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16 December, 09:46

Show that the € operator is associative. That is, using the definition of the € operator and the rules of Boolean Algebra, show that the following equality holds: ((g1, p1) € (g2, p2)) € (g3, p3) = (g1, p1) € ((g2, p2) € (g3, p3))

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  1. 16 December, 11:21
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    (g1, p1) € (g2, p2) € (g3, p3) = (g1, p1) € (g2, p2) € (g3, p3)

    = (g1, p2) € (g2, p3) € (g3, p1)

    = (g1, p3) € (g2, p1) € (g3, p2)
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