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Relations between sets proofs

  1. Mar 1, 2005 #1
    I'm trying to prove the following by contradiction: [(A^B)-(B^C)]-(A^C)'=0. A, B, C are sets. All I know is in order to prove by contradiction you simply set the above not equal to zero. But I don't know where to go from there.

    "^" means the intersection symbol.
     
  2. jcsd
  3. Mar 1, 2005 #2
    Suppose

    [tex](A \cap B - B \cap C) - (A \cap C)' \neq \emptyset[/tex].

    Then there is some element [itex]x \in ((A \cap B - B \cap C) - (A \cap C)')[/itex]. But then

    [tex]x \in A \cap B - B \cap C[/tex]
    [tex]x \notin (A \cap C)'[/tex]

    <=>

    [tex]x \in A \cap B - B \cap C[/tex]
    [tex]x \in A \cap C[/tex]

    <=>

    [tex]x \in A \cap B[/tex]
    [tex]x \notin B \cap C[/tex]
    [tex]x \in A \cap C[/tex]

    <=>

    [tex]x \in A[/tex]
    [tex]x \in B[/tex]
    [tex]x \notin B \cap C[/tex]
    [tex]x \in A[/tex]
    [tex]x \in C[/tex]

    <=>

    [tex]x \in A[/tex]
    [tex]x \in B[/tex]
    [tex]x \in C[/tex]
    [tex]x \notin B \cap C[/tex].

    Which is a contradiction.
     
  4. Mar 1, 2005 #3

    Galileo

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    Suppose it is not zero. Then there is some element x in the set on the left side.
    From the last term we infer that x is not an element of [itex](A \cap C)'[/itex], (does that ' mean the complement?) so:

    [tex]x \in A, \quad \mbox{and } x \in C[/tex],

    take it from there.
     
    Last edited: Mar 1, 2005
  5. Mar 30, 2005 #4
    Ok, let's prove it by contradiction:
    let's suppose that
    (A^B-B^C)-(A^C)'=!0, where =! means "different to". Then, there is a element namely x such that x E(belongs to) (A^B-B^C)-(A^C)', then, xE(A^B-B^C) and xE!( not belongs to)(A^C)', then, xEA^B and xE!B^C and xEA^C, then
    xEA, xEB, xEC and( xE!B or xE!C). Therefore, we have a contradiction.
     
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