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Math Help - [SOLVED] Boolean Algebra

  1. #1
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    Exclamation [SOLVED] Boolean Algebra

    Assume that B is a Boolean algebra with operations + and .
    For each true statement below, give a detailed proof. For each false statement below, write
    out its negation, then give a proof of the negation.

    (∀a, b ∈ B) (a + b = 1 ↔ b ā =
    ā)
    (∀a, b ∈ B) (a b = 0 b + ā = ā)
    (∀a, b ∈ B) (a + b = a + c → b = c)

    for the last one, i had the same proof as http://www.mathhelpforum.com/math-he...b-b-c-b-c.html
    but it just doesn't seem right to me...
    completely stumped on the first two .

    please help!!!!
    thanks!!!
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    South Coast of England
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    Boolean Algebra Proofs

    Hello starman_dx
    Quote Originally Posted by starman_dx View Post
    Assume that B is a Boolean algebra with operations + and .
    For each true statement below, give a detailed proof. For each false statement below, write
    out its negation, then give a proof of the negation.

    (∀a, b ∈ B) (a + b = 1 ↔ b ā =
    ā)
    (∀a, b ∈ B) (a b = 0 b + ā = ā)
    (∀a, b ∈ B) (a + b = a + c → b = c)

    for the last one, i had the same proof as http://www.mathhelpforum.com/math-he...b-b-c-b-c.html
    but it just doesn't seem right to me...
    completely stumped on the first two .

    please help!!!!
    thanks!!!
    Here are the proofs you need. I'll leave you to supply the names of the Laws I'm using at each stage.

    (1) a+b= 1

    \rightarrow (a+b)\cdot \bar{a} = 1\cdot\bar{a}

    \rightarrow a\cdot\bar{a} +b\cdot\bar{a} = \bar{a}

    \rightarrow 0+b\cdot\bar{a} = \bar{a}

    \rightarrow b\cdot\bar{a} = \bar{a}

    And  b\cdot\bar{a} = \bar{a}

    \rightarrow b\cdot\bar{a} + a = \bar{a}+a

    \rightarrow (b+a)\cdot(\bar{a}+a) = 1

    \rightarrow (a+b).1 = 1

    \rightarrow a+b = 1

    (2) a\cdot b=0

    \rightarrow a\cdot b + \bar{a} = 0 + \bar{a}

    \rightarrow (a+\bar{a})\cdot(b+\bar{a}) = \bar{a}

    \rightarrow 1\cdot(b+\bar{a}) = \bar{a}

    \rightarrow b+\bar{a} = \bar{a}

    And  b+\bar{a} = \bar{a}

    \rightarrow a\cdot(b+\bar{a}) = a\cdot\bar{a}

    \rightarrow a\cdot b + a\cdot \bar{a} = 0

    \rightarrow a\cdot b + 0 = 0

    \rightarrow a\cdot b = 0

    (3) is not true. The negation is \exists a, b, c \in B, (a+b = a+c) \wedge (b \ne c)

    As a counter-example, consider sets A = \{1\}, B = \{2\} and C = \{1, 2\}. Then A \cup B = A \cup C, but B \ne C.

    Grandad
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  3. #3
    Newbie
    Joined
    Feb 2009
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    you are a life saver
    thank you!!!!
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