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Math Help - set operations 2

  1. #1
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    set operations 2

    use the law of algebra of sets to that emptyset=((X union Y) intersection (X union Ycomplement)) intersect ((Xcomplement union Y) intersection (Xcomplement union Ycomplement)) please specify which you use in the derivation....


    ANDANDANDANDANDANDANDANDANDANDANDANDANDANDANDANDAN DANDAND

    what does it mean by |A * A| where A={1,2,3,4,5}

    small explainations would be helpful

    thanks in advance
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  2. #2
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    Lexington, MA (USA)
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    Hello, npm1!

    I don't know the names of your Rules and Theorems.
    I'll make up some names:

    . . \begin{array}{cccccc}\text{Rule} & \text{Na{m}e} \\ \hline \\[-3mm]<br />
A \cap A' \:=\:\emptyset & \text{[C1]} \\ \\[-3mm]<br />
A \cup \emptyset \:=\:A & \text{[C2]} \end{array}



    Use the laws of algebra of sets to prove that:

    . . (X \cup Y) \cap (X \cup Y') \cap (X' \cup Y) \cap (X' \cup Y') \;=\;\emptyset

    \begin{array}{ccccc}(X \cup Y) \cap (X \cup Y') \cap (X' \cup Y) \cap (X' \cup Y') & \text{Given} \\ \\[-3mm]<br /> <br />
\bigg[X \cup (Y \cap Y')\bigg] \cap \bigg[X' \cup (Y \cap Y')\bigg] & \text{Distr.} \\ \\[-3mm]<br /> <br />
\bigg[X \cup \emptyset\bigg] \cap \bigg[X' \cup \emptyset\bigg] & \text{[C1]} \\ \\[-3mm]<br /> <br />
X \cap X' & \text{[C2]} \\ \\[-3mm]<br /> <br />
\emptyset & \text{[C1]}<br />
\end{array}<br />


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  3. #3
    Newbie
    Joined
    Apr 2010
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    Cool thanks u

    very clear
    ((X \cup Y) \cap (X \cup Y')) \cap ((X' \cup Y) \cap (X' \cup Y')) \;=\;\emptyset (new to forum and don't how to use the math tags)

    the laws i've been given for set operations is.....
    idempotent laws
    associative laws
    commutative laws
    distributive laws
    identity laws
    Involution laws
    involution laws
    complement laws
    de morgans laws
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