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?
 A: 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.
A: 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 $x \in A$.
A: $\rm\ A\backslash B\  =\ A\cap\overline B\ \ \:$ so $\rm\ \: (A\backslash B)\cup (A\cap B)\ =\ (A\cap\overline B)\cup(A\cap B)\ =\ A\cap(\overline B\cup B)\ =\ A$
A: Let $x \in A$. Then $x \in A \backslash B$ or $x \in A \cap B$. Likewise, if $x \in A \backslash B$ or $x \in A \cap B$ then $x \in A$. 
A: Working inside a universe $X$: 
$$A = A \cap X = A \cap (B \cup (X \setminus B)) = ( A \cap B ) \cup ( A \cap (X \setminus B)) = (A \cap B) \cup (A \setminus B)$$   
A: Using your initial arguments:

$(A\setminus B) \cup (A \cap B)$ 
  
  
*
  
*$X \in A\setminus B \implies  { x \in A  \text{ and } x\not\in  B }$
  
  
  or                                    
  
  
*
  
*$X \in A \cap B \implies { x \in A \text{ and } x \in B }$
  

So what we have is $x\in A\setminus B$ non-exclusive OR $x\in  A \cap B$.
So  $ { x \in A  \text{ and } x\not\in  B } \lor { x \in A \text{ and } x \in B }$.
In Boolean logic this becomes $A \land (B\lor ¬B)=A$.
A: For all, each, any, and every $x$ we have $$(x\in A)\iff ((x\in A\land x\in B)\lor (x\in A\land x\not \in B))\iff$$ $$\iff (x\in A\cap B)\lor (x\in A \backslash B).$$
