Can this form of reflection be consistent?

Question

Is this form of reflection consistent?

First I'll begin by clarifying the notation I'm using here:

By a quantifier being relativized or bounded it means that the first occurrence of the quantified variable must be followed by $\in $ or $\subseteq$ symbols, followed by a variable symbol, like: $\forall x \in v; \exists x \in k; \forall x \subseteq v; \exists y \subseteq k$.

For any formula $\varphi$ that doesn't use the symbol "$W$" let $\varphi^{``W}$ denote any formula that is obtained by merely relativizing some of the quantifiers in $\varphi$ by "$\in W$" in such a manner that any quantifier so relativized to $W$ must have all preceding quantifiers appear bounded. To be worned is that no quantifier in $\varphi^{``W}$ can be newely bounded by other than "$\in W$", and so all quantifiers in $\varphi^{``W}$ preceding a quantifier relativized to $W$ that are not bounded by $W$ must appear as bounded originally in $\varphi$ and so retain their original bounds in $\varphi$.

Examples: Let $\varphi$ be the formula $(\forall a \subseteq A \exists y: y=a)$ , and $\varphi^{``W}$ be the formula $(\forall a \subseteq A \exists y \in W: y=a)$; however we can also have $\varphi^{``W}$ be the formula $(\forall a \in W (a \subseteq A \to \exists y \in W: y =a ))$. Another example is let $\varphi$ be the formula $(\forall a \in A \forall y \in a \exists z: z=y) $, and $\varphi^{``W}$ being the formula $(\forall a \in A \forall y \in a \exists z \in W: z=y)$, but also we can have $\varphi^{``W}$ be the formula $(\forall a \in W (a \in A \to \forall y \in W (y \in a \to \exists z \in W: z=y)))$

Reflection$^*$: $\forall \vec{v} \exists W \, (\varphi \to \varphi^{``W})$

I highly suspect that this scheme is inconsistent. Hence the question.

The point is that if we allow quantifiers in $\varphi^{``W}$ prior to those bounded by $W$ to be unbounded, then this straightforwardly enacts a paradox, simply take $\varphi$ to be $\forall x \exists y : y =x$ and $\varphi^{``W}$ to be $\forall x \exists y \in W: y=x$, then $W$ would be the universal set.

Of note is that if it turns to be consistent, then it together with Extensionality and Separation would prove all axioms of $\sf ZF-Reg.$. Also, it would directly prove Collection without resorting to any of the axioms of $\sf ZFC$.

Answer 1

Edit, Mar 3: After some discussion with Zuhair Al-Johar, the full answer (that $\mathrm{Reflection}^*$ is a theorem schema of $\mathrm{ZFC}$) appears in the last line of this answer.

Here is a partial answer. If only one open (necessarily the leftmost that is open) quantifier bounded to $W$ is allowed in $\varphi^{``W}$, then this schema is consistent relative to an inaccessible cardinal.

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Let $\mathscr L_\in$ be the first-order language with relation symbol $\in$, and without quantifiers $\forall x \subseteq v$ or $\exists y \subseteq k$. Let $\mathscr L_{\subseteq}$ be the first-order language with relation symbol $\in$ and bounded quantifiers $\forall x \subseteq v$ and $\exists y \subseteq k$. Let $\mathsf{Reflection}^*_{\in,1}$ be the schema $\mathsf{Reflection}^*$, but only considering $\varphi$ and $\varphi^{``W}$ that are $\mathscr L_\in$ formulae, and only allowing one open (necessarily the outermost open) quantifier in $\varphi^{``W}$ to be bounded by $W$. Define $\mathsf{Reflection}^*_{\mathcal P,1}$ similarly except using the language $\mathcal L_{\mathcal P}$ that is yet to be defined.

It follows from lemma 6.1 on p.57 of Kanamori's The Higher Infinite that if $\kappa$ is an inaccessible cardinal, then there is an unbounded set of $\alpha<\kappa$ such that $V_\alpha\prec V_\kappa$. For any finite subset $\vec v$ of $V_\kappa$, as the ranks of sets in $\vec v$ must be bounded below $\kappa$, there is a $V_\alpha\prec V_\kappa$ such that $\vec v\subset V_\alpha$. (Call this property the "substructure property of $\kappa$".)

Theorem 1: Assume $\kappa$ is inaccessible. Then $V_\kappa$ satisfies the $\mathsf{Reflection}^*_{\in,1}$.

Proof: Let $Q_0(y_0\in b_0)\ldots Q_k(y_k\in b_k)\psi(x,\vec y,\vec v)$ be an arbitrary $\mathscr L_\in$ formula in which $W$ does not appear, where each $Q_i(y_i\in b_i)$ is a bounded quantifier and $\vec y$ is $y_0,\ldots,y_k$, and choose a sequence of parameters $\vec v\in V_\kappa$. By the substructure property of $\kappa$, choose an $\alpha<\kappa$ such that $V_\alpha\prec V_\kappa$, $b_0,\ldots,b_k\in V_\alpha$, and $\vec v\in V_\alpha$, this $V_\alpha$ will serve as $W$ from here. It is sufficient to show that if $V_\kappa\vDash Q_0(y_0\in b_0)\ldots Q_k(y_k\in b_k)\forall x\psi(x,\vec y,\vec v)$ then $V_\kappa\vDash Q_0(y_0\in b_0)\ldots Q_k(y_k\in b_k)\forall(x\in W)\psi(x,\vec y,\vec v)$, and if $V_\kappa\vDash Q_0(y_0\in b_0)\ldots Q_k(y_k\in b_k)\exists x\psi(x,\vec v)$ then $V_\kappa\vDash Q_0(y_0\in b_0)\ldots Q_k(y_k\in b_k)\exists(x\in W)\psi(x,\vec y,\vec v)$. As any given $\varphi^{``W}$ must be in either of these latter forms, then $V_\kappa\vDash\varphi$ will imply $V_\kappa\vDash\varphi^{``W}$ as desired.

