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Axiom scheme of Ordinal Reflection: if $\phi$ is a formula that doesn't use the symbol $V$, whose parameters are among $x_1,..,x_n$; then: $$\forall x_1 \in V,\dotsc,\forall x_n \in V: \phi(On) \to \\\forall c \subseteq On \\ (c=\{\alpha \in On| \phi(\alpha)\} \lor \bigcup( c) = On \to |c|=On);$$ is an axiom.

Where: $$On=\{\alpha \in V|\operatorname{ordinal}(\alpha)\};$$ and "$|c|$" stands for cardinality of class $c$, defined as the least von Neumann ordinal class bijective to $c$.

Add this axiom on top of Ackermann's set theory.

What would be the consistency strength of the resulting theory?

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The schema (if πœ™ is a formula that doesn't use the symbol 𝑉, whose variables are among π‘₯1,..,π‘₯𝑛; then: βˆ€π‘₯1βˆˆπ‘‰,…,βˆ€π‘₯π‘›βˆˆπ‘‰:πœ™(𝑂𝑛)β†’|{π›Όβˆˆπ‘‚π‘›|πœ™(𝛼)}|=On) is provable in Ackermann's set theory plus foundation. To see this suppose
πœ™ is a formula that doesn't use the symbol 𝑉, whose variables are among π‘₯1,..,π‘₯𝑛; and βˆ€π‘₯1βˆˆπ‘‰,…,βˆ€π‘₯π‘›βˆˆπ‘‰:πœ™(𝑂𝑛). Suppose that there was an ordinal d∈On such that π›Όβˆˆπ‘‚π‘› and πœ™(𝛼) implies π›Όβˆˆd. Then On would be definable(for suitable definition of ordinal) as the least ordinal x such that πœ™(x) and d∈x. Since this is impossible, ⋃{π›Όβˆˆπ‘‚π‘›|πœ™(𝛼)}=On. Let F(x,y) be a formula which "says" x and y are ordinals and πœ™(y) and the order type of {π›Όβˆˆy|πœ™(𝛼)}=x. Suppose F(b,On) holds for some b∈On. Then On is definable, which is impossible. Therefor the order type of {π›Όβˆˆπ‘‚π‘›|πœ™(𝛼)} is On.

Ackermann's set theory plus foundation plus βˆ€π‘βŠ†π‘‚π‘›(⋃(𝑐)=𝑂𝑛→|𝑐|=𝑂𝑛) proves Con(ZF). To see this note that this theory proves the formula "L(On) satisfies ZF".

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  • $\begingroup$ still that doesn't fully answer the question. It only says that its consistency strength is higher than ZF, but to what extent? is it equivalent to a Mahlo or just below it? $\endgroup$ Dec 23 '19 at 12:23
  • $\begingroup$ Why foundation? ordinals are well-founded built-in classes. $\endgroup$ Dec 23 '19 at 12:35

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