How is the demonstration of this? $\displaystyle (A\bigtriangleup B)\bigtriangleup C = A\bigtriangleup (B\bigtriangleup c) $ knowing that $\displaystyle (A\bigtriangleup B) = (A/B)\cup(B/A) $
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See here PlanetMath: proof of the associativity of the symmetric difference operator
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