Does permitting comprehension for all (and only) contingent formulas result in paradoxes?

Question

Does permitting comprehension over all well-formed formulas that are neither contradictions nor tautologies result in paradoxes?

I have a hunch that a simple extensional set theory with the "axiom schema of contingent comprehension" is probably susceptible to a well-known paradox and there's a short proof of its inconsistency, but I'm having trouble finding one.

This question is inspired by the argument that plural logic is not susceptible to Russell's paradox given here and the set theory NFU which uses the type system of the simple theory of types only in its comprehension schema. The idea is to insist that all sets are nonempty (similar to rules about plurals in plural logic), but to identify all the different levels of our hierarchy, like NFU does.

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Let's take a set theory with the following axioms.

1. Every set has at least one element.

2. Every set has at least one non-element.

3. Extensionality. Sets with the same elements are equal.

Let us additionally insist on the following axiom schema of restricted comprehension.

$$ \text{there exists a $z$ such that $\varphi(z)$ holds and a $w$ such that $\varphi(w)$ fails} \implies \{ x : \varphi(x) \} \; \text{exists} $$

Russell's paradox begins by defining the Russell set $R = \{ x : x \not\in x \} $.

However, in this setting, using this notation in first place brings with it the additional assumption that $z \not\in z$ is contingent.

If we reject this and say that $z \not\in z$ is either a tautology or a contradiction, then we can defuse Russell's paradox by refuting one of its hypotheses.

A version of Curry's paradox has us consider the set

$$ A = \{ x : x \in x \to \varphi \} $$

However, if we insist that $z \in z$ is a contradiction, then $x \in x \to \varphi$ is not allowed as the body of an expression in set-builder notation.

Answer 1

Well, if you allow empty models, then this theory is consistent because the empty model satisfies it. There are also two different models with just two sets: one in which each set contains only itself, and another in which each set contains only the other set. But if you add an axiom saying that more than two sets exist, then I claim your axioms become inconsistent.

By Russell's paradox, you must have either $x\in x$ for all $x$ or $x\not\in x$ for all $x$. Suppose first that $x\in x$ for all $x$. By comprehension, the singleton $\{x\}$ exists for any $x$, so $\{x\}\in \{x\}$, so $x=\{x\}$. In particular, every set is a singleton. But now since there exist more than two sets, we can pick two different sets $x$ and $y$ and form the set $\{x,y\}$ (which we know exists because it is not the whole universe). This set is not a singleton, which is a contradiction.

The case where $x\not\in x$ for all $x$ is similar, you just take complements of all the sets. For any $x$, the set $V\setminus\{x\}$ exists (where $V$ is the universe). Since $V\setminus\{x\}\not\in V\setminus\{x\}$, we must have $x=V\setminus\{x\}$, so every set is the complement of a singleton. But we also know that singletons exist, and they are not complements of singletons since there exist more than two sets. This is a contradiction.

(In fact, you can deduce the second case from the first case by symmetry, since your axioms are unchanged if you swap the relation $\in$ with its negation.)