Rigorously prove If $A \setminus B = A$, then $A \cap B = \emptyset$ Rigorously prove: If $A\setminus B = A$, then $A \cap B = \emptyset$.
Logically it makes sense i.e. since if $A\setminus B$ is equal to $A$, then $B$ must be the empty set hence $A \cap B$  must be equal to the empty set since anything intersected with the empty set is the empty set.
My main question is how I can go about this to prove it formally, i.e 'mathematically/rigorously'
My attempt: 
$\left(A \setminus  B \right) \subseteq A$ (1)
$A$ is a subset of $A$ hence also a subset of $\left(A \setminus  B \right)$
So:
$A \subseteq   \left(A \setminus  B \right)$ (2)
Combining (1) and (2) we are allowed to conclude that:
$  \left(A \setminus  B \right)  = A$
Hence since $ \left(A \setminus  B \right)  = A$, then $B = \emptyset$ then rtp that $A \cap B = \emptyset$
$A \cap B$
$(x \in A \land x \in B)$
$(x \in A \land x  \in \emptyset)$
$x \in \emptyset$
Thanks in Advance
 A: 
"  Hence since $A\setminus B = A$, then $B = \varnothing$ then rtp that $A\cap B=\varnothing$"

Nowhere we reach the conclusion that $B=\varnothing$.

This is a way to do it:
Assume that $x\in A\cap B$. 
Then $x\in A=A\setminus B$ and $x\in B$. 
Then $x\notin B$ and $x\in B$ so a contradiction is found.
We conclude that the assumption $x\in A\cap B$ is false, and this for every $x$.
That means exactly that $A\cap B$ is empty.
A: Note that it is always true that 
$$A \setminus B = A \cap B^C$$
Hence, if $A \setminus B =A$, then $A \cap B^C =A$, and so:
$$A\cap B= (A \cap B^C) \cap B= A \cap (B^C \cap B)=A \cap \emptyset=\emptyset$$
A: A correct proof  is sketched in the other answer. To answer your question about whether your reasoning is correct: Not really. You're trying to show that if $A\setminus B=A$ then $\cap B=\emptyset$. So $A\setminus B=A$ is given. Hence a line saying "we may conclude $A\setminus B=A$ makes very little sense.
Not that it matters, since you're given that $A\setminus B=A$, but your proof of this fact is also wrong. You state that $A$ is a subset of $A\setminus B$, but that's not true in general.
A: Suppose $x\in A\cap B$. Then $x\in A$ and $x\in B$. But then $x\not\in A\setminus B$. Since $A\setminus B=A$, this leads to a contradiction. Done.
