# Simplify the expression by algebraic manipulation

#### Margarie

Joined Jan 19, 2018
4

(WX + Z’)(Y’ + Z) + W’(X+Y’)

I used the association property on the first operand and it leads me nowhere further from this initial expression.

This is what I did:

(WX + Z’)(Y’ + Z) + W’(X+Y’)
WXY' + WXZ + Z'Y' + Z'Z + W'(X+Y')
WXY' + WXZ + Z'Y' + W'(X+Y')

and now I can't seem to figure out what to do next. Any help is much aprecciated!

#### LesJones

Joined Jan 8, 2017
4,013
I can see one more thing that you can expand.

Les.

#### MrChips

Joined Oct 2, 2009
29,212
Have you learned Karnaugh mapping as yet?

#### Margarie

Joined Jan 19, 2018
4
Have you learned Karnaugh mapping as yet?
Yes I have. By using it I got the expression:

XW' + W'Y' +Y'Z' + XWZ + XWY'

#### WBahn

Joined Mar 31, 2012
29,154
Yes I have. By using it I got the expression:

XW' + W'Y' +Y'Z' + XWZ + XWY'
Isn't that the same as in your first post once you distribute out the last term?

One subtle point on these kinds of question is what constitutes a "minimized" expression?

Is A(B+C) more "minimal" than AB+AC?

#### Margarie

Joined Jan 19, 2018
4
I think I have just simplified it.
I used the theorems:
1) A+A'.B = A+B
2) A.B+A'.C+BC = A.B+A'.C

And so these were the steps, if it may help someone:

XW' + W'Y' +Y'Z' + XWZ + XWY'
X(W'+WZ)+Y'(W'+WX)+Y'Z' , using 1) with X(W'+WZ) assuming (A=W', A'=W, B=Z) and with Y'(W'+X) assuming (A=W', A'=W, B=X):
X(W'+Z)+Y'(W'+X)+Y'Z'
XW'+XZ+Y'W'+Y'X+Y'Z' , and using 2) with A=Z, A'=Z', B=X, C=Y':
XW'+Y'W'+XZ+Y'Z'

This was the simplest I got. Thank you guys and I hope it's correct, even tho I checked on logisim and it was equivalent.