Universal quantifier case: Assume that $V_\kappa\vDash Q_0(y_0\in b_0)\ldots Q_k(y_k\in b_k)\forall x\psi(x,\vec y,\vec v)$. As the leading nodded quantifies are absolute between $V_\kappa$ and $W$, we can continue by choosing arbitrary values of $y_i$, $0\leq i\leq k$, that witness the satisfaction. Then $V_\kappa\vDash\psi(x,\vec y,\vec v)$ holds for arbitrary parameter $x\in V_\kappa$, so if we choose arbitrary $x\in W$, $V_\kappa\vDash\psi(x,\vec v)$ holds. As $x\in W$ was arbitrary, we have that $\forall(x\in W)(V_\kappa\vDash\psi(x,\vec v)$, so $V_\kappa\vDash\forall(x\in W)\psi(x,\vec v)$. We can then recover the leading bounded quantifiers.

Existential quantifier case: Assume that $V_\kappa\vDash q_0(y_0\in b_0)\ldots Q_k(y_k\in b_k)\exists x\psi(x,\vec y,\vec v)$. Again, the leading bounded quantifiers are absolute, so choose arbitrary $\vec y$ witnessing the satisfaction. By elementarity, $W\vDash\exists x\psi(x,\vec y,\vec v)$ as well. Choose a witnessing $x\in W$. We have $W\vDash\psi(x,\vec y,\vec v)$, and by elementarity again, $V_\kappa\vDash\psi(x,\vec y,\vec v)$. As $x$ was chosen from $W$, we have $V_\kappa\vDash\exists(x\in W)\psi(x,\vec v)$. The leading bounded quantifiers can then be recovered. $\square$

So $\mathsf{Reflection}^*_{\in,1}$ is consistent, assuming an inaccessible cardinal $\kappa$ exists. But how do we deal with bounded quantifiers $\forall x \subseteq v$ and $\exists y \subseteq k$?

Let $\mathscr L_{\mathcal P}$ be the first-order language with relation symbol $\in$ and a function symbol $\mathcal P$, to be interpreted as powerset. In particular $\mathscr L_{\mathcal P}$ does not have the bounded quantifiers $\forall x \subseteq v$ and $\exists y \subseteq k$. There is a translation from $\mathscr L_\subseteq$ to $\mathscr L_{\mathcal P}$ by replacing $\forall(x\subseteq v)(\ldots)$ with $\forall x(x\in\mathcal P(v)\implies\ldots)$ and $\exists(y\subseteq k)(\ldots)$ with $\exists y(y\in\mathcal P(k)\implies\ldots)$.

In fact, in The Higher Infinite (lemma 6.1, p.57) there is a superficially stronger version of the susbtructure property of inaccessible cardinals, which allows class parameters: For any $R\subseteq V_\kappa$, the set of $\alpha<\kappa$ such that $(V_\alpha,\in,R\cap V_\alpha)\prec(V_\kappa,\in,R)$ is closed and unbounded in $\kappa$. Then we get the following strengthening of theorem 1, giving a partial answer, at least when only one quantifier bounded to $W$ appears in $\varphi{``W}$.

Theorem 2: Let $\kappa$ be an inaccessible cardinal. Then $\mathsf{Reflection}^*_{\mathcal P,1}$ is consistent with $\mathsf{ZFC}$.

Proof: Similarly to the proof of theorem 1, let $\vec Q\psi(x,\vec v)$ be an arbitrary $\mathscr L_\in$ formula in which $W$ does not appear, where $\vec Q$ is an abbreviation of the part $Q_0(y_0\in b_0)\ldots q_k(y_k\in b_k)$ as appearing before, and choose a sequence of parameters $\vec v\in V_\kappa$. Now choose an $\alpha<\kappa$ such that $(V_\alpha,\in,\mathcal P\upharpoonright V_\alpha)\prec(V_\kappa,\in,\mathcal P\upharpoonright V_\kappa)$ and $\vec v\in V_\alpha$, this $V_\alpha$ will serve as $W$. Repeat the rest of the proof of theorem 1.

This shows $V_\kappa\vDash\mathsf{Reflection}^*_{\mathcal P,1}$. As $\kappa$ is inaccessible, $V_\kappa$ is also a model of $\mathsf{ZFC}$, therefore $\mathsf{Reflection}^*_{\mathcal P,1}$ is consistent with $\mathsf{ZFC}$. $\square$

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My naive attempt to try and allow more than one quantifier bounded by $W$ by removing the "in which $W$ does not appear" condition from $\psi$ in theorem 1 did not work: the proof relies on $W$ being an unused symbol so that an interpretation $V_\alpha$ of it can be chosen. It would also over-proves, as it would prove that $V_\kappa$ models this stronger schema when in actuality it is inconsistent, for example if $\varphi$ is $\exists x\forall(y\in W)(x\in y)$ and $\varphi^{``W}$ is $\exists(x\in W)\forall(y\in W)(x\in y)$.