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Check my Boolean Algebra!! I think it's right...?

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Am I doing this right? The question is...
    (ab) + (a' + b') = 1

    2. Relevant equations
    (a) Commutative a · b = b · a a + b = b + a
    (b) Associative (a · b) · c = a · (b · c) (a + b) + c = a + (b + c)
    (c) Distributive a · (b + c) = (a · b) + (a · c) a + (b · c) = (a + b) · (a + c)
    (d) Identity a · 1 = a a + 0 = 0
    (e) Negation a + a' = 1 a · a' = 0
    (f) Double negative (a')' = a
    (g) Idempotent a · a = a a + a = a
    (h) DeMorgan’s laws (a · b)' = a' + b' (a + b)' = a' · b'
    (i) Universal bound a + 1 = a a · 0 = 0
    (j) Absorption a · (a + b) = a a + (a · b) = a
    (k) Complement of 1 and 0 1' = 0 0' = 1

    3. The attempt at a solution
    My Steps....
    (ab) + (a' + b') = (ab + a') + b' [Associative]
    = (a' + ab) + b' [Commutative]
    = (a' + a)(a' + b) + b' [Distributive]
    = 1(a' + b) + b' [Negation]
    = (a' + b) + b' [Identity]
    = a' + (b + b') [Associative]
    = a' + 1 [Negation]
    = 1 [Absorbtion]
     
  2. jcsd
  3. Oct 19, 2015 #2

    andrewkirk

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    How are you justifying the third step? It doesn't look like a straightforward application of the Distributive law to me.
     
  4. Oct 19, 2015 #3
    I'm sorry. I probably should of spaced the relevant equations better. According to the properties in my textbook, one way the distributive property can be written as is...
    a + (bc) = (a+b)(a +c)
    so I'm treating a as my 1st part and bc as my 2nd part.
    Now to my work... in the previous step, I have (a' + ab) + b' and I think I can use the distributive property stated to say
    (a' + ab) + b' = (a' + a)(a' + b) + b'
    if I consider a' as my first part and ab as my 2nd part. I hope that makes sense...
     
  5. Oct 19, 2015 #4

    andrewkirk

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    Yes, that looks OK.
     
  6. Oct 19, 2015 #5
    Ok, cool. Thanks! :smile:
     
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