Question is:
$\forall$ Sets $A, B, C$ , If $A - (B \cap C)$ = $\emptyset$, then $(A - C)\subset B$
I know this question is True, here is my proof.
Let $A , B, C$ be sets
Assume $A - (B \cap C) = \emptyset$. I will prove $A - C \subset B$. Let $x \in (A-C)$. Then, $x \in A, x \notin C$. I will prove $x \in B$ by contradiction.
Suppose $x \notin B$.
Since $A - (B\cap C) = \emptyset$, $\forall x\in A,x \in (B\cap C)$
Then, since $x \in (B \cap C)$, $x \in B$, and $x \in C$. Since we assumed $x \notin B$, $x\notin C$, we have proof by contradiction that $x \in B$.
It is okay, your formulation in the last part is just a little bit strange.
Suggestion: Suppose $x\notin B$. Then $x\notin B\cap C$ and hence $x\in A-(B\cap C)$. But $A-(B\cap C)=\emptyset$ so we have proof by contradiction that $x\in B$.
EDIT: Now I got the spare time to explain your mistakes. You wrote
Two problems: First, you already used the variable $x$ before, so this is highly confusing. Second, even if you used $y$ instead, the comma "$,$" is still irritating. It looks like you wanted to write
i.e. any element in $A$ must also be in $B\cap C$. Let's move on.
again, it is unclear what your $x$ is. Why is $x\in B\cap C$? Are you referring to
? Then this would be okay, but because of your "$\forall x$" notation earlier, this is very unclear.
Finally, you make a technical mistake in your proof. You assumed $x\in A - C$ and followed that $x\in C$. You already reached a contradiction! But not in the sense of "proof by contradiction", but you showed that $A-C$ is the empty set! Assuming at one point that $x\notin B$ is pointless, because you showed that such an $x$ does not exist. So concluding that $x\in B$ by "proof by contradiction" is highly invalid!
But luckily for you, since $A-C=\emptyset$ and $\emptyset\subseteq D$ for any set $D$, you can still conclude that $A-C\subseteq B$. with your approach.