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    I was wondering is this valid for sets A and D.

    \neg (x \in (A \cup D)) = x \in \neg (A \cup D) = x \in (\neg A \cap \neg D)

    I think no.

    But is this true instead

    \neg (x \in (A \cup D))= x \in (A^c \cap D^c)

    I still think this has problems. Hmm.

    Its trivial that(well, it seems to work when I do it in truth tables)

    \neg (x \in (A \cup D)= x \not \in A and \not \in D

    but thats just from truth tables. Anyone, know to show it without truth tables.

    (Original post by Simplicity)
    I was wondering is this valid for sets A and D.

    \neg (x \in (A \cup D)) = x \in \neg (A \cup D) = x \in (\neg A \cap \neg D)

    I think no.
    \neg (x \in (A \cup D)) = x \in \neg (A \cap D)

    Should be correct.
    When negating through, unions and intersections swap(poor way to describe what I mean...)

    Edit- Although, the notion of being in ¬(AnD) is a bit suspect anyways, you'd be better off using the sign epsilon with a line through it.
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Updated: October 9, 2009

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