Let's say that a formula in the language of set theory is *flexibly stratified* iff there exists a function $f$ from variable symbols to $\omega$ such that if $x=y$ appears in the formula, then $f(x)=f(y)$, and if $x\in y$ appears in the formula, then $f(x)<f(y)$ (this is in contrast to regular stratification which requires $f(y) = f(x) + 1$ exactly).


I'm going to define *Flexible NF* as:


* Extensionality (same as in NF)
* Flexibly Stratified Comprehension (same as in NF except allowing comprehension over "flexibly stratified" formulae)


This is clearly at least as strong as NF, since every theorem of NF is a theorem of Flexible NF.


Is Flexible NF obviously inconsistent, or obviously equiconsistent with NF?




---


My thoughts:


Could we somehow use flexible stratified comprehension to define the bijection $f: x \mapsto \{x\}$ on e.g. the universal set, and then use that in contradiction with Cantor's Theorem to get inconsistency?



Strengthening Quine's New Foundations with a more flexible stratification criterion?
Flexibly Stratified Comprehension is inconsistent. By Flexibly Stratified Comprehension, there is an s such that


$$\forall x\:\bigl(x\in s\leftrightarrow\exists y\:\bigl(\forall t\:(t\in y\leftrightarrow t\in x)\land y\notin x\bigr)\bigr).$$
If $s\in s$, then there is an $S$ with the same members as $s$ such that $S\notin s$.


But if $S\notin s$, then for all $T$ with the same members as $S$, $T\in S$. In particular $S\in S$ and thus $S\in s$.


If $s\notin s$, then for all $S$ with the same members as $s$, $S\in s$. In particular $s\in s$.


