Suppose the Set theory in question followed all axioms of ZF or ZFC, except for the axiom of regularity. Additionally the axiom schema of specification was altered from
$$\forall z\forall w_1 \forall w_2 \dots \forall w_n\exists y\forall x[x \in y \iff ((x\in z)\wedge\phi(x))]$$
$$\forall w_1 \forall w_2 \dots \forall w_n\exists y\forall x[x \in y \iff ((x\neq y)\wedge\phi(x))]$$
for the purpose of constructing sets beyond just subsets of some $z$.
Without regularity and standard specification, this immediately opens the doors to a possibility of Russel's Paradox. However, a set $R$ cannot be built (this way) to have $R\in R$, as it would have to follow $R \in R \iff ((R\neq R)\wedge\phi(R))$.
Would this avoid Russel's Paradox? Even if so, would it lead to another paradox?
Best Answer
No, this form of comprehension is inconsistent with even the ability to form singletons and unions of two sets.
Suppose $A=\{x: x\not\in x\wedge x\not=A\}$. Then I claim $A\cup\{A\}$ is the usual Russell set.
If $x\in A\cup \{A\}$ then $x\not\in x$. This is because for such an $x$, either $x\in A$ or $x=A$, and no element of $A$ contains itself by the "$x\not\in x$"-clause of the definition of $A$ and $A\not\in A$ by the "$x\not=A$"-clause of the definition of $A$.
If $x\not\in x$ then $x\in A\cup\{A\}$. Again, we reason by cases. If $x\not\in x$ then either $x\not=A$ (in which case $x\in A$) or $x=A$ (in which case $x\in\{A\}$), and either way $x\in A\cup\{A\}$.
And now we ask whether $A\cup\{A\}$ is an element of itself.