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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- Aug 19th 2007, 06:51 PMvoidstuffsymetric difference
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)

$ - Aug 19th 2007, 09:34 PMred_dog