Define: $n$-skew pair of $x,y$, denoted by $\langle x,y \rangle^n$, as: $(singleton^n(x), y)$

Define: $(-n)$-skew pair of $x,y$, denoted by $\langle x,y \rangle^{-n}$, as: $(x, singleton^n(y))$

Where $(-,-)$ is the Kuratwoski ordered pair implementation, and $n$ is a natural

where: $singleton^0(x) = x$

$singleton^{i+1}(x) = \{singleton^i(x)\}$

so $\langle x,y \rangle ^0$ and $\langle x,y\rangle^{-0}$ are both level pairs.

Define: $f \text { is }n \text{-skew injection } \equiv_{df} f \text { is injection} \land \forall p \in f (p \text { is n-skew pair})$

Define: $x \leq^* y \equiv_{df} \exists n \in \mathbb Z, \exists f (f:x \to y, f \text{ is n-skew injection)}$

Define: $x =^* y \equiv_{df} \exists n \in \mathbb Z, \exists f (f:x \to y, f \text{ is n-skew bijection)}$

Where $\mathbb Z$ is the set of Integers.

Write Cantor-Bernstein-Schroeder theorem in terms of $\leq^*,=^*$ , denoted as "skew-CBS", as:

$\forall x,y [(x \leq^* y \land y \leq^* x) \to x=^* y]$

Question: Is skew-CBS consistent with NFU?

Of course both $\leq^*$ and $=^*$ are non-stratified relations, so they work externally.