# Is Bounding Reflection consistent?

Working in the first order language of set theory.

Let $$\varphi^{*B}$$ be the formula obtained from $$\varphi$$ by merely bounding all open quantifiers in $$\varphi$$ by the symbol "$$B$$".

Here a bounded quantifier in $$\varphi$$ can only be via $$\in$$ or $$\subseteq$$ relations, otherwise it's considered open. But, the bounding by $$B$$ to form $$\varphi^{*B}$$ is restricted via $$\in$$ relation.

Bounding Reflection: if $$\varphi$$ is a formula that doesn't use the symbol "$$B$$", then: $$\forall \vec{v} \, \exists B \, (\varphi \to \varphi^{*B})$$

This can prove: pairing, union, power, infinity, and some instances of Replacement. I'm not sure if it can prove full replacement, but I don't think so.

Is this consistent with the axiom schema of Separation?

This question is related to this one , which virtually differs only in requiring bounding of open quantifiers to be closed anteriorly; that is, if an open quantifier is bounded by $$B$$ then all prior open quantifiers must be bounded by $$B$$, thereby only ensuring an initial segment of open quantifiers in $$\varphi$$ to be bounded by $$B$$, thus may be named as anterior bounding reflection. However, here a complete bounding of all open quantifiers by $$B$$ is required.

The rationale is that if this is solved to the positive, then an anterior bounded pathology to reflection would be excluded, and since this answer suggests no posterior pathology to reflection, therefore anterior bounding reflection may stand a chance of being consistent.

• Isn't your version of reflection an immediate consequence of the usual reflection theorem? So yes, it is consistent with separation, since it is a theorem of ZF. After all, if $\varphi$ is absolute to $V_\kappa$, then take $B=V_\kappa$. (Let me say also that in set theory, the usual meaning of bounded quantifier is $\exists x\in y$, not $\exists x\subseteq y$.) Feb 25 at 14:33
• To say that $\varphi$ is absolute between the set-theoretic universe $V$ and a transitive set $B$ is exactly to say $\varphi\iff\varphi^B$, where one bounds all the quantifiers by $B$. So your property, stated with a biconditional, is nearly identical to the reflection theorem. There are variations of the theorem, to be sure, depending on whether one takes $B$ merely as a transitive set or as $V_\alpha$ for some $\alpha$, but these are easy to see equivalent. Stronger forms of reflection (but still equivalent) find a class club of $\alpha$ for which $V_\alpha$ is $\Sigma_n$ correct. Feb 25 at 16:39
• But we can take the previously bounded ones to be part of B, so this is just immediate from replacement. Feb 25 at 18:41
• All the other bounds come ultimately from the variables in $\vec v$ and their elements, or from variables already chosen from B. So all the bounds are B or below. Feb 25 at 18:51
• A must be one of the v's, otherwise your axiom has parameter A, which makes no sense. Or, just take B above all the parameters appearing, including A. The ordinary reflection theorem shows this is possible. Feb 25 at 19:25

Bounding reflection is a theorem scheme of $$\sf ZFC$$.
For any formula $$\varphi$$ having all its free variables among $$v_1, \dots, v_n$$, we can get an equivalent formula "$$(\exists x_1: x_1=v_1 \land \dots \land\exists x_n: x_n=v_n) \land \varphi$$", call this formula $$\varphi^+$$. Now, working in $$\sf ZFC$$, by reflection we have some $$V_\alpha$$ such that $$\varphi^+ \iff \varphi^{+ ^{V_\alpha}}$$. According to that we'll have all parameters $$v_i \in V_\alpha$$, so is the case for all of their elements and all of their subsets, since $$V_\alpha$$ is super-transitive. Now any $$\in$$ or $$\subseteq$$ bounded quantifier in $$\varphi^{+^{V_\alpha}}$$ would have its bound either being $$V_\alpha$$ or $$v_i \cap V_\alpha$$ for a parameter $$v_i$$ or $$x \cap V_\alpha$$ for some bound variable $$x$$ in $$\varphi$$ and this $$x$$ ultimately goes back via successive membership\subset-hood relations up to either $$V_\alpha$$ or a parameter $$v_i$$ which also would go back to $$V_\alpha$$, this means that all bounds of quantifiers that are originally present in $$\varphi$$ would be elements of $$V_\alpha$$, and so their intersections with $$V_\alpha$$ would only yield them themselves. That is, we have $$\mathcal Q x \in V_\alpha: x \in v_i :\cdots$$ which would be $$\mathcal Q x \in V_\alpha \cap v_i: \cdots$$ and this would be redundant since it is equivalent to $$\mathcal Q x \in v_i: \cdots$$ Similarly, $$\mathcal Qx \in V_\alpha: x \subseteq v_i: \cdots$$ which would be $$\mathcal Qx \in V_\alpha \cap \mathcal P(v_i): \cdots$$ and this would also be redundant since it is equivalent to $$\mathcal Q x \in \mathcal P (v_i) :\cdots$$ which is $$\mathcal Q x \subseteq v_i:\cdots$$. So skipping bounding already bounded quantifiers in $$\varphi^+$$ won't affect truth values, and so we have $$\varphi^{+*^{V_\alpha}} \iff \varphi^{+^{V_\alpha}}$$. Now, from that and reflection we have $$\exists V_\alpha: \varphi^+ \to \varphi^{+*^{V_\alpha}}$$, and so $$\exists V_\alpha: \varphi \to \varphi^{+*^{V_\alpha}}$$. Since $$\varphi$$ doesn't use $$V_\alpha$$, then we have $$\varphi \to \exists V_\alpha: \varphi^{+*^{V_\alpha}}$$. Since $$\varphi^{+*^{V_\alpha}}$$ is conjunctive, we can remove the $$\overset n \bigwedge \exists x_i:x_i=v_i$$ string, so we are left with: $$\varphi \to \exists V_\alpha: \varphi^{*^{V_\alpha}}$$, pull back $$\exists V_\alpha$$ to the front, and we get: $$\exists V_\alpha: \varphi \to \varphi^{*^{V_\alpha}}$$. $$\square$$
As regards anterior bounding reflection, we use the same argument but begin with Hamkins argument by taking $$B$$ to be simply the next $$\Sigma_n$$ correct $$V_\kappa$$ above the ranks of $$\vec v$$. $$\square$$