# Thread: Simplifying Boolean Equations

1. ## Simplifying Boolean Equations

Hi

The following question i don't understand why it is incorrect.

1) (z+v'(w'+x(v+z'))'

z'v+w(v'z)+x

(wv'z+x)(w+vx)

(x+w)(x+v')(x+z)(w+r)(w+x)

(x+v')(x+z)(w+v)(w+x)

(x+v'z)(w+vx)

xw+vx+v'zw

Book's Answer is z'(v+x'w)

2) [c(u+v')]'[c'(u+v)]'

(c'+u+v')(c+u'v')

(c'+u+v')(c+u')(c+v')

v'+((c'+u)c)(c+u')

(v'+cu)(c+u')

v'c+v'u'+cu

Book's Answer is u'v'+uc

Hope someone can tell me where my mistake is.

P.S

2. In #1, I think your parenthesis placement is not careful enough. I also think you're trying to skip too many steps. Do one at a time! For example, assuming you meant
(z+v'(w'+x(v+z')))' in your first problem, I would go about it like this:

(z+v'(w'+x(v+z')))'
(z+v'(w'+xv+xz'))'
(z+v'w'+v'xv+v'xz')'
(z+v'w'+v'xz')'

Now implementing that last negation is a bit problematic. I don't think you can do it the way you tried to do it in your solution. Instead, I think you have to work like this:

(z+v'w'+v'xz')'
z'(v'w')'(v'xz')'

Then you work out the individual pieces there. Does that make sense?

Perhaps you can finish?

Your answer to #2 is similarly flawed. You're not using DeMorgan correctly.

3. ok i have redone the problems and this is what i have done, but i still cannot get the correct answer.

1)

(z+v'w'+v'xv+v'xz')'
z(v+w)(v+x'z)
z'(v+vx'+v'z+wv+wx'+wz)
z'v+z'vx'+z'v'z+wvz'+wx'z'+wzz'
z'v+z'vx'+wvz'+wx'z'
z'v+wvz'+z'x'v+z'x'w
z'v+z'x'(v+w)
z'v+z'x'v+z'x'w

2)

(cu+cv')((c+u')(c+v'))
(cu+cv')(cc+cv'+u'c+u'v')
cuc+cucv'+u'ccu+u'v'cu+cv'c+cv'+cv'uc+cv'u'v'
cu+cuv'+cv'+cv'cu'
cu+cv'

4. No, no, no. The way you are taking the negation of all the stuff in the parentheses (i.e., going from Step One to Step Two in Problem #1) is incorrect. It should be as follows:

(z+v'w'+v'xv+v'xz')'
(z+v'w'+v'xz')' (the v'xv term is always false, because it's contradictory; hence you can leave it out)
z'(v'w')'(v'xz')'
z'(v+w)(v+x'+z) Here's where you made your mistake: you need v+x'+z, not v+x'z.

Your solution to #2 seems very problematic as well. Slow down, man! You skip way too many steps. You need to be more careful with DeMorgan's laws.

Try this for the first few steps:

[c(u+v')]'[c'(u+v)]'
[cu+cv']'[c'u+c'v]'
(cu)'(cv')'(c'u)'(c'v)'
(c'+u')(c'+v)(c+u')(c+v')
Then continue... (try foiling next).

5. ok i got question 1

For question 2 i wrote the incorrect equation its meant to be [c(u+v')']'[c'(u+v)]'

[c+u'v]'[c'u+c'v]'
[c(u'v)]'[(c+u')(c+v')]
[c'+u+v'][(c+cv'+u'c+u'v')]c'u'
c'u'v'+uc+v'c+v'u'c+v'u'
v'u'+v'c+uc

6. Hello Paymemoney
Originally Posted by Paymemoney
ok i got question 1

For question 2 i wrote the incorrect equation its meant to be [c(u+v')']'[c'(u+v)]'

[c+u'v]'[c'u+c'v]'
[c(u'v)]'[(c+u')(c+v')]
[c'+u+v'][(c+cv'+u'c+u'v')]c'u'
c'u'v'+uc+v'c+v'u'c+v'u'
v'u'+v'c+uc
You're still trying to do too many steps once, and writing nonsense as a result. Take your time.

In fact, you have ended up with a correct expression, but the working I have shown in red is incorrect. I have really looked at any working after the first two lines

Here's one possible solution. I've indicated in green which negation sign(s) I'm removing each time.

[c(u + v')']'[c'(u + v)]'

= [c' +(u + v')][c + (u + v)']

= [c' + u + v'][c + u'v']

= cc' + c'u'v' + cu + uu'v' + cv' + u'v'

= c'u'v' + cu + cv' + u'v', noting that cc' = uu' = 0

= u'v'(1 + c) + cu + cv'

= u'v' + cu + cv', noting that 1 + c = 1

To get the answer in the book, note that cv' can be written cv'(u + u'), since u + u' = 1. So we get:

u'v' + cu + cv'(u + u')

= u'v' + cu + cuv' + cu'v'

= u'v'(1 + c) + cu(1 + v')

= u'v' + cu

7. How do you get the plus sign in front of the c when you have already canceled the negative signs (two negative signs = positive)

[c+(u+v')']'???

8. Hello Paymemoney
Originally Posted by Paymemoney
How do you get the plus sign in front of the c when you have already canceled the negative signs (two negative signs = positive)

[c+(u+v')']'???
I'm not sure that I understand your question. Which 'plus sign in front of the c'? In the first few lines of working there is no c with a plus sign in front.

And nowhere in my working does this expression
[c+(u+v')']'
appear.

If you're going to make a success of Boolean Algebra, you'll have to learn to be meticulously accurate in writing down exactly what you mean. There's no room for even the smallest error.

Perhaps you mean: how did I simplify the first bracket in the first line of my working by removing its last negative sign? In other words, how did I do:
[c(u + v')']'[...]

=[c' + (u + v')][...]
If that is what you meant, then replace (u + v') by w. So we get:
[c(w)']'[...]

=[c' + ((w)')'][...]

=[c' + w]