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Working in bi-sorted FOL, where lower cases stand for sets and upper cases for classes; add all axioms of ZF written completely in lower case, and add an axiom of Extensionality over all classes, and the following axioms:

$\forall x \ \exists Y: Y=x$

$\forall X \ \forall Y: X \in Y \to \exists z: z=X$

$\forall \vec{Z} \ \exists X \ \forall y \ (y \in X \iff \phi(y, \vec{Z}))$; if $\phi(y, \vec{Z})$ is a formula not using $``X"$.

Define $V$ as the class of all sets.

Is there a clear inconsistency with the following principle?

Define recursively: $J_0``x = J(x) \\ J_{n+1} `` x= \{J_n`` y \mid y \in x \}$

Internalization schema:$n=1,2,... \\ \forall \text { infinite } x,y \ \exists J \ ( J: x \rightarrowtail y \land \forall s: J_n ``s \in V )$

Where: $\text {infinite}(x) \iff \neg \exists \alpha \in \mathbb \omega_0 \exists f (f: x \rightarrowtail \{\beta \mid \beta < \alpha\})$; $``↣"$ is short for "injection"

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  • $\begingroup$ Since you didn't include the axiom of choice, how are you defining "infinite"? Also, what is $\rightarrowtail$? $\endgroup$
    – Mark Saving
    Commented Oct 24, 2021 at 21:52
  • $\begingroup$ @MarkSaving, it's Tarski infinite, I will update the post. Thanks! $\endgroup$
    – Zuhair Al-Johar
    Commented Oct 25, 2021 at 12:05
  • $\begingroup$ Can't you take x,y to be aleph numbers, e.g. aleph1, aleph0, and then by induction on aleph1 show that $J$ is a set, which will lead to inconsistency? e.g. If $J:\omega_1↣\omega$, let $j_0=\emptyset$ and for $\alpha\in\omega_1$ let $j_{\alpha+1}=j_\alpha\cup\{(\alpha, J(\alpha))\}$, and at limit points take the union. Every stage is a set, as well as the union of all stages. That union is injective iff $J$ is injective, which is impossible $\endgroup$
    – Holo
    Commented Oct 25, 2021 at 22:20
  • $\begingroup$ @Holo, No! because set Replacement is fully written in set variables (lower case) , so you cannot use the upper case symbol $J$. $\endgroup$
    – Zuhair Al-Johar
    Commented Oct 26, 2021 at 4:34
  • $\begingroup$ @ZuhairAl-Johar Where did I use replacement on $J$? $j_0$ is a set because it is the empty set, if $α∈ω_1$ then $J(α)∈ω$ so it is also a set, hence $(α,J(α))$ (from normal ZF arguments) is a set, so $j_{α+1}$ is a set. And if $β∈ω_1$ is limit, we can define the functional relation $φ$ such that $φ(x,y,β)⇔(x∈β\text{ and }y=j_{x})\text{ or }(x∉β\text{ and }y=0)$, this is a formula only on sets, then there exists a set $b$ such that $a∈b$ if and only if there exists $c∈β$ such that $φ(a,c,β)$, that is, $a=j_{α}$ for some $α∈β$. $\endgroup$
    – Holo
    Commented Oct 26, 2021 at 17:43

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