Unrestricted comprehension with non-circular formulas? Define a non-circular formula to be a formula $\varphi$ in which we can assign a level $|x| \in \mathbb{N}$ to each variable $x$ in $\varphi$ (whether it's free or bound) such that whenever $\varphi$ contains $x \in y$, $|x| < |y|$.
Then, for any non-circular formula $\varphi(x)$ with one free variable $x$, we add a non-circular comprehension axiom saying that $\{x : \varphi(x)\}$ exists.
Is ZFC - comprehension + non-circular unrestricted comprehension consistent?
 A: This is rather close to Quine's theory New Foundations (NF) and its relatives.
NF consists of stratified comprehension (the standard term for what you call "non-circular unrestricted comprehension") and extensionality. Many basic principles (e.g. Union and Pairing) are derivable from these. In particuar, if we drop Choice, Separation, Replacement, and Foundation from ZF and add stratified comprehension we get NF.
There is however a surprisingly deep issue here: NF disproves the axiom of choice. So it's crucial that we get rid of choice as well.
To make matters worse, it's currently unknown whether NF is consistent (even under rather strong set-theoretic assumptions); a claimed consistency proof has been given by Randall Holmes and several variations (at least in presentation), but as far as I know it hasn't been universally accepted yet. Meanwhile NFU, gotten by dropping Extensionality from NF, is known to be consistent, as are many of its variants (including NFU+AC). 
A: From the Pairing axiom of ZF, we know that for every set $x$ there exists a set $y:=\{x\}(:=\{x,x\})$ such that $x\in y$. Therefore, 
$$ a:=\{\,x_{\color{red}1}\mid \exists y_{\color{red}2}\colon x_{\color{red}1}\in y_{\color{red}2}\,\}$$
is a set in your theory and is the set of all sets. Consequently $a\in a$, whch contradicts the Axiom of Foundation in ZF. More precisely, Foundation states that $b:=\{a\}$ contains an element disjoint from $b$ - but the only possible such element is $a$ and that has $a$ as common element with $b$. 
So your theory is inconsistent.
