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Thread: Boolean Simplify : (x+y)'(x'+y') = ?

  1. #1
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    Boolean Simplify : (x+y)'(x'+y') = ?

    Hello,

    I have a question about simplifying this expression:

    Here is my progress, but i cannot seem to get the answer --> x'y'

    (x+y)'(x'+y')

    = (x'y') (x'+y')
    = (x'y' * x') + (x'y' * y')
    = (x''y') + (x'y'')
    = (xy') + (x'y)
    = ........................

    Where have i gone wrong ?

    Thanks.
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  2. #2
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    Re: Boolean Simplify : (x+y)'(x'+y') = ?

    x'x' does not equal x

    x'x' = x'
    Thanks from anpk05
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  3. #3
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    Re: Boolean Simplify : (x+y)'(x'+y') = ?

    Quote Originally Posted by anpk05 View Post
    I have a question about simplifying this expression:
    Here is my progress, but i cannot seem to get the answer --> x'y'
    (x+y)'(x'+y') = (x'y') (x'+y')
    = (x'y' * x') + (x'y' * y')
    = (x''y') + (x'y'')
    = (xy') + (x'y)
    $ \begin{align*}(x+y)'(x'+y')&=(x'y')(x'+y')\\&=(x'y 'x'+x'y'y')\\&=(x'x'y'+x'y'y')\\&=(x'y'+x'y')\\&=( x'y')\end{align*}$
    Last edited by Plato; Sep 5th 2017 at 04:46 PM.
    Thanks from anpk05
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  4. #4
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    Re: Boolean Simplify : (x+y)'(x'+y') = ?

    Quote Originally Posted by abender View Post
    x'x' does not equal x

    x'x' = x'
    Thanks for that, i don't know how i came up with that. Since i knew x * x = x.
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  5. #5
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    Re: Boolean Simplify : (x+y)'(x'+y') = ?

    Quote Originally Posted by Plato View Post
    $ \begin{align*}(x+y)'(x'+y')&=(x'y')(x'+y')\\&=(x'y 'x'+x'y'y')\\&=(x'x'y'+x'y'y')\\&=(x'y'+x'y')\\&=( x'y')\end{align*}$
    How does (x'y' + x'y' ) = (x'y') ?
    Please explain

    Thanks
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  6. #6
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    Re: Boolean Simplify : (x+y)'(x'+y') = ?

    The property that i know is, (x+y)' = (x'y')

    But how can (x'y' + x'y') = (x'y') ? hmm......
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  7. #7
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    Re: Boolean Simplify : (x+y)'(x'+y') = ?

    Quote Originally Posted by anpk05 View Post
    The property that i know is, (x+y)' = (x'y')

    But how can (x'y' + x'y') = (x'y') ? hmm......
    Suppose z is (x'y').

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

    (x' AND y') OR (x' AND y') = (x' AND y')
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