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Boolean Algebra

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that the left equals the right.

    a((bc)'d+b)+a'b=(a+b)(b+d)



    3. The attempt at a solution

    ab'd+adc'+ab+a'b distribution
    ab'd+adc'+b complementary
    I don't know what to do next, or what i should look for to give me a clue as to how to progress.
     
  2. jcsd
  3. Sep 11, 2013 #2

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    A Karnaugh map (K-map) will prove useful.

    (By the way, you can reduce both sides of the equation, not just the left side. :wink:)
     
  4. Sep 11, 2013 #3
    we aren't supposed to use the karnaugh map yet, and they only want us to manipulate the left side.
     
  5. Sep 11, 2013 #4

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    There are a couple of uses of the redundancy law that you can use.

    x + x'y = x + y

    After that, you can use it again in a different way,

    x + xy = x

    And you'll reach the minimal solution.

    But since the right hand side of the equation is not at its minimal form, you'll have to actually go backwards after that to get the (a+b)(b+d). That is, if you are not allowed to simplify the right hand side.
     
  6. Sep 11, 2013 #5
    So, I would basically want the minimal on the left (before going a little backwards) to be ab+ad+b+ad?
     
  7. Sep 11, 2013 #6

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    It's even simpler than that! :smile:

    Notice that you have "ad" terms in there twice. You can just get rid of one of them (x + x → x). :wink: (But don't get rid of both them, of course).

    And then there's still more you can do. There's a "b" all by itself, and then there's another term with "b" in it. Use the redundancy rule,
    x + xy → x.
     
  8. Sep 11, 2013 #7
    Got it! Thank you very much.
     
  9. Sep 11, 2013 #8

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    Oh, wait. Yes, since you're going backward, you'll eventually want to get it into a form that you can factor into (a + b)(b + d) such as, ab + ad + bb + bd.

    The point of my last post is that ab+ad+b+ad is still not minimal. That's all I meant.
     
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