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?
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?
In my notation $ A^c = \{ x : x \notin A\} $. Then we must prove that
\begin{equation} [ (A \cap B^c )^c \cap (B \cap A^c)^c ] \cap A = \emptyset. \end{equation} Notice that \begin{equation} (A \cap B^c)^c = (A^c \cup B) \end{equation} 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}
$\begin{align}&x\in((A\setminus B)^{\mathsf c}\,\setminus\, (B\setminus A)^{\mathsf c})\cap A\\\implies&x\in(A\setminus B)^{\mathsf c}\land x\notin(B\setminus A)^{\mathsf c}\land x\in A\\ \implies&x\notin(A\setminus B)\land x\in (B\setminus A)\land x\in A\\ \implies&x\in (B\setminus A)\land x\in A\\ \implies&x\in B\land x\notin A\land x\in A\\ \implies& x\notin A\land x\in A\\ \implies &\perp \end{align}$
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*} $
The key fact with such set equations is that they can usually be translated into corresponding logical formulae.
For example:
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.