# What is the limit to iterating class comprehension, reflection and limitation of size?

In posting about a reflection principle coupled with a limitation of size axiom over Ackermann set theory, the answer is that the theory is blown up to a Mahlo cardinal.

I'm here just wondering if this method can be iterated, and what is the maximal that it can reach to via this iteration process.

For example lets define a theory $$\mathsf{K}^{+}(V_{\lambda})$$ in the language of $$FOL(=,\in, V_1, V_2,..,V_{\lambda})$$ as long as $$\lambda$$ is some specific recursive ordinal having some specific ordinal notation, i.e. as long as $$\lambda < \omega_1^{CK}$$

Now the idea is that each theory $$\mathsf{K}^{+}(V_{\lambda})$$ has axioms of Extensionality, Class comprehension axiom schema for $$V_{\alpha}$$, a reflection axiom scheme for $$V_{\alpha}$$, and limitation of size axiom for $$V_{\alpha}$$, for each $$\alpha < \lambda$$, also we have the axiom schema:

if $$\alpha < \beta$$, then: $$\forall x (x \subset V_{\alpha} \to x \in V_{\beta})"$$ is an axiom.

More specifically the formula of class comprehension for $$V_{\alpha}$$ is:

$$\forall x_1,..,x_n \subseteq V_{\alpha} \exists x \forall y (y \in x \leftrightarrow y \in V_{\alpha} \wedge \varphi(y,x_1,..,x_n))$$, where $$\varphi(y,x_1,..,x_n)$$ is a formula that do not use primitives $$V_{\beta}$$ when $$\beta>\alpha$$.

While the formula of reflection schema for $$V_{\alpha}$$ would be written as:

$$\forall x_1,..,x_n \in V_{\alpha} \\ [\exists y (\varphi(y,x_1,..,x_n)) \to \exists y \in V_{\alpha}(\varphi(y,x_1,..,x_n)) ]$$ where $$\varphi(y,x_1,..,x_n)$$ doesn't use any primitive symbol $$V_{\beta}$$ as long as $$\beta \geq \alpha$$.

Now what is the limit to the consistency strength of the $$\mathsf{K}^{+}(V_{\lambda})$$ theories?

I claim $$K^+(V^\lambda)$$, for any $$\lambda$$ (I switched the notation so not to be confused with the von Neumann universe) is equiconsistent with the schema "$$ORD$$ is Mahlo" (Not to be confused with full stationarity, which could be called "$$Ord$$ is Mahlo" if you really wanted to distinguish them). First off each $$V^\lambda$$ is a Grothendieck universe, and so of the form $$V_\kappa$$ for inaccessible $$\kappa$$. Second of all, $$V_\kappa\prec W$$, where $$W=\{x|x=x\}$$. To see this, suppose $$\exists x(\phi(x,x_0...x_n))$$ where $$\phi(x,x_0...x_n)$$ is absolute.
Then $$\exists x(\phi(x,x_0...x_n))$$ if and only if $$\exists x\in V^\lambda(\phi(x,x_0...x_n))$$ if and only if $$\exists x\in V^\lambda(\phi^{V^\lambda}(x,x_0...x_n))$$ if and only if $$V^\lambda\vDash\exists x(\phi(x,x_0...x_n))$$. Note that at no point do we use reflection for a formula that uses $$V^\lambda$$. Therefore $$K^+(V^\lambda)$$ proves that there exists a reflection cardinal, and so the consistency strength of $$K^+(V^\lambda)\ge$$ the consistency strength of "$$ORD$$ is Mahlo."
Then, suppose $$ORD$$ is Mahlo. Then there exists a proper class of reflecting cardinals, and if we take $$V^\lambda=V_\kappa$$ for reflecting $$\kappa$$, we get $$K^+(V^\lambda)$$. Therefore the consistency strength of "$$ORD$$ is Mahlo"$$\ge$$ the consistency strength of $$K^+(V^\lambda)$$, and so the consistency of "$$ORD$$ is Mahlo"$$=$$ the consistency strength of $$K^+(V^\lambda)$$.
• I just was under the impression that $K^+(V^2)$ already interprets "$Ord$ is Mahlo", I mean the second tier of this theory already proves the consistency of $Ord$ is Mahlo. The reason is because the proof in the linked answer already establishes $V^1$ as a a model of $\sf ZF + M$. Where $\sf M$ is the schema presented in the linked answer. So there exists $\kappa$ where $V^1=V_\kappa$ where $\kappa$ is a Mahlo. So I thought that we are already beyond Mahlo cardinals. – Zuhair Al-Johar Jul 5 at 9:30
• I believe the reason is that, while $V_\kappa$ might satisfy any individual axiom of "$ORD$ is Mahlo," you can't prove it satisfies all of them. – Master Jul 5 at 15:49
• Can you explicitly write $ORD$ is mahlo formally – Zuhair Al-Johar Jul 5 at 16:18
• It is a schema. If $C=\{\alpha|\phi(\alpha,p)\}$ is club, there there is some regular $\kappa\in C$. It is not a single assertion. Here is a link to Cantors attic: cantorsattic.info/ORD_is_Mahlo – Master Jul 5 at 16:20
• I see what you mean. I myself made a mistake in my comment, I meant $V^2$ when I said $V^1$. So again I thought that Packomov's answer established that $V^2$ is a model of $ZFC+M$, this means that $V^2=V_\kappa$ where $\kappa$ is a Mahlo, so the set of all ordinals in $V^2$ is a Mahlo cardinal, and so it proves the consistency of $ORD$ is a Mahlo, so it is already stronger than $ORD$ is a Mahlo, so as I said we are already way beyond that. So there must be something wrong with your argument? – Zuhair Al-Johar Jul 5 at 18:15