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Math Help - Urgent pleaseeeeeee

  1. #1
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    Lightbulb Urgent pleaseeeeeee

    Use the laws of Boolean algebra to show that the identity (xy V x'y')' = x'y V xy' holds. Give all necessary steps and state explicitly which laws you are using. You may use de Morganís laws without deriving them yourself. I have no idea of how to even start or which of the laws to use. Please help!!!
    Last edited by juliie; February 3rd 2013 at 02:05 AM.
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    Super Member ILikeSerena's Avatar
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    Re: Urgent pleaseeeeeee

    Quote Originally Posted by juliie View Post
    Use the laws of Boolean algebra to show that the identity (xy V x'y')' = x'y V xy' holds. Give all necessary steps and state explicitly which laws you are using. You may use de Morganís laws without deriving them yourself. I have no idea of how to even start or which of the laws to use. Please help!!!
    Hi juliie!

    The 2nd De Morgan law states: (a \vee b)' = a' \wedge b'.

    Now let a = xy and b=x'y'.
    Then
    (xy \vee x'y')' = (a \vee b)' = a' \wedge b'

    Now back substitute those expressions for a and b:

    a' \wedge b' = (xy)' \wedge (x'y')'


    This is the first step.
    Can you apply the 1st law of the De Morgan on (xy)' and also on (x'y')'?
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    Re: Urgent pleaseeeeeee

    Hey thanks for that! Problem is I don't understand how to use the laws at all, I tried but its not making sense to me
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    Re: Urgent pleaseeeeeee

    That's okay.
    What did you try?
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    Re: Urgent pleaseeeeeee

    Sorry but I don't understand why we apply the law to (xy)' and (x'y')' I don't know how to show that they equal each other through algebra, I did it through the table but using the laws I missed the lecture and don't have an idea of how to do it.
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    Super Member ILikeSerena's Avatar
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    Re: Urgent pleaseeeeeee

    Well, you can find the laws here.

    Can you find the 1st law of De Morgan?
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    Re: Urgent pleaseeeeeee

    yes its this one (¨x)∧(¨y) = ¨(x∨y)
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    Super Member ILikeSerena's Avatar
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    Re: Urgent pleaseeeeeee

    Oops. I mixed them up myself. I meant the 2nd law of De Morgan.

    It says: (\neg x) \vee (\neg y)= \neg(x \wedge y)

    or if we put it the other way around: \neg(x \wedge y) = (\neg x) \vee (\neg y)


    The expression (xy)' is actually the same as \neg(x \wedge y).
    It is just a different notation.

    Can you now apply the 2nd law of De Morgan to (xy)'?
    And also to (x'y')'?
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    Re: Urgent pleaseeeeeee

    so (xy)' = (x') V (y')

    but (x'y')' I don't know what to do ?
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    Re: Urgent pleaseeeeeee

    The same thing.

    Since (xy)' = (x') V (y')

    (x'y')' = ((x')(y'))' = ((x')') V ((y')') = (x'') V (y'')


    There is another law called double negation.
    Can you find it?
    And can you apply it to this last expression (twice)?
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    Re: Urgent pleaseeeeeee

    ¨¨x = x which is the same as (x')' I think.

    and so its (x'') V (y'') = x V y ?
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    Super Member ILikeSerena's Avatar
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    Re: Urgent pleaseeeeeee

    Yes! That is correct.

    So (xy \vee x'y')' = (xy)' \wedge (x'y')' = (x' \vee y') \wedge (x \vee y)

    The logical next step is to try the law of distributivity.
    Care to try?
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    Re: Urgent pleaseeeeeee

    distributive law is x∧(y∨z) = (x∧y)∨(x∧z)

    so would it be ((x' ∧ y')(x)) V ((x' ∧ y')(y))

    or does that not make sense?
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  14. #14
    Super Member ILikeSerena's Avatar
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    Re: Urgent pleaseeeeeee

    Quote Originally Posted by juliie View Post
    distributive law is x∧(y∨z) = (x∧y)∨(x∧z)

    so would it be ((x' ∧ y')(x)) V ((x' ∧ y')(y))

    or does that not make sense?
    Yep. That's basically it. Except you have replaced occurrences of ∨ by ∧, which you shouldn't have.
    It should be: ((x' ∨ y')(x)) V ((x' ∨ y')(y)).


    Can you repeat the distributive law in the left part, and also in the right part?
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    Re: Urgent pleaseeeeeee

    Quote Originally Posted by ILikeSerena View Post
    Yep. That's basically it. Except you have replaced occurrences of ∨ by ∧, which you shouldn't have.
    It should be: ((x' ∨ y')(x)) V ((x' ∨ y')(y)).


    Can you repeat the distributive law in the left part, and also in the right part?

    so does that mean I do that law for all 3 parts of (xy \vee x'y')' = (xy)' \wedge (x'y')' = (x' \vee y') \wedge (x \vee y)
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