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I've noticed that most of the work in Ackermann set theory is primarily concerned with constructing sets in $V$, the rest of the classes are just excess material, carrying no comprehension over them. There is a try of Muller in which he strengthen the class existence principle of Ackermann into Separation over classes, the resultant theory is $A$, and adding Regularity $R$, and Choice $C$, he gets into $ARC$, a theory claimed [see here] to serve as a foundation of both category and set theory, and thus for most of mathematics.

This gave me the idea of reflecting-out of $V$ principle, since Ackermann's set theory can be interpeted in systems using reflection [see here] , so if to any of the two systems appearing in that posting (with reflection in them re-named as reflection in $V$), we add the following principle:

Reflection out of $V$ schema: if $\varphi$ is a sentence in $FOL(=,\in)$, i.e. doesn't use the symbol $V$, and $\varphi^V$ is the bounded by $V$ sentence of $\varphi$, i.e. the sentence obtained by merely bounding every quantifier in $\varphi$ by $V$, then: $ \varphi^V \to \varphi $, is an axiom.

In other words we are reversing the reflection process, so we are concluding things about classes in general by reflecting from the inside of $V$ to outside it. By that, all set axioms (i.e. sentences in the language of set theory that are satisfied in $V$), would generalize over all classes. This way we easily get to interpret Muller's theory.

Question: is there an obvious inconsistency with a theory that both uses reflection in $V$ and reflection out of $V$ principles?

