$A \cap B= \emptyset \implies A\setminus B=A$ 
*

*Given: $A \cap B=  \emptyset $

*Prove: $A\setminus B=A$.

*My thoughts:


$x \in  \emptyset  \Rightarrow x \in A \cap   \emptyset  $ and $ x \in A \cap B \Rightarrow x \in A \cap B \cap   \lnot B $ and $ x \in A \cap B \Rightarrow x \in (A \cap  \lnot B) \cap B$  and $ x \in A \cap B \Rightarrow x \in (A\setminus B) \cap B $ and $ x \in A \cap B \Rightarrow A\setminus B=A
$
 A: 
Def. let be $A,B$ sets: $$A\setminus B:=\{x|x \in A \wedge x \notin B\} $$

We prove following:

The$_0$. let be $A,B$ sets: $$ A \setminus (A \cap B)=A \setminus B$$ Proof: $$ \begin{align} x \in A \setminus (A \cap B) &\leftrightarrow
 x \in A\wedge (x \notin (A\cap B)) \\ &\leftrightarrow x \in A \wedge
 (x \notin A \vee x \notin B) \\ &\leftrightarrow (x \in A \wedge x
 \notin A) \vee (x \in A \wedge x \notin B) \\ &\leftrightarrow x \in A
 \setminus A \vee x \in A \setminus B \\ &\leftrightarrow x \in
 \emptyset \vee x \in A \setminus B \\ &\leftrightarrow x \in \emptyset
 \cup A \setminus B \\ &\leftrightarrow x \in A\setminus B
 \end{align}$$

Now we prove your theorem:

The$_1$. let be $A,B$ sets: $$A \cap B=\emptyset \to A \setminus B=A$$
  Proof: we use The$_0$. therefore: $$ A \setminus B= A \setminus (A \cap B)= A \setminus \emptyset= A$$

A: It is hard to follow what is going on here. You should adopt a systematic approach.
To prove that $S_1 = S_2$, where $S_1$ and $S_2$ are sets, you must prove that $S_1 \subseteq S_2$ and $S_2 \subseteq S_1$. 
This then means that you prove that whenever $x \in S_1$ then also $x \in S_2$ and whenever $x \in S_2$ then also $x \in S_1$.
A: If $x \in A \setminus B$ then certainly $x \in A$ , so $A \setminus B \subseteq A$.
If $x \in A$, then $x \notin B$, otherwise $x \in A \cap B = \emptyset$, contradiction.  So $x \in A \setminus B$ by definition, hence $A \subseteq A \setminus B$, so we have equality of sets.
If you like "equation style" proofs ($X$ denotes the "universe" with respect to which we take complements): 
$$A  =A \cap X = A \cap (B \cup B^c) = (A\cap B) \cup (A \cap B^c) = \emptyset \cup (A \cap B^c)  = A \cap B^c= A \setminus B$$
A: Without words, arguments tend to be pretty hard to follow. Here's how I'd do it:
It's clear that $A \setminus B \subseteq A$. So let's show the converse. Assume $x \in A.$ We need to that show $x \in A \setminus B$, in other words $x \in A$ and $x \notin B$. Well, we already know $x \in A$. So all we need to do is show $x \notin B$. Assume toward a contradiction that $x \in B$. Then $x \in A \cap B$. So $x \in \emptyset$, a contradiction. We conclude that $x \notin B$, as required. This completes the proof.
A: If you don't want to use subset relationships, you can use co-implications:
$$ x \in A\setminus B \iff x \in A \  \wedge x \not\in B \iff x \in A$$
where the last co-implication holds since $A \cap B = \emptyset$ and so, if $x \in A$ it cannot be in $B$.
