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Thread: théorème des ensembles

  1. #1
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    Unhappy théorème des ensembles

    please help me to do this exercice !

    http://i23.servimg.com/u/f23/11/67/15/58/88_bmp10.jpg

    thanx
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  2. #2
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    Lexington, MA (USA)
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    Hello, marocker!

    I think I've got the second part . . .


    Given: .$\displaystyle \begin{array}{ccc}H_1 & = & A \cap B' \cap C'
    \\ H_2 & = & (A \cap B) \cup C \end{array}$

    Find: .$\displaystyle [1]\;H_1\cup H_2\qquad [2]\;H_1 \cap H_2$

    $\displaystyle [2]\;\;H_1 \cap H_2 \;=\;(A \cap B' \cap C') \cap [(A \cap B) \cup C]$

    Distribute: .$\displaystyle A \cap B' \cap C' \cap \overbrace{[(A \cup C) \cap (B \cup C)]} $

    Rearrange terms: .$\displaystyle \underbrace{[A \cap (A \cup C)]} \cap\, [\underbrace{(B' \cap C')} \cap (B \cup C)] $

    DeMorgan's Law: . . . . $\displaystyle A \quad\;\cap\quad\;\underbrace{[(B \cup C)' \cap (B\cup C)]} $

    Then: . . . . . . . . . . . . .$\displaystyle \underbrace{A \qquad\quad\cap\qquad\quad \emptyset} $

    Therefore:. . . . . . . . . . . . . . . $\displaystyle \emptyset$

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  3. #3
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    and for another question ?
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