This is a possible salvage for the failed attempt in this posting.

The salvage here is to require that every subformula $ \psi(y)$ of $ \phi$ having no parameter other than those in $ \phi$ , must satsify the antecdent of comprehension. To write this formally, it is:

**Comprehension:** If $ \phi$ is a formula in the first order language of set theory (i.e.;$ \sf FOL(=,∈))$ , in which the symbol $ “x”$ doesn’t occur free, and if $ \psi_1(y),..,\psi_n(y)$ are *all* subformulas of $ \phi$ in which $ y$ is free, and having no parameter that is not a parameter of $ \phi$ ; then: $ $ [\bigwedge_{i=1}^n \big{(}\exists y ((\psi_i(y))^=) \wedge \exists y ((\neg \psi_i(y))^=) \big{)} \to \exists x \forall y (y \in x \iff \phi)] $ $ ; is an axiom.

**Axiom of Multiplicity:** $ \forall x,y \ \exists z (z \neq x \land z \neq y)$

/

I personally think this is complex a little bit, I highly doubt its consistency though. Yet if there is a chance that this is consistent, then it would actually prove all axioms of $ \sf NF$ , since full Extensionality is assumed here.