Prove A = (A\B) ∪ (A ∩ B) I have to demonstrate this formulae:
Prove $A = (A\setminus B) ∪ (A ∩ B)$
But it seems to me that it is false.

$(A\setminus B) ∪ (A ∩ B)$ 
  
  
*
  
*$X \in A\setminus B \implies  { x ∈ A  \text{ and } x ∉ B }$
  
  
  or                                    
  
  
*
  
*$X ∈ A ∩ B \implies { x ∈ A \text{ and } x ∈ B }$
  



so: 
$x ∈ A ∩ B$
so:
  $A ≠ (A\setminus B) ∪ (A ∩ B)$

Did I solve the problem or I am just blind?

To show that two sets are equal, you show they have the same elements. 
Suppose first $x\in A$. There are two cases: Either $x\in B$, or $x\notin B$. In the first case, $x\in A$ and $x\in B$, so $x\in A\cap B$ (by definition of intersection). In the second case, $x\in A$ and $x\notin B$, so $x\in A\setminus B$ (again, by definition).
This shows that if $x\in A$, then $x\in A\cap B$ or $x\in A\setminus B$, i.e., $x\in (A\setminus B)\cup(A\cap B)$. 
Now we have to show, conversely, that if $x\in (A\setminus B)\cup(A\cap B)$, then $x\in A$. Note that $x\in(A\setminus B)\cup(A\cap B)$ means that either $x\in A\setminus B$ or $x\in A\cap B$. In the first case, $x\in A$ (and also, $x\notin B$). In the second case, $x\in A$ (and also, $x\in B$). In either case, $x\in A$, but this is what we needed.
In summary: We have shown both $A\subseteq (A\setminus B)\cup(A\cap B)$ and $(A\setminus B)\cup(A\cap B)\subseteq A$. But this means the two sets are equal.
