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Thread: 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 $\displaystyle (AB)^c = A^c \cup B^c $ by using De Morgan's first law: $\displaystyle (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.
    $\displaystyle \begin{gathered}
    \left( {A^c \cup B^c } \right)^c = \left( {A^c } \right)^c \left( {B^c } \right)^c = AB \hfill \\
    \left( {AB} \right)^c = A^c \cup B^c \hfill \\
    \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 $\displaystyle (A \cap B)^c = A^c \cup B^c $ by using De Morgan's first law: $\displaystyle (E \cup F)^c = E^c \cap F^c $

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

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

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

    Take the complementary of both sides :

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

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


    This is the second law.
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