Suppose $x \in A \bigcap \bar{B}$ then is $x \notin (A \bigcap B)$ a contradiction? I'm having trouble convincing myself this isn't right. We know for sure $x \in A$ by supposition. $x \notin (A \bigcap B)$ says $x$ is not in $A$ and $x$ not in $B$. To me this means that if either of these cases fails a contradiction is evoked. Is my logic incorrect? Why?
 A: Saying $x\notin (A\cap B)$ means

it is not the case that $x$ belongs to both $A$ and $B$

so either it does not belong to $A$ or it does not belong to $B$.
Say your universe is formed by squares and triangles that can be colored white or black, so you have a certain amount of white squares, black squares, white triangles and black triangles.
If $A$ is the set of white figures and $B$ the set of triangles, saying that a figure is not a white triangle (that is, belongs to $A\cap B$) means it is a square (of any color) or it is black.
You are mistaking this by saying that something that's not a white triangle must be a black triangle; however, a white square is not a white triangle (likewise, a non “white triangle” can be a black square or a black triangle).
Thus if $x\in A\cap \bar{B}$, it belongs to $A$ and it does not belong to $B$. Hence it qualifies as not being an element of $A\cap B$.
If you want to see it algebraically, assuming $x\in A\cap \bar{B}$ and $x\in A\cap B$ implies $x\in (A\cap\bar{B})\cap(A\cap B)=A\cap B\cap\bar{B}$, which is the empty set. This is a contradiction, so if $x\in A\cap\bar{B}$ we deduce that $x\notin(A\cap B)$.
In conclusion, we can say that

$x\notin A\cap B$ is a consequence of $x\in A\cap \bar{B}$

or

$x\in A\cap\bar{B}$ implies $x\notin A\cap B$

or

$A\cap\bar{B}\subseteq A\cap B$

A: $A\cup B$ means elements which are in $A$ or $B$ or both. So for an element to not be in the union it shouldn't be in any of $B$ and $A$.
$A\cap B$ means the elements which are in both $A$ and $B$. So for a element to not be intersection it would suffice if the element isn't in anyone of the sets. Both ways it would not be in the intersection.
A: Piggybacking off egreg-
Let $x \in A \ \cap B^c$. Then $x \in B^c$.
1.) By contradiction assume $x \in (A \cap B)$.
2.) Then $x \in B$.
3.) So $x \in B$ and $x \in B^c$. The required contradiction.
