McCluskey tabulation method we get x'z+xy .

Ratch

Last edited: Jul 20, 2008
12. ### Ratch New Member

Mar 20, 2007
1,068
3
Stacey,

The best way is to simplify it.

F = A'BC + ABC + AB + AC' {multiplying (A' +A) by the first term}

Continuing further F = A'BC + ABC + ABC + ABC' + ABC'+AB'C'

= AB'C + ABC +ABC' + AB'C'

SOP = 5 7 6 4

Using a 3 variable K-map or the Quine-McCluskey tabulation method we get F = A . As you can see, no matter what values we assign to B or C, the value is always A .

Now we have.

Ratch

13. ### mps New Member

Feb 4, 2010
8
0

Now,
AB'(D'+C'D)+AB'C [A+B'C=(A+B')(A+C)]
AB'[(D+D')(D'+C')]+AB'C
AB'[D'+C']+AB'C
AB'(C'+C)+AB'D'
AB'+AB'D'
AB'(1+D') [1+A'=A]
=AB'

14. ### mps New Member

Feb 4, 2010
8
0

1) Remember that
A'+B'=(AB)'
(A+B)'=A'B'
Now
(A + B)' + (A' + B')'
[(A+B)(A'+B')]'
[AB+A'B]'
B'

2) y(wz' + wz) + xy
z+z'=1
yz+xy
y(x+y)

3) xy + x'z + yz
xy+x'z+(x+x')yz
x'z(1+y)+xy(1+z)
[1+x=1]
x'z+xy