Question: Suppose the Set theory in question followed all axioms of ZF or [ZFC](https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory), **except** for the **axiom of regularity**. Additionally the **axiom schema of specification** was altered from

$$\forall z\forall w\_1 \forall w\_2 \dots \forall w\_n\exists y\forall x[x \in y \iff ((x\in z)\wedge\phi(x))]$$

to [unrestricted comprehension](https://en.wikipedia.org/wiki/Axiom_schema_of_specification#Unrestricted_comprehension)

$$\forall w\_1 \forall w\_2 \dots \forall w\_n\exists y\forall x[x \in y \iff ((x\neq y)\wedge\phi(x))]$$

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](https://en.wikipedia.org/wiki/Russell%27s_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?

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.
The above "unrestricted comprehension" schema is equivalent to the statement that there is a single element and that element is empty.

Note that the schema holds in the one element domain where the one element is empty.

There is an empty set.
Proof:There is a Y such that x∈Y⟺≠Y∧x≠x. Such a Y must be empty.

Call a set b with exactly 2 elements a 2-set if one element of b is empty and b∉b.

If there are distinct elements, then there is a set which a 2-set or a set whose only element is empty.
Poof: Suppose a is empty and a≠b. Then there is a Y such that x∈Y⟺x≠Y∧(x=a∨x=b). Y is not empty and Y∉Y. If b∈Y then Y is a 2-set. If b∉Y then a is the only element of Y.

Let F(x) be the formula ∀t(t∉x∨x∉t). We observe that if F(s) for all s∈x, then F(x).

Suppose that b is not empty. Then there is a set Y whose only element is b.  

Proof: There is a Y such that x∈Y⟺x≠Y∧x=b. Y is not empty because if it were then b∈Y. Therefore Y is a set whose only element is b.

Suppose there are distinct elements. Then there is a non-empty set Y such that F(Y).

Proof: If there is a set b whose only element is empty, then F(b). Suppose there is no such set. There is a W such that "x∈W⟺x≠W∧(x is not a 2-set)". W is not empty because there is a 2-set and a set whose only element is a 2-set. Suppose c is empty. There is a set Y such that x∈Y⟺x≠Y∧(x=c∨x=W). Y is a 2-set and F(Y).

There is an A such that x∈A⟺x≠A∧F(x). Then F(A).

There is only one element.  

Proof: Suppose there are distinct elements. Then there is a non-empty element b such that F(b). But then there is a Y whose only element is b. One of b and Y must be in A. There is a B such that x∈B⟺x≠B∧(∃t(t∈x)∧((x∈A∨x=A))). A≠B since B only has non-empty elements and so A∈B. But F(B) and therefore B∈A. But this contradicts F(A).