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Math Help - De Morgan's second law

  1. #1
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    De Morgan's second law

    Hi,

    how do I prove De Morgan's second law  (AB)^c = A^c \cup B^c by using De Morgan's first law:  (E \cup F)^c = E^c F^c

    Thanks!
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  2. #2
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    I do not know the why you have them numbered.
    Here is a possiblity.
    \begin{gathered}<br />
  \left( {A^c  \cup B^c } \right)^c  = \left( {A^c } \right)^c \left( {B^c } \right)^c  = AB \hfill \\<br />
  \left( {AB} \right)^c  = A^c  \cup B^c  \hfill \\ <br />
\end{gathered}
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  3. #3
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    Hello,
    Quote Originally Posted by lds09 View Post
    Hi,

    how do I prove De Morgan's second law  (A \cap B)^c = A^c \cup B^c by using De Morgan's first law:  (E \cup F)^c = E^c \cap F^c

    Thanks!
    Start from de Morgan's first law for A^c and B^c :

    (A^c \cup B^c)^c=(A^c)^c \cap (B^c)^c

    (A^c \cup B^c)^c=A \cap B (by noting that (A^c)^c=A

    Take the complementary of both sides :

    \underbrace{\bigg((A^c \cup B^c)^c\bigg)^c}_{=A^c \cup B^c}=\bigg(A \cap B\bigg)^c

    A^c \cup B^c=(A \cap B)^c


    This is the second law.
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