Prove that A-B=A ⇔ A⋂B=Φ

Question

Another question from my 11th grade Mathematics textbook. I actually proved it but I still want to confirm if it's right or not. Please check it out :

In order to prove that A-B=A ⇔ A⋂B=Φ, we will first prove that A-B=A ⇒ A⋂B=Φ and then we'll prove 
that A⋂B=Φ ⇒ A-B=A
First part : Proving that  A-B=A ⇒ A⋂B=Φ :-
    Let x be an arbitrary element of A
    So, x ∈ A
    Since A=A-B, so x ∈ A-B
    So, x ∈ A and x ∉ B
    This means that for an arbitrary element of A, it is not an element of B
    So, A and B are both disjoint sets
    So, they have no element in common
    So, A⋂B=Φ
    So, A-B=A ⇒ A⋂B=Φ

Second part : Proving that A⋂B=Φ ⇒ A-B=A :-
    A-B = A-(A⋂B)
    Since A⋂B = Φ, so, A-B = A-Φ = A
    So,  A⋂B=Φ ⇒ A-B=A

So, A-B=A ⇔ A⋂B=Φ
Hence, proved

Let me know about any flaws in this proof, even the slightest, especially if the proof can be made shorter and if I have included anything unnecessary in it.

Thanks :)

Answer 1

Here's a proof for the second part: suppose that $A \cap B = \emptyset$. It is clear that $A - B \subseteq A$, so we show that $A \subseteq A - B$. Let $x$ be an arbitrary element of $A$.

• So, $x \in A$.
• Since $A \cap B$ is empty, $x \in A$ and $x \notin A \cap B$.
• So, $x \in A$ and it does not hold that ($x \in A$ and $x \in B$).
• Equivalently, $x \in A$ and it does not hold that $x \in B$.
• So, $x \in A - B$.

So, $A \subseteq A-B$, which is to say that $A - B = A$ as was desired.