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Thread: Equality proof

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

    Can someone help me to solve this.

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

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

    $\displaystyle x \in A \wedge x \notin (A\backslash B) \Leftrightarrow x \in A \wedge x \in B$
    $\displaystyle x \in A \wedge x \notin (x \in A \wedge x \notin B) \Leftrightarrow x \in A \wedge x \in B$
    $\displaystyle x \in A \wedge (x \notin A \wedge x \in B) \Leftrightarrow x \in A \wedge x \in B$
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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 $\displaystyle A\backslash (A\backslash B) = A \cap B$

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

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

    $\displaystyle
    x \in [A\backslash (A\backslash B)] \Leftrightarrow
    [x \in A] \wedge [x \notin (A\backslash B)] \Leftrightarrow $$\displaystyle
    [x \in A] \wedge [x \in B] \Leftrightarrow
    x \in (A \cap B)
    $


    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 $\displaystyle x \notin A$ in $\displaystyle
    [x \in A] \wedge [x \notin (A\backslash B)] \Leftrightarrow $$\displaystyle
    [x \in A] \wedge [x \in B]
    $

    Shouldn't be $\displaystyle x \notin (A\backslash B)$ equal to $\displaystyle x \notin A\wedge x \in B$?
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  4. #4
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    $\displaystyle x \in A \setminus (A \setminus B)$ iff $\displaystyle x \in A \wedge x \not\in (A \setminus B)$ iff $\displaystyle x \in A \wedge {\sim}(x \in A \setminus B)$ iff $\displaystyle x \in A \wedge {\sim}(x \in A \wedge x \not\in B)$ iff $\displaystyle x \in A \wedge (x \not\in A \vee x \in B)$ iff $\displaystyle (x \in A \wedge x \not\in A) \vee (x \in A \wedge x \in B)$ iff $\displaystyle F \vee x \in A \cap B$ iff $\displaystyle x \in A \cap B$.
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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 $\displaystyle \neg (x \notin A \wedge x \in B)$ means that beside negation of $\displaystyle \in $ or $\displaystyle \notin $ means that even $\displaystyle \wedge $ turns into $\displaystyle \vee $?

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

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