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Math Help - Boolean Algebra Problem

  1. #1
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    Boolean Algebra Problem

    I've just started learning boolean algebra at university and am stuck by the following problem:
    Prove a.b + a'.c = (a+c).(a'+b) // Corrected from previous typo

    Could somebody please give me a push in the right direction.

    Many Thanks
    Last edited by StaryNight; October 12th 2010 at 09:25 AM.
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  2. #2
    A Plied Mathematician
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    As written, your equation is false. Set a = 0, b = 0, c = 1. Then ab + a'c = 0 + 1 = 1. However, ac(a'+b) = 0(a'+b) = 0.

    Perhaps you have a typo in there somewhere?

    [EDIT]: OP corrected ac(a'+b) to (a+c)(a'+b) for the RHS.
    Last edited by Ackbeet; October 12th 2010 at 09:28 AM. Reason: OP corrected.
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  3. #3
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    Sorry, I meant to type

    Prove a.b + a'.c = (a+c).(a'+b)
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  4. #4
    A Plied Mathematician
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    That I'll buy. What techniques are available to you?
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  5. #5
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    It basically says to do it using 'Boolean Algebra'. We've studied DeMorgan's theorem and K-maps.
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  6. #6
    A Plied Mathematician
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    Ok. I would probably go with the K-map method. In order to do that, you're going to need to expand out the RHS so that it's in disjunctive normal form. What do you get when you do that?
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  7. #7
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    I've managed to prove it by firstly writing a truth table and finding the inverse fuction and then applying De Morgan and using absorbtion. However, from the question I was supposed to prove the identity entirely algebraically.
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  8. #8
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    Here's one approach. Start with

    a'b'c+a'bc+abc'+abc, and simplify.

    However, it is also true that

    a'b'c+a'bc+abc'+abc=(a'b'c'+a'bc'+ab'c'+ab'c)'.

    Simplify this RHS.

    It's the step

    a'b'c+a'bc+abc'+abc=(a'b'c'+a'bc'+ab'c'+ab'c)'

    that you must justify. It follows, ultimately, because of the law of the excluded middle. You've basically exhausted all 8 possible truth assignments of the variables.

    I'm not sure how much sense that makes in my mind, but I know what I'm trying to say.
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