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Is this a good definition for $A\neq B$ ?

I define $A=B$ to be

$\forall x[(x\in A)\leftrightarrow (x\in B)]\label{1}\tag{1}$

If ($\ref{1}$) is my definition for "$=$", then its negation is ($\ref{2}$)

$\exists x[(x\notin A)\oplus(x\notin B)]\label{2}\tag{2}$

My question: Is ($\ref{2}$) a good definition for $A\neq B$ ? If it is not, which out of ($\ref{1}$) and ($\ref{2}$) is wrong?

($\ref{2}$) seems to say that the elements of A are not guaranteed to be the same as B. It does not appear to say that they will never be the same no matter which element we pick. This is what confused me about the definition, for I have sometimes thought of $\neq$ as meaning that they are never the same.

As a bonus question; would the negation of ($\ref{3}$) have any meaning?

$(A=B)\leftrightarrow(\forall x[(x\in A)\leftrightarrow (x\in B)])\label{3}\tag{3}$

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    The simplest way to negate (1) gives $\exists x [(x \in A) \oplus (x \in B)]$, but of course, (2) is equivalent. In general, $A \neq B$ means $\neg(A = B)$. – 2017-02-23
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    @FabioSomenzi But is that an appropriate definition? Will it get me through all instances of the use of = as something to refer to? Also, thanks for the simplification. I have tried to make it s habit not to negate the sub-statements of a formula involving $\leftrightarrow$ when I negate the whole thing but sometimes I forget. – 2017-02-23
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    @FabioSomenzi: The _simplest_ way to negate (1) is to stick a $\neg$ in front of it and do nothing further. – 2017-02-23
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    Concerning (3), as a formula of the language it is a tautology, because after substituting the definition of $A=B$, it has the form $P \leftrightarrow P$. Its negation is well defined; it's just always false. – 2017-02-23
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    Bonus question : if we have in out theory $T$ the **definition** (1), this is an axiom of our theory : $T \vdash A=B \leftrightarrow \ldots$. Thus, its negation is a *false* sentence of the theory. – 2017-02-23
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    @HenningMakholm "gives" and "is" are different verbs in my neck of the woods. – 2017-02-23
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    @MauroALLEGRANZA I am not well enough informed to be able to read T⊢A=B↔…(because of the ⊢ symbol). I should point out here that I have not yet looked into the subject of "Theories" in logic. Thanks for making me curios enough to look into it though. – 2017-02-23
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    You are welcome :-). The gist is simple: as you said, a def is a bi-conditional. Usually, we introduce them in a theory: in your def you are using $\in$; this means that you are working in set theory (a theory where the symbol $\in$ is present). If you are working in e.g. first-order arithmetic, the symbol $\in$ is not part of the language: thus, you cannot define $n=m$ that way. – 2017-02-23
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    @FabioSomenzi: And the procedure I describe **gives** $\neg \forall x(x\in A\leftrightarrow x\in B)$. – 2017-02-23
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    @HenningMakholm It sure does, but why do you ignore the context of my sentence? I have a high-enough opinion of your competence that I can't come up with a satisfactory explanation. – 2017-02-23
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    @FabioSomenzi: Which content is it you think I'm ignoring? I'm arguing that the OP should _not_ be asked, expected, or advised to to any deep level rewriting on the formula while negating it. Just stick a $\neg$ in font and _let that be the definition_ of $\neq$. Saying that he should do anything more than that _is what I'm opposing_. – 2017-02-23
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    @HenningMakholm thanks for clarifying. I agree about the definition of $\neq$. (See the second half of my first comment.) What happens next depends on the circumstances. For instance, when translating LTL to Buechi automata, reducing a formula to negation normal form may be not only advisable, but required to avoid automaton complementation. The context for my comment, however, is that when you negate an equivalence by replacing $\leftrightarrow$ with $\oplus$ there is no need to negate the operands. The OP's response suggest that that point went across. – 2017-02-23

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The sane thing is to define $A\ne B$ to mean $\neg(A=B)$.

If you unfold the definition of $A=B$, this becomes $$ A\ne B \quad\text{ means }\quad \neg \forall x(x\in A\leftrightarrow x\in B) $$ There are various things this is logically equivalent to, but that's not a matter for the definitions themselves -- it's just something you might need to appeal to when you reason with those definitions.

Your (2) just says that there is at least one thing that is in one set and is not in the other. That's a perfectly good characterization of disequality between sets.

For example, $\{1,2,3\}$ and $\{2,3,4\}$ are different sets, even through $2$ and $3$ are in both of them, and $6$ or $\pi$ is in neither.