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Math Help - simplifying using set algebra...

  1. #1
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    simplifying using set algebra...

    I can seem to do most of them however when the expression is a little long i just seem to make it longer and more confusing rather than simplifying it.

    Q1)  (A - B^c) \cup (B \cap (A \cap B)^c)

    Q2)  [(A \cap B)^c \cap A] \cap B^c

    Can someone please explain how to simplyfy these =/
    And sorry dont really know the syntax for all the maths (the c is meant to be compliment).

    thanks
    Last edited by Plato; November 6th 2009 at 03:25 AM. Reason: Fix LaTeX
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  2. #2
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    I'll solve the second for you, see if you can solve the first using the same methods.

    We want to simplify ((A\cap B)^c)\cap A)\cap B^c

    Recall that x \in X^c \Leftrightarrow x \notin X

    Therefore (A\cap B)^c\cap A = \{x : x \notin A \cap B , x \in A\} = A - B

    \Rightarrow ((A\cap B)^c)\cap A)\cap B^c = (A-B)\cap B^c =  \{x : x \in A, x \notin B, x \notin B\} = \{x:x\in A, x \notin B\} = A-B
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by lost1 View Post

    Q1)  (A - B^c) \cup (B \cap (A \cap B)^c)

    Defunkt has already given you an outline on how to proceed with these proofs. Something that may make the first easier is that X-Y=X\cap Y'...so saying A- B^c\Leftrightarrow A\cap B
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  4. #4
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    Quote Originally Posted by Defunkt View Post
    I'll solve the second for you, see if you can solve the first using the same methods.

    We want to simplify ((A\cap B)^c)\cap A)\cap B^c

    Recall that x \in X^c \Leftrightarrow x \notin X

    Therefore (A\cap B)^c\cap A = \{x : x \notin A \cap B , x \in A\} = A - B

    \Rightarrow ((A\cap B)^c)\cap A)\cap B^c = (A-B)\cap B^c =  \{x : x \in A, x \notin B, x \notin B\} = \{x:x\in A, x \notin B\} = A-B
    Hmm I see what you are doing there. However with the examples done in my lecture notes they only simplify based on the 'rules of set algebra' and ask you to name each law at each step (distributive law, etc). The way you have done it you kind of define each part of the expresion as a set and then find a simpler expression from that. And I like this method as it seems very logical to do (rather than what i do which is just try and simplify based on one of the laws and not really knowing where im heading). But how would i do the same qustion using the other method of just using the 'laws of set algebra'?

    thanx
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