How do I show a set $A = (A\setminus B)\cup (A\cap B)$ for discrete math? How would I go about showing that for any set A and B,
$$A = (A\setminus B)\cup (A\cap B)?$$
I don't really understand how to show this. Here's my interpretation:
$$A = (A\setminus B)\text{ or } (AB)?$$
$\cup$ means either or both sides? $\cap$ means both $A$ and $B$?
How do I show that
$$A = (A\setminus B) \cup (A\cap B)?$$
Thanks!
 A: $$\begin{align} x \in (A-B) \cup (A\cap B) &\iff x \in A- B, \;{\bf \text{  or }}\; x \in A \cap B \\ \\ &\iff \underbrace{(x \in A {\bf\text{ and }} x \notin B)}_{\large x \in \,A-B} \text{ or } \underbrace{(x \in A {\bf\text{ and }} x \in B)}_{\large x\in\, A\cap B}\\ \\ &\iff x \in A \text{ and } \underbrace{(x \in B \text{ or } x \notin B)}_{\text{TRUE}}\tag{DL} \\ \\ &\iff x \in A\end{align}$$
This relies only on definitions: set union, set-minus $(A - B = A\setminus B)$, set intersection, along with the distributive law (DL): $$(p \text{ and } q) \text{ or }(p \text{ and } r) \equiv p \text{ and } (q \text{ or } r)$$
A: Consider the mechanism of a "truth table" where "true" means "is contained in set X" and "false" means "is not contained in set X":
A B | A\B | A intersect B
-------------------------
T T |  F  |      T
T F |  T  |      F
F T |  F  |      F
F F |  F  |      F

The remaining statements should be easy to finish.
Note that your interpretation of the symbols is basically correct; $\cup$ is called "union" and means "is in either set", while $\cap$ is called "intersection" and means "is in both sets".  The $\setminus$ "set subtraction" operator is also written as the intersection with the complement of the right-hand set, so we have $A\setminus B=A\cap B^c$.
A: You can draw a Venn diagram but it would not be a formal proof, more of a way to visualize the problem for yourself. If you want to proof the equation you should only use the rules of set theory page 5 and 6. 
You can then justify each step by saying which rule you apply. For example a first step could be to apply the distributive law then work your way forward, should not be that hard.
Here is the derivation:
$$
(A \setminus B) \cup (A \cap B) = ((A \setminus B) \cup A) \cap ( (A \setminus B) \cup B) )\\
= (A) \cap ( (A \setminus B) \cup B)) \\
= (A) \cap ( A \cup B) \\
= A
$$
A: First, to adress your question of what $\cup$ and $\cap$ mean. They are operators joining two sets into one set. An element $x$ is in $A\cap B$ if it is BOTH in $A$ and in $B$. Mathematically written that means
$$x\in A\cap B \Longleftrightarrow x\in A \wedge x\in B$$
where the operator $\wedge$ means (and is read as) "and". Similarly, an element is in $A\cup B$ if it is in $A$ or if it is in $B$:
$$x\in A\cup B \Longleftrightarrow x\in A \vee x\in B$$
Where $\vee$ means "or".
Another operation on sets used in your task is set difference, $A\setminus B$. An element is in thus set if it is in $A$ and it is $not$ in $B$, meaning
$$x\in A\setminus B \Longleftrightarrow x\in A \wedge x\notin B.$$
Now that operations are clear, all you have to do is to prove two sets are equal. For any $2$ sets $C$ and $D$, the sets are equal if they have the same elements, meaning every element of $C$ is in $D$ and every element of $D$ is in $C$. This means that to prove $C=D$, you must prove $$x\in C \Longleftrightarrow x\in D$$ for all $x$. Try it!
