Element method on set inequality proof-verification With respect to the following question:
Prove that For all sets $A,B,C$,  if $(A \cap B) - C= \varnothing$, then $A-B = A-C$
Does the following proof verify it?

Specifically, does the existence of element $x$ in $A$ and $B$ and $C$ also prove that $A-B = A-C$
 A: Let us correct your proof.
First of all, the definition of $A-B$ is the set of elements contained in $A$ which are not in $B$. Therefore, if $X-Y = \phi$, it does not mean $X=Y$, rather that all elements of $X$ are also in $Y$, or that $X \subset Y$.
With this in mind, our aim is to show the following:$A-B=A-C$, given $(A\cap B) - C = \phi$, which translates to $A \cap B \subset C$.
I have no clue what you did after that, but let me proceed any way.
For this, let $x \in A - C$. Then, $x \in A, x \notin C$. Therefore, $x \notin A \cap B$, therefore $x \notin B$, therefore $x\in A - B$. So $A - C \subset A-B$.
Now, let $x \in A -B$. As it turns out, $A-B \nsubseteq A-C$ in general. The logic is simple. If $x \in A - B$, then $x \notin A \cap B$. But then, $A \cap B$ is only a subset of C, it is not equal to $C$. Therefore, we cannot conclude from here that $x \notin C$, so $x \in A-C$.
For example, let $A = \{ 1,2,3\}, B = \{ 2,3,4\}, C = \{ 1,2,3,7\}$. Then, $A - B = \{1\}$, and $A-C = \phi$, so they are not equal. We can see that $A \cap B = \{2,3\}$, which is a strict subset of $C$.
