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Math Help - Simplify using the set rules of inference

  1. #1
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    Simplify using the set rules of inference

    Simplify the following: A, B, and C are subsets of the Universe.

    \overline{[((A-B)\cup(A \cap C)) \cap\overline{(A \cap B)}}
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  2. #2
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    Hello, aaronrj!

    I think I've got it . . .


    A, B,C are subsets of the Universe.

    Simplify: . \overline{\left[\,(A-B)\cup(A \cap C)\,\right] \:\cap \:\overline{(A \cap B)}}
    Under the long negation bar, we have:

    \left[\,(A - B) \cup (A \cap C)\,\right] \:\cap \:\overline{(A \cap B)}

    . . =\; \left[\,(A \cap \overline{B}) \cup (A \cap C)\,\right] \:\cap\:(\overline{A} \cup \overline{B}) . . Def. of set subtraction, DeMorgan's Law

    . . =\;\left[\,A \cap (\overline{B} \cup C)\,\right] \cap (\overline{A} \cup \overline{B}) . . . . . . Distributive Property

    . . =\;A \cap \left[\,(\overline{B} \cup C) \cap (\overline{A} \cup \overline{B})\,\right] . . . . . . Associative Property

    . . =\; A \cap \left[\,(\overline{B} \cup C) \cap (\overline{B} \cup \overline{A})\,\right] . . . . . . Commutative Property

    . . =\;A \cap \left[\,\overline{B} \cup (C \cap \overline{A})\,\right] . . . . . . . . . .Distributive Property

    . . =\; (A \cap \overline{B}) \,\cup \, \left[\,A \cap (C \cap \overline{A})\,\right] . . . . . .Distributive Property

    . . =\;(A \cap \overline{B}) \,\cup\,\underbrace{\left[\,A \cap \overline{A} \cap C\,\right]}_{\text{This is false}} . . . . . . .Comm., Assoc. Properties

    . . =\qquad A \cap \overline{B} . . . . . . . . . . . . . . .
    Don't know the name of this property.


    Replacing the long negation bar, we have:

    . . . .  \overline{(A \cap \overline{B})} \;\;=\;\;\overline{A} \cup B . . . . . . . . . DeMorgan's Law

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