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Math Help - Simplifying Boolean Algebra

  1. #1
    Junior Member shinhidora's Avatar
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    Simplifying Boolean Algebra

    So I'm preparing for my exam on monday and I found this exercise hanging around in my math book, been staring at it for some time now but still no clue where to start...

    Simplify the given Boolean formula:

    x' (yzu + z' + u') + (x(y'+u))'

    Edit: not sure if this was Advanced Algebra, so I just put it with Algebra
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  2. #2
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    Answer?

    Was the answer:
    x(yuz +1 +2u+y) +z ???
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  3. #3
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by Henryt999 View Post
    Was the answer:
    x(yuz +1 +2u+y) +z ???
    Since it's Boolean (0 & 1) algebra, I think it's quite impossible there's a 2 in the answer...
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  4. #4
    Moo
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    Hi

    x' (yzu + z' + u') + (x(y'+u))'<br />

    First, (x(y'+u))'=x'+yu'=x'(y+y')+yu'=x'y+x'y'+yu'

    So x' (yzu + z' + u') + (x(y'+u))'=x' (yzu + z' + u')+x'y+x'y'+yu' =x'(yzu+z'+u'+y')+x'y+yu'

    Finally, note that (yzu)'=z'+u'+y'. So yzu+z'+u'+y'=1


    ---> x' (yzu + z' + u') + (x(y'+u))'=x'+x'y+yu'
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  5. #5
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    Hello shinhidora
    Quote Originally Posted by shinhidora View Post
    So I'm preparing for my exam on monday and I found this exercise hanging around in my math book, been staring at it for some time now but still no clue where to start...

    Simplify the given Boolean formula:

    x' (yzu + z' + u') + (x(y'+u))'

    Edit: not sure if this was Advanced Algebra, so I just put it with Algebra
    Using De Morgan's Law twice:
    (x(y'+u))' = x'+(y'+u)'
    =x'+yu'
    So
    x'(yzu+z'+u') + (x(y'+u))'=x'(yzu+z'+u')+x'+yu'
    =x'+yu', since x'p+x' = x'(p+1)=x' where p = (yzu+z'+u')
    Grandad
    Last edited by Grandad; January 9th 2010 at 07:53 AM. Reason: Corrected typo
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  6. #6
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by Moo View Post
    Hi

    x' (yzu + z' + u') + (x(y'+u))'<br />

    First, (x(y'+u))'=x'+yu'=x'(y+y')+yu'=x'y+x'y'+yu'

    So x' (yzu + z' + u') + (x(y'+u))'=x' (yzu + z' + u')+x'y+x'y'+yu' =x'(yzu+z'+u'+y')+x'y+yu'

    Finally, note that (yzu)'=z'+u'+y'. So yzu+z'+u'+y'=1


    ---> x' (yzu + z' + u') + (x(y'+u))'=x'+x'y+yu'
    Thx Moo

    But was just wondering...

    x'+x'y+yu' = x' + yu' ?

    Edit: Thx Grandad aswell
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  7. #7
    Moo
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    Quote Originally Posted by shinhidora View Post
    Thx Moo

    But was just wondering...

    x'+x'y+yu' = x' + yu' ?

    Edit: Thx Grandad aswell
    Yes

    Sorry, I'm not too used to Boolean algebra
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