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Math Help - How do I prove this..

  1. #1
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    How do I prove this..

    Hi,

    I have an exam coming up tomorrow and am having trouble trying to prove this theorem:

    Question: Using DeMorgans Theorem, show that:



    I just haven't got a clue how to tackle this one. Do I only need to use DeMorgans or are there other laws I have to use first??

    Thanks In Advance..
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  2. #2
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    Hello, atwarwithmaths!

    We use DeMorgan's Theorem just once.
    Then we use a variety of other laws:
    . . (1)\;A \cap \overline{A} \:=\:\emptyset,\quad (2)\;A \cup \emptyset \:=\:A,\quad\text{Distributive Laws},\quad\text{Commutative Laws}


    Using DeMorgans Theorem, show that:

    (A \cap \overline{B}) \cup (\overline{A} \cap B) \;=\;(A \cup B)\cap (\overline{A \cap B})
    On the right, we have:

    \overline{A \cap B}) \;=\;(A \cup B) \:\cap\: (\overline{A} \cup \overline{B})\qquad\text{(DeMorgan)}" alt="(A \cup B) \:\cap \\overline{A \cap B}) \;=\;(A \cup B) \:\cap\: (\overline{A} \cup \overline{B})\qquad\text{(DeMorgan)}" />

    . . . . . . . . . . . . = \;\bigg[(A \cup B) \cap \overline{A}\,\bigg] \;\cup \;\bigg[(A \cup B) \cap \overline{B}\,\bigg] \qquad\text{(Distr.)}

    . . . . . . . . . . . . = \;\bigg[(A \cap \overline{A}) \cup (B \cap \overline{A})\bigg] \;\cup \;\bigg[(A \cap \overline{B}) \cup (B \cap \overline{B})\bigg]\qquad\text{(Distr.)}

    . . . . . . . . . . . . =\;\bigg[\,\emptyset \cup (B \cap \overline{A})\bigg] \;\cup \;\bigg[(A \cap \overline{B}) \cup \emptyset\,\bigg]\qquad\text{(Law 1)}

    . . . . . . . . . . . . A \cap \overline{B}) \qquad\text{(Law 2)} " alt="= \;(B \cap \overline{A}) \;\cup\A \cap \overline{B}) \qquad\text{(Law 2)} " />


    . . . . . . . . . . . . \overline{A} \cap B)\qquad\text{(Comm.)} " alt="= \;(A \cap \overline{B})\:\cup\\overline{A} \cap B)\qquad\text{(Comm.)} " />

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