Prove $A $ \ $B $ = $A \cap B^c $ I see the use of $A $ \ $B $ = $A \cap B^c $ being used in bigger problems but how do you prove this? Is the proof as simple as:
$A $ \ $B $ $\iff$
$ x \in (A \setminus B) \iff 
 x\in A \cap \lnot(x\in  B)  \iff A \cap B^c  $ 
Thanks for your help!

Your suggested proof is syntactically incorrect, because $A\setminus B$ is a term, not a proposition. Its type, if you want to talk about programming or type theory, is "set" rather than "boolean".
The same goes for $A\cap\lnot(x\in B)$.
You have the idea. But my advice to you is not to try and "cut it short and cool with lots of math symbols". Write in English, or whatever your native tongue is, a full and wordy proof. Readability beats suave looks any given day in mathematics. Especially on introductory level.
If you do want to do it in a single $\iff$ chain, you can still do it, but you have to remember the check your types. Intersect only sets with other sets, and be sure that the "value" before setting another $\iff$ in the chain has to be a boolean value.
Let me help you get started:


*

*$x\in A\setminus B$ if and only if

*$x\in A$ and $x\notin B$ if and only if

*$x\in A$ and ...

