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Math Help - Set Theory equality proof

  1. #1
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    Set Theory equality proof

    Hello,

    I'm having difficulties with proving the following:
    A \ (B\C) = (A\B) UNION (A INTERSECTION C)

    I know it's correct, I can draw Venn diagrams that show it, but have no idea how to proof it using words (or other statements).

    Any help will be appreciated.

    Thanks!
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  2. #2
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    Re: Set Theory equality proof

    Quote Originally Posted by BscStud View Post
    I'm having difficulties with proving the following:
    A \ (B\C) = (A\B) UNION (A INTERSECTION C)

     \begin{align*}A\setminus(B\setminus  C) &=A\cap\overline{B\cap\overline{C}} \\&= A\cap (\overline{B}\cup C)\\&=(A\cap\overline{B})\cup(A\cap C)\end{align*}

    Can you finish?
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  3. #3
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    Re: Set Theory equality proof

    Quote Originally Posted by Plato View Post
     \begin{align*}A\setminus(B\setminus  C) &=A\cap\overline{B\cap\overline{C}} \\&= A\cap (\overline{B}\cup C)\\&=(A\cap\overline{B})\cup(A\cap C)\end{align*}

    Can you finish?
    The concept is very clear to me, though only because I know negation from other study fields. In the course I'm undertaking now we did not study anything relating to negation of sets thus I'm afraid I cannot use it in profing the statement above (I know... University rules).

    Is there any other way to proof it without negation? I do not need nor want a complete solution, just how to start.

    Thank you very much for your respond.
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  4. #4
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    Re: Set Theory equality proof

    Quote Originally Posted by BscStud View Post
    The concept is very clear to me, though only because I know negation from other study fields. In the course I'm undertaking now we did not study anything relating to negation of sets thus I'm afraid I cannot use it in profing the statement above (I know... University rules).

    Is there any other way to proof it without negation? I do not need nor want a complete solution, just how to start.

    Thank you very much for your respond.
    That is totally absurd.

    You can use the "pick-a-point" approach.
    t\in A\setminus(B\setminus  C)\text{ iff }t\in A \text{ and }t\notin(B\setminus  C)

    t\notin(B\setminus  C)\text{ iff }t\notin B\text{ or }t\in C.

    You then put those together.
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  5. #5
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    Re: Set Theory equality proof

    Trust me, I had a week long argue over this with my professor, mainly because I'm a long time computer programmer but they don't let us use any previous knowledge to solve basic tasks resulting in a lower-standards solutions.

    This approach is what I was looking for, thanks a lot Plato. You've been a big help.
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