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Math Help - Equality proof

  1. #1
    Senior Member OReilly's Avatar
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    Equality proof

    Can someone help me to solve this.

    Prove equality A\backslash (A\backslash B) = A \cap B

    I have tried to solve it something like this, but I really don't know.

    x \in A \wedge x \notin (A\backslash B) \Leftrightarrow x \in A \wedge x \in B
    x \in A \wedge x \notin (x \in A \wedge x \notin B) \Leftrightarrow x \in A \wedge x \in B
    x \in A \wedge (x \notin A \wedge x \in B) \Leftrightarrow x \in A \wedge x \in B
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by OReilly
    Can someone help me to solve this.

    Prove equality A\backslash (A\backslash B) = A \cap B

    I have tried to solve it something like this, but I really don't know.

    x \in A \wedge x \notin (A\backslash B) \Leftrightarrow x \in A \wedge x \in B
    x \in A \wedge x \notin (x \in A \wedge x \notin B) \Leftrightarrow x \in A \wedge x \in B
    x \in A \wedge (x \notin A \wedge x \in B) \Leftrightarrow x \in A \wedge x \in B
    Isn't your first line all you require?

    <br />
x \in [A\backslash (A\backslash B)] \Leftrightarrow<br />
 [x \in A] \wedge [x \notin (A\backslash B)] \Leftrightarrow <br />
[x \in A] \wedge [x \in B] \Leftrightarrow <br />
x \in (A \cap B)<br />


    RonL
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  3. #3
    Senior Member OReilly's Avatar
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    This is what is confusing me.
    What happened to x \notin A in <br />
 [x \in A] \wedge [x \notin (A\backslash B)] \Leftrightarrow <br />
[x \in A] \wedge [x \in B]<br />

    Shouldn't be x \notin (A\backslash B) equal to x \notin A\wedge x \in B?
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  4. #4
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    x \in A \setminus (A \setminus B) iff x \in A \wedge x \not\in (A \setminus B) iff x \in A \wedge {\sim}(x \in A \setminus B) iff x \in A \wedge {\sim}(x \in A \wedge x \not\in B) iff x \in A \wedge (x \not\in A \vee x \in B) iff (x \in A \wedge x \not\in A) \vee (x \in A \wedge x \in B) iff F \vee x \in A \cap B iff  x \in A \cap B.
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  5. #5
    Senior Member OReilly's Avatar
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    rgep's answer has confused me a little.

    I have one question.

    Does expression \neg (x \notin A \wedge x \in B) means that beside negation of  \in or  \notin means that even  \wedge turns into  \vee ?

    If that is true than answer is understable to me, because I can form tautology then.

    Then it would be
    \begin{array}{l}<br />
 x \in A \wedge x \notin (A\backslash B) \Leftrightarrow x \in A \wedge x \in B \\ <br />
 x \in A \wedge \neg (x \in A \wedge  \notin B) \Leftrightarrow x \in A \wedge x \in B \\ <br />
 x \in A \wedge (x \notin A \vee  \in B) \Leftrightarrow x \in A \wedge x \in B \\ <br />
 \end{array}<br />
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  6. #6
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    Does expression \neg (x \notin A \wedge x \in B) means that beside negation of \in or \notin means that even \wedge turns into \vee?
    Yes: those are de Morgan's laws: \neg(A \vee B) \equiv (\neg A) \wedge (\neg B) and \neg(A \wedge B) \equiv (\neg A) \vee (\neg B).
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  7. #7
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    Thats all I needed. Thanks!
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