Prove that (A \ B) ∩ (B \ A) = ∅ using mostly formal logic

[Ozma]

Let $A,B$ be subsets of a set $X$. I wanted to know if the following proof of $(A\setminus B) \cap (B\setminus A) = \varnothing$ is correct from a logical point of view: I know there are simpler proofs, but I wanted to improve my skill with formal logic.

Proof: $\exists x, (x \in A \wedge x \notin A \wedge x \in B \wedge x \notin B)$ is false for each $x \in X$, so $\nexists x, (x \in A \wedge x \notin A \wedge x \in B \wedge x \notin B)$ is true for each $x \in X$. Hence, by associativity and commutativity of logical disjunction, we deduce that $\nexists x, ((x \in A \wedge x \notin B) \wedge (x \in B \wedge x \notin A))$ is true for each $x \in X$. So, $\nexists x, ((x \in A \setminus B) \wedge (x \in B \setminus A))$ is true for each $x \in X$; finally, $\nexists x, (x\in (A \setminus B) \cap (B \setminus A))$ is true for each $x \in X$ and so $(A \setminus B) \cap (B \setminus A) = \varnothing$ is true.

Is this formally correct?

 

[blamocur]

I don't see any serious flaws in there.

One minor objection: I find phrases like "$\nexists x, P(x)$ ... is true for each $x\in X$" somewhat awkward/contradictory: if there is no such $x$ for which preidicate $P(x)$ is true, then "is true for each $x$" makes no sense to me.