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

[Ozma]

Let [imath]A,B[/imath] be subsets of a set [imath]X[/imath]. I wanted to know if the following proof of [imath](A\setminus B) \cap (B\setminus A) = \varnothing[/imath] 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: [imath]\exists x, (x \in A \wedge x \notin A \wedge x \in B \wedge x \notin B)[/imath] is false for each [imath]x \in X[/imath], so [imath]\nexists x, (x \in A \wedge x \notin A \wedge x \in B \wedge x \notin B)[/imath] is true for each [imath]x \in X[/imath]. Hence, by associativity and commutativity of logical disjunction, we deduce that [imath]\nexists x, ((x \in A \wedge x \notin B) \wedge (x \in B \wedge x \notin A))[/imath] is true for each [imath]x \in X[/imath]. So, [imath]\nexists x, ((x \in A \setminus B) \wedge (x \in B \setminus A))[/imath] is true for each [imath]x \in X[/imath]; finally, [imath]\nexists x, (x\in (A \setminus B) \cap (B \setminus A))[/imath] is true for each [imath]x \in X[/imath] and so [imath](A \setminus B) \cap (B \setminus A) = \varnothing[/imath] is true.

Is this formally correct?

 

[blamocur]

I don't see any serious flaws in there.

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