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Thread: Simple set proofs

  1. #1
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    Simple set proofs

    I find these problems easy enough to understand, I just have trouble with proofs that make sense.

    For example, if I wanted to prove:

    $\displaystyle (C \cup D) \setminus (C\cap D)=(C\setminus D)\cup(D\setminus C)$

    Can I write:

    $\displaystyle (C \cup D)-(C\cap D)=\{x:x\in C$ or $\displaystyle x\in D\}\wedge \{x\in C:x\not\in D\}\wedge \{x\in D:x\not\in C\}$

    $\displaystyle \Rightarrow ( \forall x \in C)(x \in (C\cup D^c ))$ and $\displaystyle ( \forall x \in D)(x \in (D\cup C^c ))$

    Hence,

    $\displaystyle (C \cup D)\setminus (C\cap D)= (C\cup D^c ) \cup (D\cup C^c ) = (C\setminus D)\cup(D\setminus C)$
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  2. #2
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    If you can use a non 'pick-a-point' proof, here is one.
    $\displaystyle \begin{array}{rcl}
    {\left( {C \cup D} \right)\backslash \left( {C \cap D} \right)} & \equiv & {\left( {C \cup D} \right) \cap \left( {C \cap D} \right)^c } \\
    {} & \equiv & {\left( {C \cup D} \right) \cap \left( {C^c \cup D^c } \right)} \\
    {} & \equiv & {\left( {C \cap D} \right) \cup \left( {D \cap C^c } \right)} \\
    {} & \equiv & {\left( {C\backslash D} \right) \cup \left( {D\backslash C} \right)} \\

    \end{array} $
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  3. #3
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    Yeah, I can do that much, but I think I would be accused of only "showing" it. Is the attempt I made above even an actual proof?
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  4. #4
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    Quote Originally Posted by MathSucker View Post
    Yeah, I can do that much, but I think I would be accused of only "showing" it. Is the attempt I made above even an actual proof?
    Well a proof is proof. But follow your instructor’s instructions.

    If $\displaystyle x\in(C\cup D)\setminus (C\cap D)$ means that $\displaystyle x\in C\text{ or }x\in D\text{ and }x\notin(C\cap D).$
    So the last implies $\displaystyle x\notin C\text{ or }x\notin D$.
    From there takes cases.
    Last edited by Plato; Feb 2nd 2011 at 03:53 PM.
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    delete
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  6. #6
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    I am struggling to understand the following.
    Quote Originally Posted by MathSucker View Post
    $\displaystyle (C \cup D)-(C\cap D)=\{x:x\in C$ or $\displaystyle x\in D\}\wedge \{x\in C:x\not\in D\}\wedge \{x\in D:x\not\in C\}$
    First, $\displaystyle \land$ usually denotes conjunction, which joins propositions, not sets. (Speaking of this, it's not good to use both $\displaystyle \land$ and "and", as well as - and $\displaystyle \setminus$.) Second, I don't see how the right-hand side immediately follows from the definition of the left-hand side.
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