# (xy' + w'z)(wx' + yz') boolean simplification

Discussion in 'Homework Help' started by 4thLaw_of_Robotics, Mar 26, 2008.

1. ### 4thLaw_of_Robotics Thread Starter New Member

Mar 26, 2008
2
0
Hi. This is driving me nuts. I have a text book with several problems such as the one below, but the book only has like two examples total. That does not seem to be enough to really demonstrate good methodology for such a challenging topic.

Here is the problem.
F(w, x, y, z) = (xy' + w'z)(wx' + yz') please simplify and show the rules you apply.

Here is what I tried.
F(w, x, y, z) = (xy' + w'z)(wx' + yz') [given]
F(w, x, y, z) = xy'wx + xy'y'z + w'wzx + w'yz'z [distributive]
F(w, x, y, z) = y'w + xz' + zx'+ w'y [inverse]

Now i am stuck...it seems to me that it can be further simplified, I just can't see what rule to use to do it.

Also, while I'm at it.. I would like to double check this one.
F(x, y, z) = x'y + xyz' +xyz [given]
F(x, y, z)= x'y + xy(z'+z) [distributive]
F(x, y, z)= x'y + xy(1) [inverse]
F(x, y, z)= x'y +xy [identity]
F(x, y, z)= y(x'+x) [distributive]
F(x, y, z)= y [inverse, then identity again]

I have the feeling that I overcomplicated that one. Any tips?

2. ### veritas Active Member

Feb 7, 2008
167
0
i think there's a couple typo's in your first step. Distributing the terms through, I get:
w.x'.x.y' + w.w'.x'.z + y.z'.x.y' + y.z'.w'.z

All of the terms then cancel because A.A' = 0

*edit*
you can also use de morgan's law to go:
( (x.y')' . (w'.z)' )( (w.x')' . (y.z')' )
(x'.y.w.z')(w'.x.y'.z)
x'.x.y'.y.z'.z.w'.w

and you get the same result

3. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
I agree, your distributive step is wrong.

Dave

4. ### 4thLaw_of_Robotics Thread Starter New Member

Mar 26, 2008
2
0
Thanks. I don't know why I didn't notice that mistake.

I guess my ratios were off.

A = Ek / -(s/c)

Accuracy = (E)xpertise with respect to knowledge / -(sleep/ coffee)