Proving equality of sets How do I prove the following equations (I am new to statistics and not sure where to begin even after trying to figure it out):
(a) $A - B = A - A \cap B = A \cup B - B$
(b) $A \mathbin{\Delta} B = A \cup B - A \cap B$
 A: First make sure you know precisely what the involved symbols mean. Then, you can prove the equality of two sets $S$ and $T$ by showing $S \subseteq T$ and $T \subseteq S$.
In your first example, you have to show that $A - B = A - A \cap B$. The first step is to prove that $A - B \subseteq A - A \cap B$. To this end, take an arbitrary element $x \in A - B$. By definition of the set difference, we know that $x \in A$ and $x \notin B$. Since $x \notin B$, we also have $x \notin A \cap B$. Therefore, $x \in A - A \cap B$. Since $x$ was an arbitrary element of $A - B$, we can conclude that $A - B \subseteq A - A \cap B$. Then, prove that $A - A \cap B \subseteq A - B$ similarly.
A: 
Proof$_1$: $A\setminus(B\cap C)=(A\setminus B) \cup (A\setminus C)$
  \begin{align*} x\in A\setminus(B\cap C) &\leftrightarrow x \in A \wedge x \notin (B \cap C)\\ &\leftrightarrow x \in A \wedge (x \notin B \vee x \notin C)\\ &\leftrightarrow (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C) \\ &\leftrightarrow x \in (A \setminus B) \vee x \in (A\setminus C) \\ &\leftrightarrow x \in (A\setminus B) \cup (B\setminus C) \end{align*}
  therefore Proof$_{1_1}$: $A \setminus (A \cap B)=(A\setminus A)\cup (A\setminus B)=\emptyset \cup (A\setminus B)=A\setminus B$
Proof$_2$:    $(A \cup B)\setminus B=A \setminus B$
  \begin{align*} x \in(A \cup B)\setminus B &\leftrightarrow x \in (A\cup B) \wedge x \notin B \\ &\leftrightarrow (x \in A \vee x \in B)\wedge x \notin B\\ &\leftrightarrow (x \in A \wedge x \notin B) \vee (x \in B \wedge x \notin B) \\ &\leftrightarrow x \in (A\setminus B) \vee x \in (B\setminus B) \\  &\leftrightarrow x \in ((A\setminus B) \cup \emptyset )\\ &\leftrightarrow x \in (A\setminus B) \end{align*}
Proof$_3$: $(A \cup B) \setminus (A \cap B)=(A \setminus B) \cup (B\setminus A) =: A \bigtriangleup B$
  \begin{align*}  (A \cup B)\setminus (A\cap B) &= ((A \cup B) \setminus A) \cup ((A \cup B) \setminus B) \text{ (by Proof} _1)\\ &=(B \cup A) \setminus A) \cup ((A \cup B) \setminus B) \\ &= (B \setminus A) \cup (A \setminus B) \text{ (by Proof} _2)\\ &=(A \setminus B) \cup (B\setminus A) \end{align*}

