Prove that $A\subseteq B\Longleftrightarrow A\cap B = A$ In set theory logic mathematics. How would i do the proof for: $A\subseteq B\Longleftrightarrow A\cap B = A$
 A: First Part:
Suppose  A⊆B.  Then if for any x belonging to A, then x belongs to B.
Now suppose that x belongs to (A∩B).  So, x belongs to A, and x belongs to B also.  Thus, x belongs to B.  Since x comes as arbitrary, for any x if x belongs to (A∩B), x belongs to A also.  
Suppose that x belongs to A.  Since A⊆B, x belongs to B also.  Thus, since x belongs to A, as well as B, x belongs to (A∩B).  Since x comes as arbitrary, for any x, if x belongs to A, then x belongs to (A∩B).  This paragraph and the last paragraph imply that (A∩B)=A.
Consequently, if A⊆B, then (A∩B)=A.  End of first part.
Second Part:
Suppose that (A∩B)=A.  Suppose that x belongs A.  Since (A∩B)=A, x belongs to (A∩B).  Since x belongs to (A∩B), x belongs to B also.  Note that x comes as arbitrary.  Thus, if x belongs to A, then x belongs to B.  By definition of ⊆ it follows that A⊆B.
Consequently, if (A∩B)=A, then A⊆B.  End of second part.
Both parts combine to imply that {[A⊆B]⟺[(A∩B)=A]}.
A: If $A \subseteq B$, then the $\Rightarrow$ part is true, since all elements of $A$ are also elements of $B$ - proceed next by using the definition of $\cap$. If $A\cap B=A$, then all the elements of $A$ are also elements of $B$, by definition of $\cap$. Thus, $A\cap B=A\implies A\subseteq B$, i.e., $A\subseteq B\iff A\cap B=A$.
A: First suppose that $A\subseteq B$.


*

*Let $x\in A \cap B \implies x\in A $
So, $A\cap B\subseteq A$  

*Now, let $x\in A $. Since $A\subseteq B, x\in A\cap B \implies A\subseteq A\cap B$
Since $A\subseteq A\cap B$ and $A\cap B \subseteq A, A\cap B = A$

Now suppose that $A\cap B = A$
Let $x \in A \implies x \in A \cap B \implies x \in B$
Since $x \in A \implies x \in B, A\subseteq B$   

$$\therefore A\subseteq B\iff A\cap B=A$$
