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Thread: symetric difference

  1. August 19th 2007, 06:51 PM #1
    voidstuff
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    Talking symetric difference

    How is the demonstration of this?


    <br /> (A\bigtriangleup B)\bigtriangleup C = A\bigtriangleup (B\bigtriangleup c)<br />







    knowing that
    <br /> (A\bigtriangleup B) = (A/B)\cup(B/A)<br />
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  2. August 19th 2007, 09:34 PM #2
    red_dog
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    PlanetMath: proof of the associativity of the symmetric difference operator
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