ZF without regularity with "reflexive" comprehension.
Can we successfully defuse known paradoxes (and produce a consistent theory) by using a comprehension schema that limits comprehension to well-formed formulas of the form $v \in v \lor \varphi$ and $v \in v \land \varphi$?
I was thinking about the set of all reflexive sets $\{ x : x \in x \}$ and was wondering if it was as badly behaved as the Russell set. On the face of it it isn't, since $\{x : x \in x \} \in \{x : x \in x\}$ appears to be capable of being given either truth value without resulting in inconsistency. Then I had the idea to try to use $x \in x$ as a guard, rather than $x \in A$ as the ZF comprehension schema uses.
This version of the comprehension axiom that's intended to defuse Moh Shaw-Kwei's paradox and Russell's paradox while being intentionally incompatible with the comprehension schema for ZF and for NF(U).
This comprehension schema is very bizarre, operations like intersection and union cannot be defined using it and reflexive sets are constantly getting added in or quantified over.
Only well-formed formulas of the form $v \in v \land \varphi$ or $v \in v \lor \varphi$ where $v$ is the variable symbol used in the comprehension.
So, this comprehension schema gives us the universal set $U$ and the empty set $\varnothing$.
$$ U = \{ x : x \in x \lor \top \} $$
$$ \varnothing = \{ x : x \in x \land \bot \} $$
Here's an explanation of why Russell's paradox doesn't go through, let $@$ represent $\land$ or $\lor$.
We define the Russell set as follows.
$$ R = \{ x : x \in x \mathop@ x \not\in x \} $$
However, if we picked $@$ to be $\land$, then $R$ is the empty set. If we picked $@$ to be $\lor$, then $R$ is the universal set.
The universal set $U$ in this set theory is reflexive, but this doesn't appear to immediately cause a paradox.
For Moh Shaw-Kwei's paradox, we examine the following set
$$ A = \{ x : x \in x \mathop@ x \in x \to \varphi \} $$
Which is equivalent to
$$ A = \{ x: x \in x \mathop@ (x \not\in x \lor \varphi) $$
$A$ is equivalent to the universal set if $@$ is or, and is equivalent to the following if $@$ is and.
$$ \{ x: (x \in x \land x \not\in x) \lor (x \in x \land \varphi) \} $$
Which is
$$ \{ x : x \in x \land \varphi \} $$
So, then, we can ask if $A$ is reflexive
$$ A \in A \leftrightarrow A \in A \land \varphi $$
This does not appear to result in a paradox, this statement only fails when $A$ contains itself, but $\varphi$ is false.
For concreteness, here are all the axioms I am using. Most of them are copied verbatim from ZF. Some of the axioms of ZF have been left out.
Capital letters are free variables; lowercase letters are bound variables.
I take the unmodified axiom of extensionality, the toy set theory in this question does not have urelements.
The right direction is a trivial theorem, but I'm adding it anyway.
$$ (\forall x . x \in A \leftrightarrow x \in B) \leftrightarrow A = B $$
I'm excluding the axiom of regularity. The set theory in question is not well-founded.
I take the axiom of pairing.
$$ \exists c \mathop. \forall x \mathop. x \in c \leftrightarrow (x = A \lor x = B) $$
I take the axiom of union. I'm not sure whether limiting comprehension so dramatically means that we lack the unary intersection operator or not.
$$ \exists w \mathop. \forall x \mathop. (x \in w) \iff x \,(\in^2)\, A $$
I take the unrestricted axiom of replacement for all well-formed formulas with parameters $\varphi(\,)$ provided $\exists! w \mathop. \varphi(w, A) $.
This is the usual axiom schema of replacement in ZF.
I take the axiom of infinity.
$$ \exists i \mathop. \varnothing \in i \land (\forall x \in i \mathop. (x \cup \{x\}) \in i) $$
I take the axiom of the powerset.
$$ \exists y \mathop. \forall z \mathop. z \in y \leftrightarrow (\forall w \mathop. w \in z \to w \in X) $$
