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 set equality you show $\supset$, $\subset$ respectively.
$\subset$:
Let $x \in A$. Then $x$ either in $A \cap B$ or in $A \cap B^c = A - B$, so $x \in (A \cap B) \cup (A - B)$.
$\supset$:
Let $x \in (A \cap B) \cup (A - B)$. Then either $x$ in $ A \cap B$ or x in $A \cap B^c$. But in both cases $x \in A$, therefore $x \in A$.
