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Thread: Proving using the set equality theorem

  1. #1
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    Proving using the set equality theorem

    Prove that (A - B) (B - A) = (A B) -(A B) using the set equality theorem

    The best way to answer this is ?
    hmmm
    I know I have to use x.

    so
    x ∈ A
    x
    ∈ A ∪ B
    hmmm need to explain how the x is used in the
    (A - B) (B - A) and (A B) -(A B)
    and also say a y is in both .....
    am I on the right lines...?
    any help would be appreciated thanks..
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  2. #2
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    Re: Proving using the set equality theorem

    (A - B) all belong to A and not B
    ∪ union
    (B - A) all belong to B and not A
    isn't the first part just really saying its A ∪ B with the disjoint part left out ?
    second part
    (A ∪ B) - (A ∩ B)
    is what I just mentioned but how do I prove this using elements of x etc
    are the subsets of each other ?

    ......
    Last edited by bee77; Jul 26th 2017 at 09:55 PM.
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  3. #3
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    Re: Proving using the set equality theorem

    Quote Originally Posted by bee77 View Post
    Prove that (A - B) (B - A) = (A B) -(A B) using the set equality theorem
    $ \begin{align*}(A\setminus B)\cup(B\setminus A)&=(A\cap B^c)\cup(B\cap A^c)\\&=([(A\cap B^c)\cup B]\cap[(A\cap B^c)\cup A^c] \\&=(A\cup B)\cap (A^c \cup B^c)\\&=(A\cup B)\cap (A \cap B)^c\\&=(A\cup B)\setminus(A\cap B) \end{align*}$
    Last edited by Plato; Jul 27th 2017 at 05:56 AM.
    Thanks from bee77
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  4. #4
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    Re: Proving using the set equality theorem

    Proposition: $(A\setminus B) \cup (B\setminus A) = (A\cup B) \setminus (A\cap B)$
    Proof:
    First, we will prove that $(A\setminus B)\cup (B\setminus A) \subseteq (A\cup B)\setminus (A\cap B)$.
    Let $x \in (A\setminus B)\cup (B\setminus A)$. If $x \in A\setminus B$, then $x \in A\cup B$. If $x \in B\setminus A$ then $x\in A\cup B$. Therefore, $x\in A\cup B$ regardless of which set it originally came from. We just need to verify that it is not in $A\cap B$.
    If $x \in A\setminus B$ then $x \notin B$, so $x \notin A\cap B$. If $x \in B\setminus A$ then $x \notin A$, so $x\notin A\cap B$. Therefore, $x \in (A\cup B) \setminus (A\cap B)$. Since this is true for any arbitrary $x$, we have $(A\setminus B)\cup (B\setminus A) \subseteq (A\cup B)\setminus (A\cap B)$.

    Next, let's prove the reverse. Let $x\in (A\cup B) \setminus (A\cap B)$. Since $(A\cup B) \setminus (A\cap B) \subseteq (A\cup B)$, we know $x \in A$, $x\in B$, or both.
    Suppose $x \in A$. Since $x \notin (A\cap B)$, we know that $x \notin B$, so $x\in A\setminus B$.
    Suppose $x\in B$. Since $x \notin (A\cap B)$, we know that $x \notin A$, so $x\in B\setminus A$.
    Since this is true for any arbitrary element $x$, this proves that $(A\cup B) \setminus (A\cap B) \subseteq (A\setminus B) \cup (B\setminus A)$.

    Since we proved dual containment, it must be that the two sides are equal. Q.E.D.
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  5. #5
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    Re: Proving using the set equality theorem

    Thanks Plato ,
    I will carefully go through your answer bit by bit and write it down in english as I have just started discrete mathematics working with sets...really appreciate it .
    cheers
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  6. #6
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    Re: Proving using the set equality theorem

    Thanks SlipEternal ,
    wow in great detail , I was trying to make each side a subset of the other but got confused ,I'll go through your answer and try my best to get it.
    Thanks a million again!!!! looks like you're very solid with sets .
    cheers
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