Zuhair Al-Johar
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  • What about a schema with a constant symbol $c$ asserting that $c$ is some $V_\alpha$, and $V_\alpha$ is elementary in $V$? This is a mild theory. – Monroe Eskew Dec 18 '18 at 11:57
  • @MonroeEskew your point is not clear? what do you want to say? – Zuhair Al-Johar Dec 18 '18 at 14:06
  • Fix some $V_\alpha$ satisfying ZFC and pretend it is $V$ and the higher-rank sets are hyperclasses. Possibly $V_\alpha \prec V$. In such a situation, we have reflection out of $V$. Also everything true in $V$ reflects to $V_\alpha$. We can also have this on a club of $\beta$, so that also things can reflect below $\alpha$ simultaenously. – Monroe Eskew Dec 18 '18 at 14:47
  • See https://mathoverflow.net/a/103779/1946 for more discussion. Also, this theory arises in many other questions on MO. See https://mathoverflow.net/search?q=user%3A1946+Feferman+theory. – Joel David Hamkins Dec 18 '18 at 14:48
  • @MonroeEskew do your comments apply to adding the reflection out schema on top of the referred theory in the post that has the limitation of size axiom? this has a mahlo cardinal as its model, (see link in the head post), so is this also mild in your opinion? – Zuhair Al-Johar Dec 18 '18 at 18:20
  • Yep! Mahlos are everywhere. – Monroe Eskew Dec 18 '18 at 18:26
  • @MonroeEskew so from these comments of yours and Hamkins referred posts I understand that the theories I'm referring to here (containing reflection in and reflection out schemes on top of the other axioms) are all consistent! but anyhow the axiomatization that I've presented here is by far much less complex than the ones you (and Hamkins) are expressing. This is in itself interesting, at least to me. – Zuhair Al-Johar Dec 18 '18 at 18:35
  • I am confused. The reflection schema states roughly $V\prec W$, where $W={x|x=x}$. As a consequence, $\phi^V\rightarrow \phi$, unless I have mistaken something. – Master Jun 07 '19 at 21:31
  • @Master, there is a restriction on $V$ as mentioned in the axiom, so its not $\phi^V$ for whatever $\phi$. – Zuhair Al-Johar Jun 08 '19 at 03:41
  • You mean the part it says $\forall x_0...x_n\in V$? Doesn't all that mean is parameters are restricted to $V$? – Master Jun 08 '19 at 16:37
  • @Master, I meant the part that $\varphi$ doesn't use the symbol $V$. – Zuhair Al-Johar Jun 08 '19 at 16:48
  • Both reflection and counter-reflection are restricted to $L(=,\in)$. "if $\varphi(y, x_1,..,x_n)$ is a formula in $FOL(=,\in)$","if $\varphi$ is a sentence in $FOL(=,\in)$, i.e. doesn't use the symbol $V$." – Master Jun 08 '19 at 16:50
  • @yes that's what I mean. as far as counter-reflection is concerned. – Zuhair Al-Johar Jun 08 '19 at 16:55
  • Wait, but if every $FOL(=,\in)$ formulas is reflected by refection, and the same is true by counter reflection, then what is the difference? – Master Jun 08 '19 at 17:41
  • are you saying that counter-reflection is redundant? – Zuhair Al-Johar Jun 08 '19 at 17:47
  • Yes, it seems counter reflection is just reflection from a different perspective. Kind of like the schema $\forall x\in V(\phi(x,x_0...x_n))\rightarrow \forall x(\phi(x,x_0...x_n))$ is the same as $\exists x(\phi(x,x_0...x_n))\rightarrow \exists x\in V(\phi(x,x_0...x_n))$. There both absoluteness from a different direction. – Master Jun 08 '19 at 18:04
  • @Master, I don't think so, anyhow I'm not sure. – Zuhair Al-Johar Jun 08 '19 at 19:36
  • Well I still don't see what you could derive from counter-reflection that you can't from reflection, so I don't see the point of this question, unless I am missing something really obvious. – Master Jun 08 '19 at 19:38
  • @Miller's separation over classes. That was the main point behind counter-reflection. – Zuhair Al-Johar Jun 08 '19 at 20:00
  • Doesn't class comprehension proof class separation? I.e. $\phi(x)\leftrightarrow x\in Z\land \psi(x)$? – Master Jun 08 '19 at 20:20
  • @Master, No! class comprehension only proves separation over the class $V$, but it doesn't prove separation over higher classes, for example separation over the class $P(V)$ or its power, etc... – Zuhair Al-Johar Jun 08 '19 at 20:32
  • Given any predicate $\phi(x)$, $V\vDash (\lnot\exists X\lnot(\exists Y(Y={x\in X|\phi(x)})))$. Now, by induction we can see that then the universe $W={x|x=x}$ satisfies this, and we do not even have to use reflection out of $V$ to see this, and so separation is satisfied. – Master Jun 08 '19 at 20:53
  • @Master, I don't know what do you mean when you say "by induction"? how do you use "induction" to generalize what is inside $V$ to go beyond $V$? I used counter-reflection to do that, but apparently you have something else in mind. – Zuhair Al-Johar Jun 08 '19 at 21:07
  • We can prove by induction $\phi^V\leftrightarrow \phi$ for every formula $\phi$. It is clear that $(x\in y)^V\leftrightarrow (x\in y)$, and if $\phi$,$\psi$ are absolute so is $\lnot\phi$ and $\phi\land \psi$. Then if $\phi$ is absolute, $\exists x\in V(\phi)\rightarrow \exists x(\phi)$ and by reflection $\exists x(\phi)\rightarrow \exists x\in V(\phi)$, and so $\exists x(\phi)\leftrightarrow \exists x(\phi)^V$. – Master Jun 08 '19 at 21:12
  • @Master, what is $(x \in y)^V$, the way how I used $\phi^V$ is to mean $\phi$ is a "sentence" where its quantifiers are bound by $V$, the way how I see it is that $x \in y$ is not a sentence? it contains two free variables. – Zuhair Al-Johar Jun 08 '19 at 21:26
  • $(x\in y^V)$ is the sentence $(x\in y)$ with all its quantifiers bounded; as it has no quantifiers, it is identical to $(x\in y)$ . What I am trying to show is that for every formula $\phi$, $\phi^V\leftrightarrow \phi$. I am trying to do this by induction: First for atoms, and then the connectives and negation, and then the existential quantifier. – Master Jun 08 '19 at 21:31
  • @Master, Ah! I see what you want to do. I need to think about it. Its too late at night now. I'll continue this tomorrow! – Zuhair Al-Johar Jun 08 '19 at 21:38
  • @Master, $\exists x\in V(\phi)\leftrightarrow \exists x(\phi)$ only when $\phi$ has the needed qualifications, this will cut the induction flow when $V$ is included in $\phi$. So this won't work. – Zuhair Al-Johar Jun 09 '19 at 13:25
  • I see. So for example, $\forall x(\forall y(\phi))^V\rightarrow \forall x(\forall y(\phi))$, which is one of the advantages of counter reflection? – Master Jun 09 '19 at 16:02
  • @Master, EXACTLY! – Zuhair Al-Johar Jun 09 '19 at 18:24
  • However, isn't this theory not very strong. Consider the mode $V_\kappa$, where $\kappa$ is inaccessible. Any $V_\alpha\prec V_\kappa$ would satisfy counter reflection and reflection, and so this is theory is consistent relative to large cardinals? – Master Jun 09 '19 at 18:33
  • YES! correct. I think this is right. It is not very strong. The strong theory is another one, not this. – Zuhair Al-Johar Jun 09 '19 at 18:36
  • this is the stronger one: https://mathoverflow.net/questions/319319/can-ackermann-set-theory-find-a-natural-interpretation-in-light-heavy-class-dich – Zuhair Al-Johar Jun 09 '19 at 19:02

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