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Boolean algebra - distribution

  1. Sep 12, 2016 #1

    Rectifier

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    The problem
    I am trying to show that ##a'c' \vee c'd \vee ab'd ## is equivalent to ## (a \vee c')(b' \vee c')(a' \vee d) ##

    The attempt
    ## (a \vee c')(b' \vee c')(a' \vee d) \\ (c' \vee (ab'))(a' \vee d)##
    The following step is the step I am unsure about. I am distributing the left parenthesis over the right.
    ## (c' \vee (ab'))(a' \vee d) \\ c'a \vee c'd \vee ab'a' \vee ab'd##

    I am almost there but the term ##ab'a'## differs. I suspect that you can remove that term since it does not afect the output because it is always false. What do you say about that?
     
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  3. Sep 12, 2016 #2

    micromass

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    Can you show ##ab'a' = 0##?
     
  4. Sep 12, 2016 #3

    Rectifier

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    Unfortunately, I cant, but I know it is true for all values of 1 and 0 for a and b. Since ##a## is the opposite of ##a'## this means that ##a \wedge a'## will always be 0 and it does not matter which value b has . I can write a truth table but I am on my phone so its a bit hard.
     
  5. Sep 12, 2016 #4

    micromass

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    What are your axioms for Boolean algebra?
     
  6. Sep 12, 2016 #5

    Rectifier

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    Are you thinking about ##x \cdot x' = 0## ?
     
  7. Sep 12, 2016 #6

    micromass

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    I don't know, how did you define a boolean algebra?
     
  8. Sep 12, 2016 #7

    Rectifier

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    In boolean algebra variables can only have two values.
     
  9. Sep 12, 2016 #8

    micromass

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    Either you tell me how you defined a boolean algebra, or I'm out of this thread.
     
  10. Sep 12, 2016 #9

    Rectifier

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    I am sorry that my answer didn't fit you but I cant come up with anything better. In any case, I am thankful for the help I have received so far.
     
  11. Sep 12, 2016 #10

    micromass

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    How are you supposed to prove anything about a Boolean algebra if you're not given its definition or anything??
     
  12. Sep 12, 2016 #11

    Rectifier

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  13. Sep 12, 2016 #12

    micromass

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    Those are axioms for a structure that is called a bounded lattice. It's not the axioms for a Boolean algebra. Usually, there are 10 axioms for a Boolean algebra. They can be found here: https://en.wikipedia.org/wiki/Boolean_algebra_(structure)#Definition

    Note that in this case, it is an axiom that ##x\wedge x' = 0##. This is why I asked for your definition, since you might adopt different (but equivalent) ways of doing things.
     
  14. Sep 12, 2016 #13

    Rectifier

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    I didnt think they were different :). Thank you for you help kind stranger!

    So now I can basically remove that ab'a'. Halleluja, I am done with that problen.
     
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