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Thread: prooving set identities

  1. #1
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    prooving set identities

    This is an example solution I am reading but I can't understand the 3rd transformation.

    $\displaystyle (A-B) \cup(B-A) =(A \cup B) - (A \cap B)$
    I understand changing the LHS to

    $\displaystyle (A-B) \cup(B-A) = (A\bigcap B)\cup(B\cap A')$ < ---a (this)

    using set diff representation

    (The notes I am reading use this notation where sometimes the intersection operator is "big", does it have an alternate meaning when it is big, or simply a typo?)

    If it is just a typo, then I needn't worry about it.

    but the next step I don't understand:

    (to this) b ---> $\displaystyle (A \cup (B \cap A')) \cap (B' \cup (B \cap A'))$
    the notes claim this is using distributivity,
    but I can't see how they arrive at that equation using this rule

    $\displaystyle A \cap (B \cup C)) = (A \cap B) \cup (A \cap C)$ <-- this I understand,

    but I can;t see how they went from a to b
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  2. #2
    Senior Member oldguynewstudent's Avatar
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    Quote Originally Posted by dunsta View Post
    This is an example solution I am reading but I can't understand the 3rd transformation.

    $\displaystyle (A-B) \cup(B-A) =(A \cup B) - (A \cap B)$
    I understand changing the LHS to

    $\displaystyle (A-B) \cup(B-A) = (A\bigcap B)\cup(B\cap A')$ < ---a (this)

    using set diff representation

    (The notes I am reading use this notation where sometimes the intersection operator is "big", does it have an alternate meaning when it is big, or simply a typo?)

    If it is just a typo, then I needn't worry about it.

    but the next step I don't understand:

    (to this) b ---> $\displaystyle (A \cup (B \cap A')) \cap (B' \cup (B \cap A'))$
    the notes claim this is using distributivity,
    but I can't see how they arrive at that equation using this rule

    $\displaystyle A \cap (B \cup C)) = (A \cap B) \cup (A \cap C)$ <-- this I understand,

    but I can;t see how they went from a to b

    In my texts $\displaystyle \bigcap $ is usually an intersection of sets like

    $\displaystyle A_1 \cap A_2 \cap A_3 \cap ... \cap A_i$

    I believe this is a typo and should read $\displaystyle (A-B) \cup(B-A) = (A\cap B')\cup(B\cap A')$ then getting from a to b is straightforward.
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  3. #3
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    sorry, I guess the use of $\displaystyle \bigcap$ is a typo in the text i am studying, as the rest of the intersection operators are $\displaystyle \cap$ smaller size.

    Actually, leaving out the ' complement symbol was my typo (still getting used to not only "sets" but "Latex" too.

    It reads
    $\displaystyle (A-B) \cup(B-A) = (A \bigcap B')\cup(B\cap A')
    $
    in the text.
    So i still need an explanation of how to get from
    $\displaystyle (A-B) \cup(B-A) = (A \cap B')\cup(B\cap A')$

    to $\displaystyle (A \cup (B \cap A')) \cap (B' \cup (B \cap A'))$

    as i cant see how it relates to the distributivity example I have (in the original post)
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  4. #4
    Senior Member oldguynewstudent's Avatar
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    Quote Originally Posted by dunsta View Post
    sorry, I guess the use of $\displaystyle \bigcap$ is a typo in the text i am studying, as the rest of the intersection operators are $\displaystyle \cap$ smaller size.

    Actually, leaving out the ' complement symbol was my typo (still getting used to not only "sets" but "Latex" too.

    It reads
    $\displaystyle (A-B) \cup(B-A) = (A \bigcap B')\cup(B\cap A')
    $
    in the text.
    So i still need an explanation of how to get from
    $\displaystyle (A-B) \cup(B-A) = (A \cap B')\cup(B\cap A')$

    to $\displaystyle (A \cup (B \cap A')) \cap (B' \cup (B \cap A'))$

    as i cant see how it relates to the distributivity example I have (in the original post)
    This is just the distributive law of set identities on the RHS:

    $\displaystyle A \cup (B \cap C) = (A \cap B) \cup (A \cap C) $

    In the above identity replace A with $\displaystyle A \cap B'$ etc.
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