# Sets properties for simplification

Using the set properties, I have to demonstrate that

$[ (A - B)^\mathsf{c} - (B - A)^\mathsf{c}] \cap A = \emptyset.$

So far I've seen some logic properties, but never applied to sets. Could you guys help me?

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What means $\neg A$? Does this means $\{ x : x \notin A \}?$ –  user29999 Dec 6 '12 at 21:22
Complement of a set, like in here: basic-mathematics.com/complement-of-a-set.html I don't know how to put that with LaTeX. –  Zed Dec 6 '12 at 21:23
I don;t know this symbol ¬ for sets. I think it is used in logic statements. –  B. S. Dec 6 '12 at 21:24
I've just edited. Is it clearer now? –  Zed Dec 6 '12 at 21:28

The definition of $A - B$ is $\{x \mid x \in A \text{ and } x \notin B\}$, which can be rewritten as $A \cap B^c$. This fact allows us to write \begin{align*} [(A - B)^\mathsf{c} - (B - A)^\mathsf{c}] \cap A &= [(A \cap B^\mathsf{c})^\mathsf{c} - (B \cap A^\mathsf{c})^\mathsf{c}] \cap A\\ &= [(A \cap B^\mathsf{c})^\mathsf{c} \cap (B \cap A^\mathsf{c})] \cap A\\ &\subseteq A^\mathsf{c} \cap A\\ &= \emptyset. \end{align*}

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Thank you very much. –  Zed Dec 6 '12 at 21:57

\begin{align}&x\in((A\setminus B)^{\mathsf c}\,\setminus\, (B\setminus A)^{\mathsf c})\cap A\\\implies&x\in(A\setminus B)^{\mathsf c}\land x\notin(B\setminus A)^{\mathsf c}\land x\in A\\ \implies&x\notin(A\setminus B)\land x\in (B\setminus A)\land x\in A\\ \implies&x\in (B\setminus A)\land x\in A\\ \implies&x\in B\land x\notin A\land x\in A\\ \implies& x\notin A\land x\in A\\ \implies &\perp \end{align}

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Thank you very much. –  Zed Dec 6 '12 at 22:00

The key fact with such set equations is that they can usually be translated into corresponding logical formulae.

For example:

• $x \in A \cap B$ if and only if $x \in A \wedge x \in B$.
• $x \in A^\mathsf{c}$ if and only if $\neg (x \in A)$
• $x \in A - B$ if and only if $x\in A \wedge x \not\in B$.

Applying these rules repeatedly ought to transform your statement about sets into a logical statement. Membership of the left hand side corresponds to a statement which can't possibly be true, so the set has no members. Two sets are equal exactly when they have the same members, so that means it is the empty set.

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Thank you very much. –  Zed Dec 6 '12 at 22:03

In my notation $A^c = \{ x : x \notin A\}$. Then we must prove that

$$[ (A \cap B^c )^c \cap (B \cap A^c)^c ] \cap A = \emptyset.$$ Notice that $$(A \cap B^c)^c = (A^c \cup B)$$ Then \begin{eqnarray} [(A^c \cup B) \cap (B^c \cup A)] \cap A & =&\left \{ [(A^c \cup B) \cap B^c] \cup [(A^c \cup B) \cap A]\right \} \cap A \\ &=& \left \{ [A^c \cap B^c] \cup [(A^c \cup B)] \right \} \cap A\\ &=& \left \{ [A^c \cap B^c] \cap A \right \} \cup \left \{ [(A^c \cup B)] \cap A \right \}\\ &=& \emptyset \cup \emptyset \\ &=& \emptyset. \end{eqnarray}

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Thank you very much. –  Zed Dec 6 '12 at 22:01
You're welcome. –  user29999 Dec 6 '12 at 22:04