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Math Help - Need confirmation on set identity problem

  1. #1
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    Need confirmation on set identity problem

    I believe I may have solved this properly, but I have some nagging doubts about my process. Here is the question.

    Show whether the set equality (A-B) \cup (B-A) = (A \cup B) - (A \cap B) is true, using set identities.

    My process has led me to believe it is false, but I can't shake the feeling that I made a mistake somewhere. Here is my work.

    (A-B) \cup (B-A) = (A \cup B) - (A \cap B)
    (A \cap \overline{B}) \cup (B \cap \overline{A}) = (A \cup B) \cap (\overline{A \cap B})
    (A \cap \overline{B}) \cup (B \cap \overline{A}) = (A \cup B) \cap (\overline{A} \cup \overline{B}) De Morgan's Law
    (A \cap \overline{B}) \cup (\overline{A} \cap B) = (A \cup B) \cap (\overline{A} \cup \overline{B}) Commutative Law
    (A \cup \overline{A}) \cap (\overline{B} \cup B) = (A \cap \overline{A}) \cup (B \cap \overline{B}) Distributive Laws (Twice) (this is where I think I screwed up)
    U \cap U = \emptyset \cup \emptyset Complement Laws (multiple times)
    U = \emptyset Identity Laws

    I do not know whether or not my application of the Distributive Laws in that case was correct. Could someone please check this?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Runty View Post
    I believe I may have solved this properly, but I have some nagging doubts about my process. Here is the question.

    Show whether the set equality (A-B) \cup (B-A) = (A \cup B) - (A \cap B) is true, using set identities.

    My process has led me to believe it is false, but I can't shake the feeling that I made a mistake somewhere. Here is my work.

    (A-B) \cup (B-A) = (A \cup B) - (A \cap B)
    (A \cap \overline{B}) \cup (B \cap \overline{A}) = (A \cup B) \cap (\overline{A \cap B})
    (A \cap \overline{B}) \cup (B \cap \overline{A}) = (A \cup B) \cap (\overline{A} \cup \overline{B}) De Morgan's Law
    (A \cap \overline{B}) \cup (\overline{A} \cap B) = (A \cup B) \cap (\overline{A} \cup \overline{B}) Commutative Law
    (A \cup \overline{A}) \cap (\overline{B} \cup B) = (A \cap \overline{A}) \cup (B \cap \overline{B}) Distributive Laws (Twice) (this is where I think I screwed up)
    U \cap U = \emptyset \cup \emptyset Complement Laws (multiple times)
    U = \emptyset Identity Laws

    I do not know whether or not my application of the Distributive Laws in that case was correct. Could someone please check this?
    Too much stuff going on there. Just show us what you did in the distributive step...show both distributions.
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  3. #3
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    (A \cap \overline{B}) \cup (\overline{A} \cap B) = (A \cup B) \cap (\overline{A} \cup \overline{B}) Starting point
    (A \cup \overline{A}) \cap (\overline{B} \cup B) = (A \cup B) \cap (\overline{A} \cup \overline{B}) Distributive step #1
    (A \cup \overline{A}) \cap (\overline{B} \cup B) = (A \cap \overline{A}) \cup (B \cap \overline{B}) Distributive step #2
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  4. #4
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    \begin{gathered}<br />
  \left( {A \cup B} \right) \cap \left( {\overline A  \cup \overline B } \right) \hfill \\<br />
  \left[ {A \cap \left( {\overline A  \cup \overline B } \right)} \right] \cup \left[ {B \cap \left( {\overline A  \cup \overline B } \right)} \right] \hfill \\<br />
  \left[ {A \cap \overline B } \right] \cup \left[ {B \cap \overline A } \right] \hfill \\ <br />
\end{gathered}
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  5. #5
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    Quote Originally Posted by Plato View Post
    \begin{gathered}<br />
  \left( {A \cup B} \right) \cap \left( {\overline A  \cup \overline B } \right) \hfill \\<br />
  \left[ {A \cap \left( {\overline A  \cup \overline B } \right)} \right] \cup \left[ {B \cap \left( {\overline A  \cup \overline B } \right)} \right] \hfill \\<br />
  \left[ {A \cap \overline B } \right] \cup \left[ {B \cap \overline A } \right] \hfill \\ <br />
\end{gathered}
    So I was wrong about it being false. Thank you.
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