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⟺𝑥≠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).