Help with proving boolean algebra identity

Discussion in 'Homework Help' started by pghaffari, Oct 1, 2011.

  1. pghaffari

    Thread Starter New Member

    Oct 1, 2011
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    Prove: a' + a(a'b + b'c)' = a' + b + c'

    This is what I get so far for the LHS

    a' + a(ab'bc') --> a' + ab'bc' .. So we know b'b is always 0 so does that mean we can just ignore it? ---> a' + ac' --> a' + c' .. but i am missing a b somewhere..


    also problem 2:

    (a'b' + c)(b+a)(b' + ac) = a'bc

    my question is.. can you foil boolean algebra out like in regular algebra? like for this example can i do:

    a'b'b + a'ab' + cb + ac ... etc? or no??

    please advise! thanks

    :eek:
     
    Last edited: Oct 1, 2011
  2. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    831
    bb' is always 0 so you can ignore it in an OR function
    I would perform a different set of operations on your left-hand side first, then apply DeMorgans from there. Note that c = c'' by the way!
     
  3. pghaffari

    Thread Starter New Member

    Oct 1, 2011
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    0
    What else can i do for problem 1?

    a' + a(a'b + b'c)' --> can I use this law: (a + a'b ) = a+b , so that it becomes

    a' + (a'b + b'c)' ? even if i did it this way, I get same thing:

    a' + (ab'bc') --> which makes the inside always 0 and since its an AND function the whole thing is 0? so its just a' .. what am i doing wrong?
     
  4. pghaffari

    Thread Starter New Member

    Oct 1, 2011
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    Ok I got the problems up. Thanks.

    How about this new one:

    wxy + w'x(yz+yz') + x'(zw+zy') + z(x'w' + y'x) = xy+z ... lol so confusing!!!

    this is what i have at the end..

    xy + zw' + zx' + y'x' + y'z

    How do i reduce this any further??
     
    Last edited: Oct 1, 2011
  5. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    Since you haven't posted your full operation progression and I have come to a different conclusion, I will assume that you have done something wrong in between.

    Please post all of your operations.
     
  6. pghaffari

    Thread Starter New Member

    Oct 1, 2011
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    0
    wxy + w'x(yz+yz') + x'(zw+zy') + z(x'w'+y'x)
    wxy + w'x(zy+y) + x'(zw+y') + z(xy' + w')
    w'xyz + wx'z + x'y' + xy'z + w'z
    xy + z(w'(xy+q) + wx' + xy') + x'y'
    xy + z(w' + wx' + xy') + x'y' -> xy + w'z + wx'z + wy'z + x'y'
    xy + w'z + wx'z + y' (xz+x') -> xy + w'z + wx'z +y'(x'+z)
    xy + z(x'w+w') + y'(x'+z) -> xy + z(w'+x') + y'(x'+z)
    xy + zw' + zx' +y'x' + y'z
    xy + z' + zw' + y'x' + y'z
    xy+z+w'z+x'y' + z
    xy+z+wz'+x'y'
    xy+z+w+y'

    stuck here
     
  7. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    You might have understood something wrong, but all of your parentheses in the second line are wrong.

    yz+yz'=y(z+z')=y*1=y
    Similarly
    zw+zy'=z(w+y')
    and
    x'w'+y'x
    cannot be simplified.

    Care to take another look into your theory and give it another shot?
     
  8. pghaffari

    Thread Starter New Member

    Oct 1, 2011
    13
    0
    wxy + w'x(y) + x'(z(w+y')) + z(x'w' + y'x)
    xy(w+w') + x'wz + x'y'z + x'w'z + xy'z
    xy + z(x'y' + xy') + z(x'w + y'w')
    xy + z(y'(x'+x)) + z(x'(w+w'))
    xy + y'z + x'z
    xy + z(x' + y')

    im so close.. where di i go wrong? i have to get final result xy + z....

    AH i got it nevermind!! it just becomes xy + z(xy)' ==> xy + z !!
     
    Last edited: Oct 5, 2011
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