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Boolean Algebra prove

  1. Nov 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Untitled.jpg

    2. Relevant equations

    Boolean Algebra

    3. The attempt at a solution
    I use the distri to change the A+C'.D and Demorgan.
    Should i use dis and Demorgan firstly like below?
    02.jpg
    Which property I should use in the next step?
    If the first part is wrong,which property i should use first?
    Sorry for the small pictures!
    Thanks
     
  2. jcsd
  3. Nov 20, 2013 #2

    jedishrfu

    Staff: Mentor

    Continue to use demorgan until you get down into the simplest form then apply algebra reduction rules.

    So are you trying to prove:

    (A + C'D)' + A + (CD + AB)' = 1

    Just want to make sure we understand the problem as the bar above terms in your images looks like it might include the A term.
     
    Last edited: Nov 20, 2013
  4. Nov 21, 2013 #3
    Sorry for the vague picture.The bar not include in the A term.
    Should i use the Demorgan in the whole question firstly including the A.
    As the picture shows,then,i continue to use Distributive or Absorption,but failed.
     
  5. Nov 21, 2013 #4

    jedishrfu

    Staff: Mentor

    SHow your work and we can help you along the way.

    In your final expression do you have any terms like this?

    X + X' + ...... =?= 1
     
  6. Nov 22, 2013 #5
    I continue to calculate following the picture show on the The attempt at a solution

    First:
    A'(C+D')+(A+C'+D')(A+A'+B') Use distributive
    A'(C+D')+(A+C'+D')(1+B')
    But i was wondering whether i can change to 1+B' to 1,because the list only show that 1+A=1

    Second:
    A'(C+D')+A+A'C'+B'C'+A'D'+B'D'
    A'(C+D')+A+A'(C'+D')+B'(C'+D')
     
  7. Nov 22, 2013 #6

    jedishrfu

    Staff: Mentor

    It seems you are almost there. The last expression

    A'(C+D')+A+A'(C'+D')+B'(C'+D')

    can be expanded into

    A'C + A'D' + A + A'C' + A'D' + B'C' + B'D'

    and then combine some terms to get:

    A'(C + C') + A + A'D' + B'C' + B' D'

    Do you see how to finish it?

    HINT: remember that X' + X = 1 and that 1+ANYTHING = 1 in boolean algebra
     
  8. Nov 22, 2013 #7
    a'+ a + a'd' + b'c' + b' d'
    1+a'd' + b'c' + b' d'
    1
     
  9. Nov 22, 2013 #8

    jedishrfu

    Staff: Mentor

    Thats great!

    It floored me too when I got to the second to last step and thought how am I going to reduce that and then I saw the A + A' terms
    and realized the other terms just didn't matter.
     
  10. Nov 22, 2013 #9

    Thanks you!!!
     
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