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Let stratified $\sf Z$ have all axioms of $\sf Z$-$\sf Reg.$ with Infinity stipulated by existence of a Dedekind infinite set, and with Separation restricted to use only stratified formulas, where those are defined after Quine's New Foundations. AC is not assumed.

We add a new primitive unary function symbol $U$ to the language of set theory.

The idea is to have a hierarchy $$\cdots \subset U_{-2} \subset U_{-1} \subset U_0 \subset U_1 \subset U_2 \subset \cdots$$, where each stage (rank) $U_{i+1}$ is the powerset of its predecessor $U_{i}$

I call this "inverted hierarchy" because ranks can have an infinite descent.

Axiom I: $\forall i \in \mathbb Z: U_i \subset U_{i+1} \land U_{i+1} = \mathcal P (U_i) $

We call each $U_i$ an ultra-rank, denoted "$u$.rank"

Axiom II: $\forall x \exists i \in \mathbb Z: x \in U_i$

Axiom III: $\forall i \in \mathbb Z \exists x: x=U_i$

Axiom VI: $ \forall i \in\mathbb Z: \forall \alpha \, ( V_\alpha \in U_i)$

Where $V_\alpha$ is defined in the usual manner, i.e. as the $\alpha^{th}$ iterative power of $\varnothing$.

If the above is consistent can we have an external automorphism that shifts $u$.ranks one-step downwardly? That is, there is an external automorphism $j$ such that for some $i \in \mathbb Z$, we have: $U_{j(\alpha) + 1} = U_\alpha$, for all $\alpha \in \mathbb Z, \alpha < i $.

Let stratified $\sf Z$ have all axioms of $\sf Z$-$\sf Reg.$ with Infinity stipulated by existence of a Dedekind infinite set, and with Separation restricted to use only stratified formulas, where those are defined after Quine's New Foundations. AC is not assumed.

We add a new primitive unary function symbol $U$ to the language of set theory.

The idea is to have a hierarchy $$\cdots \subset U_{-2} \subset U_{-1} \subset U_0 \subset U_1 \subset U_2 \subset \cdots$$, where each stage (rank) $U_{i+1}$ is the powerset of its predecessor $U_{i}$

I call this "inverted hierarchy" because ranks can have an infinite descent.

Axiom I: $\forall i \in \mathbb Z: U_i \subset U_{i+1} \land U_{i+1} = \mathcal P (U_i) $

We call each $U_i$ an ultra-rank, denoted "$u$.rank"

Axiom II: $\forall x \exists i \in \mathbb Z: x \in U_i$

Axiom III: $\forall i \in \mathbb Z \exists x: x=U_i$

Axiom VI: $ \forall i \in\mathbb Z: \forall \alpha \, ( V_\alpha \in U_i)$

Where $V_\alpha$ is defined in the usual manner, i.e. as the $\alpha^{th}$ iterative power of $\varnothing$.

[NOTE]: $\mathbb Z$ is internal. The $+$ operator on $\mathbb Z$ is defined so that $i+1$ is one type higher than $i$ in stratification. For example we take $\mathbb Z$ to be the set of all acyclic infinitely iterative singletons $\{\{\{\cdots \space \cdots\}\}\}$, fix one of its elements to be $0$ and define $i+1 = \{i\}$. Needed axioms to support such internal grasp of $\mathbb Z$ to be added. $U$ can be used in Separation. The arguments of $U$ receive equal type indices during stratification, so $U_i$ and $i$ receive the same type.

If the above is consistent can we have an external automorphism that shifts $u$.ranks one-step downwardly? That is, there is an external automorphism $j$ such that for some $i \in \mathbb Z$, we have: $U_{j(\alpha) + 1} = U_\alpha$, for all $\alpha \in \mathbb Z, \alpha < i $.

Let stratified $\sf Z$ have all axioms of $\sf Z$-$\sf Reg.$ with Infinity stipulated by existence of a Dedekind infinite set, and with Separation restricted to use only stratified formulas, where those are defined after Quine's New Foundations. AC is not assumed.

We add a new primitive unary function symbol $U$ to the language of set theory.

The idea is to have a hierarchy $$\cdots \subset U_{-2} \subset U_{-1} \subset U_0 \subset U_1 \subset U_2 \subset \cdots$$, where each stage (rank) $U_{i+1}$ is the powerset of its predecessor $U_{i}$

I call this "inverted hierarchy" because ranks can have an infinite descent.

Axiom I: $\bigl<U_i \subset U_{i+1} \ ; \ U_{i+1} = \mathcal P (U_i)\bigr>_{i \in \mathbb Z}$

We call each $U_i$ an ultra-rank, denoted "$u$.rank"

Axiom II: $\forall x \exists i \in \mathbb Z: x \in U_i$

Axiom III: $\forall i \in \mathbb Z \exists x: x=U_i$

Axiom VI: $ \forall i \in\mathbb Z: \forall \alpha \, ( V_\alpha \in U_i)$

Where $V_\alpha$ is defined in the usual manner, i.e. as the $\alpha^{th}$ iterative power of $\varnothing$.

If the above is consistent can we have an external automorphism that shifts $u$.ranks one-step downwardly? That is, there is an external automorphism $j$ such that for some $i \in \mathbb Z$, we have: $U_{j(\alpha) + 1} = U_\alpha$, for all $\alpha \in \mathbb Z, \alpha < i $.

