# Is Cantor-Bernstein-Schroeder theorem for skew cardinality, consistent with NF?

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.