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Math Help - Simplifying Boolean Equations

  1. #1
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    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
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  2. #2
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    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.
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  3. #3
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    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'
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  4. #4
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    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).
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  5. #5
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    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
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  6. #6
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    Hello Paymemoney
    Quote Originally Posted by Paymemoney View Post
    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

    Grandad
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  7. #7
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    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')']'???
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  8. #8
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    Hello Paymemoney
    Quote Originally Posted by Paymemoney View Post
    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]

    Grandad


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  9. #9
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    yeh that was what i was trying to ask, i understand where my mistake was.
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