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Thread: Proving DeMorgan Law?

  1. #1
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    Proving DeMorgan Law?

    Prove that (A $\displaystyle \cup$ B)c = Ac$\displaystyle \cap$Bc (Demorganís Law)
    a)
    1) Let x be an arbitrary element of (A$\displaystyle \cup$B)c
    2) x $\displaystyle \notin$ (A$\displaystyle \cup$B) (By Definition of Complement)
    3) x $\displaystyle \notin$ A but x $\displaystyle \in$ Ac (Definition of Union And Definition of Complement)
    4) x $\displaystyle \notin$ B but x $\displaystyle \in$ Bc (Definition of Union And Definition of Complement)

    5) x $\displaystyle \in$ Ac, x $\displaystyle \in$ Bc <-> x $\displaystyle \in$Ac$\displaystyle \cap$Bc

    6) But x is an arbitrary element of (AUB)c

    7) Thus (A $\displaystyle \cup$ B)c $\displaystyle \subseteq$ Ac$\displaystyle \cap$Bc

    b) And Conversely
    1) Let y be an arbitrary element of Ac $\displaystyle \cap$ Bc

    2) y $\displaystyle \in$ Ac, y $\displaystyle \in$ Bc (By Definition of intersection)

    3) y $\displaystyle \notin$ A, y $\displaystyle \notin$ B (Definition of Complement)

    4) y $\displaystyle \notin$ A U B (From b(3))

    5) y $\displaystyle \in$ (A U B)c

    6) But y is an arbitrary element of Ac $\displaystyle \cap$ Bc

    7) Thus Ac $\displaystyle \cap$ Bc $\displaystyle \subseteq$ (A $\displaystyle \cup$ B)c

    From a7 and b7 we have

    (A $\displaystyle \cup$ B)c = Ac $\displaystyle \cap$ Bc

    What is wrong in above statements? any help please?
    Thank You.

    Note: Solution is also atatched in doc file.
    Attached Files Attached Files
    Last edited by mrsenim; Feb 27th 2010 at 12:37 AM. Reason: Sorry I needed to post this
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