Let stratified $\sf Z$ have all axioms of $\sf Z$-$\sf Reg.$ with Infinity stipulated by existence of a Dedekind infinite set, and with Separation restricted to use only stratified formulas, where those are defined after Quine's New Foundations. AC is not assumed.

We add a new primitive unary function symbol $U$ to the language of set theory.

The idea is to have a hierarchy $$\cdots \subset U_{-2} \subset U_{-1} \subset U_0 \subset U_1 \subset U_2 \subset \cdots$$, where each stage (rank) $U_{i+1}$ is the powerset of its predecessor $U_{i}$

I call this "inverted hierarchy" because ranks can have an infinite descent.

Axiom I: $\forall i \in \mathbb Z: U_i \subset U_{i+1} \land U_{i+1} = \mathcal P (U_i) $

We call each $U_i$ an ultra-rank, denoted "$u$.rank"

Axiom II: $\forall x \exists i \in \mathbb Z: x \in U_i$

Axiom III: $\forall i \in \mathbb Z \exists x: x=U_i$

Axiom VI: $ \forall i \in\mathbb Z: \forall \alpha \, ( V_\alpha \in U_i)$

Where $V_\alpha$ is defined in the usual manner, i.e. as the $\alpha^{th}$ iterative power of $\varnothing$.

If the above is consistent can we have an external automorphism that shifts $u$.ranks one-step downwardly? That is, there is an external automorphism $j$ such that for some $i \in \mathbb Z$, we have: $U_{j(\alpha) + 1} = U_\alpha$, for all $\alpha \in \mathbb Z, \alpha < i $.

Let stratified $\sf Z$ have all axioms of $\sf Z$-$\sf Reg.$ with Infinity stipulated by existence of $\omega$, and with Separation restricted to use only stratified formulas, where those are defined after Quine's New Foundations. AC is not assumed.

Define: $y = \mathcal P^{-1}(x) \iff x= \mathcal P(y)$

Define: $x= \bigcap_{s \in S} s \iff x=\{y \mid S \neq \varnothing \land \forall s \in S (y \in s)\}$

Now, lets try coin what we mean by rank or stage of the inverted iterative hierarchy, which we'd label as inverted rank.

$X$ is an inverted rank, if and only if, there is an ordinal $\alpha$ and there is a bijective function $f$ whose domain is $\alpha$ that sends $0$ to $X$ and that sends each $\beta+1 \in \alpha$ to $ \mathcal P^{-1}(f(\beta))$, and sends each limit $\lambda \in \alpha$ to $ \bigcap_{\kappa < \lambda} f(\kappa)$.

We phrase that as: $X$ is an inverted rank of degree $\alpha$.

For brevity "inverted rank" shall be denoted "i.rank".

Axiom I: $X \text { is an i.rank } \land Y \text{ is an i.rank} \to X \subseteq Y \lor Y \subseteq X$

Axiom II: $\forall x \exists Y: Y \text{ is an i.rank } \land x \in Y$

Axiom III: $\alpha \text { is an ordinal } \to \exists X: X \text{ is an i.rank of degree } \alpha$

Axiom VI: $\forall A \exists x \forall y \, (y \in x \leftrightarrow y \in A \land y \text { is an i.rank } \}$

Now, can this theory be consistent? I suspect the last axiom might cause problems, but even if so, can the theory without it be consistent? I mean this theory goes against the norms of the known cumulative hierarchy, where no descending power sequence like that is possible. So, does it stand a chance of being consistent whether with or without the last axiom.

If the above is consistent can we have an external automorphism from an inverted rank to its predecessor inverted rank?

Let stratified $\sf Z$ have all axioms of $\sf Z$-$\sf Reg.$ with Infinity stipulated by existence of a Dedekind infinite set, and with Separation restricted to use only stratified formulas, where those are defined after Quine's New Foundations. AC is not assumed.

We add a new primitive unary function symbol $U$ to the language of set theory.

The idea is to have a hierarchy $$\cdots \subset U_{-2} \subset U_{-1} \subset U_0 \subset U_1 \subset U_2 \subset \cdots$$, where each stage (rank) $U_{i+1}$ is the powerset of its predecessor $U_{i}$

I call this "inverted hierarchy" because ranks can have an infinite descent.

Axiom I: $\bigl<U_i \subset U_{i+1} \ ; \ U_{i+1} = \mathcal P (U_i)\bigr>_{i \in \mathbb Z}$

We call each $U_i$ an ultra-rank, denoted "$u$.rank"

Axiom II: $\forall x \exists i \in \mathbb Z: x \in U_i$

Axiom III: $\forall i \in \mathbb Z \exists x: x=U_i$

Axiom VI: $ \forall i \in\mathbb Z: \forall \alpha \, ( V_\alpha \in U_i)$

Where $V_\alpha$ is defined in the usual manner, i.e. as the $\alpha^{th}$ iterative power of $\varnothing$.

If the above is consistent can we have an external automorphism that shifts $u$.ranks one-step downwardly? That is, there is an external automorphism $j$ such that for some $i \in \mathbb Z$, we have: $U_{j(\alpha) + 1} = U_\alpha$, for all $\alpha \in \mathbb Z, \alpha < i $.