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In Metamath, "every set variable is a set" is a theorem. What to make of a statement like "not exists x ( x not equal x )"? This can be proved starting with "for all x ( x = x )", applying double negation, then quantifier negation. In "not exists x ( x not equal x )", the set variable x is a set just because it is a set variable, but also it is not a set because of what the sentence says (and the sentence is a theorem). Which view is right? What should I make of this?

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    If $x$ doesn't exist, it can have any properties you like. It can be a pink unicorn flying a helicopter. – 2017-01-19
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    The inconsistency that seems a problem to me is not that x not equal x is impossible (like the unicorn), but that the set variable x both is and isn't a set (as a set it both is and isn't an element of the universe, for instance). We wouldn't say that the set variable doesn't exist, because look, there is an x on the paper in the middle of the formula. But the sentence says that the set which the set variable represents does not exist. Let me know if I have missed the point of your comment. – 2017-01-19
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    In order to be really sure we're all talking about the same things, we may need to dig into the particular definitions and theory used in Metamath. If http://metamath.org/ is the home page, that's easy enough to find, but a little guidance from you on how to find a page that says "every set variable is a set" would be very helpful. – 2017-01-19
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    Every set variable is a set is asserted in this theorem: http://us.metamath.org/mpegif/vex.html – 2017-01-19
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    A short way of showing the other part is to get a nonexistent set from a false "property" http://us.metamath.org/mpegif/nex.html, like x not equal x above. Sorry for not including the references at first, and thanks for your interest! – 2017-01-19

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One difference I notice between the theorems named vex and nex is that $x$ is a free variable in one but is a bound variable in the other.

Specifically, vex states that $$ \vdash x \in V. $$ In this statement, $x$ is a free variable. So I think a more careful statement of the theorem in mathematical English might be, "Every free set variable is a set."

Once you put a quantifier in front of a variable, for example in $ \vdash \lnot \exists x \varphi, $ the variable becomes bound and the rules of the quantifier govern what is possible. In particular, when we use $\lnot \exists,$ saying that something does not exist, many properties (or lack of properties) that would cause trouble if the thing existed can no longer cause trouble. So, for example, $x$ must be a set if there is any such thing as $x,$ but if there is no such thing, we can just as easily call it a set and not a set at the same time.

I'm reminded of an old joke:

A customer in a butcher shop asks the price of roast beef, and on hearing the answer, complains:

"Why is the price so high? The price is a dollar lower at the shop across the street."

"Why don't you go buy your roast beef there, in that case?"

"I can't, they're all out of roast beef."

"Well, if I were out of roast beef, I could lower my price by a dollar."

Perhaps someone with deeper insight into logical notation can explain the meaning of the variables in the vex and nex theorems further.

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    Great joke and I hadn't noticed the difference, so this helps. I will not checkmark this answer yet in case someone can add more. – 2017-01-19
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You have actually recognized the problem introduced by Russell's paradox that motivates some set theorists to study set theory in terms of a modal structure obtained by its many possible models. The problem is the distinction between what a quantifier is "supposed to mean" and how Russell's paradox interferes with the representation of classical quantifiers in the square of opposition.

If you examine the formal sentences associated with the square of opposition in the link,

https://plato.stanford.edu/entries/square/#ModRevSqu

you can see the problem with "Every set is not an element of itself" and interpreting the universal quantifier in terms of sets. Free variables carry implicit universal quantification on the basis of what quantifiers are "supposed to mean".

It is best to remember that metamath is intended to execute logical transformations. So, you may view the system as either assuming that set theory has a model or that it is treating set theory in a purely formal matter.

If you would like to consider the problem of negative existentials further, look at the link,

https://plato.stanford.edu/entries/existence/#FreRusExiNotProInd