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Working in first order logic with equality and membership $``\sf FOL(=,\in)"$

Let $\phi x$ be a formula in which only $x$ occur free, and never bound.

Let $\psi_i x\vec{z}$ be a formula in which only symbols $``x,z_1,..,z_n"$ occur free, and never bound, such that:

$\sf FOL(=,\in)$ $ \vdash \forall \vec{z} \forall x: \psi_i x\vec{z} \to \phi x $

Let $\sf T$ be a theory that extends $\sf FOL(=, \in)$, with only the following axioms:

1. $\exists x. \phi x$

2. $\forall \vec{z} (\exists x. \psi_{i \leq n} x\vec{z}) $

The idea is that $\sf T$ only says that there exists an object that fulfills $\phi$, and stipulate comprehension axioms assuring the existence of objects satisfying particular finite set of formulas $\psi_1,..,\psi_n$ which are in turn provable to always satisfy $\phi$ in just the background language of $\sf T$.

My question is that given the above conditions, is there a known system that is thought to be a consistent extension of first order logic in which the following is provable? $$\sf Con(\forall x. \phi x) \land Con( T) \\\to Con(T+ \forall x. \phi x)$$.

My guess is to the negative, but I don't know of a counter-example.

Working in first order logic with equality and membership $``\sf FOL(=,\in)"$

Let $\phi x$ be a formula in which only $x$ occur free, and never bound.

Let $\pi_i x \vec{z}$ be the formula $\forall y (y \in x \leftrightarrow \psi_i y\vec{z})$ where $\psi_i y \vec{z}$ is a formula in which only symbols $``y,z_1,..,z_n"$ occur free, and never bound; such that:

$\sf FOL(=,\in)$ $ \vdash \forall \vec{z} \forall x: \pi_i x \vec{z} \to \phi x $

Let $\sf T$ be a theory that extends $\sf FOL(=, \in)$, with only the following axioms:

1. $\exists x. \phi x$

2. $\forall \vec{z} (\exists x. \pi_i x \vec{z}), _{ i=1,..,m} $

The idea is that $\sf T$ only says that there exists an object that fulfills $\phi$, and stipulate $m$-many naive comprehension axioms each assuring the existence of a set of all objects satisfying a formula among $\psi_1,..,\psi_m$ formulas, and all those sets in turn are provable to satisfy $\phi$ in just the background language of $\sf T$.

My question is that given the above conditions, is there a known set theory that is provably consistent relative to some extension of ZF in which the following is provable? $$\sf Con(\forall x. \phi x) \land Con( T) \\\to Con(T+ \forall x. \phi x)$$.

My guess is to the negative, but I don't know of a counter-example.

Working in first order logic with equality and membership $``\sf FOL(=,\in)"$

Let $\phi x$ be a formula in which only $x$ occur free, and never bound.

Let $\psi_i x\vec{z}$ be a formula in which only symbols $``x,z_1,..,z_n"$ occur free, and never bound, such that:

$\sf FOL(=,\in)$ $ \vdash \forall \vec{z} \forall x: \psi_i x\vec{z} \to \phi x $

Let $\sf T$ is a theory that extends $\sf FOL(=, \in)$, with only the following axioms:

1. $\exists x. \phi x$

2. $\forall \vec{z} (\exists x. \psi_{i \leq n} x\vec{z}) $

The idea is that T only says that there exists a set that fulfills $\phi$, and stipulate a comprehension axioms assuring the existence of sets satisfying particular finite set of formulas $\psi_1,..,\psi_n$ which are in turn provable to always satisfy $\phi$ in just the background language of T.

My question is that given the above conditions, is there a known system that is thought to be a consistent extension of first order logic in which the following is provable? $$\sf Con(\forall x. \phi x) \land Con( T) \\\to Con(T+ \forall x. \phi x)$$.

My guess is to the negative, but I don't know of a counter-example.

Working in first order logic with equality and membership $``\sf FOL(=,\in)"$

Let $\phi x$ be a formula in which only $x$ occur free, and never bound.

Let $\psi_i x\vec{z}$ be a formula in which only symbols $``x,z_1,..,z_n"$ occur free, and never bound, such that:

$\sf FOL(=,\in)$ $ \vdash \forall \vec{z} \forall x: \psi_i x\vec{z} \to \phi x $

Let $\sf T$ be a theory that extends $\sf FOL(=, \in)$, with only the following axioms:

1. $\exists x. \phi x$

2. $\forall \vec{z} (\exists x. \psi_{i \leq n} x\vec{z}) $

The idea is that $\sf T$ only says that there exists an object that fulfills $\phi$, and stipulate comprehension axioms assuring the existence of objects satisfying particular finite set of formulas $\psi_1,..,\psi_n$ which are in turn provable to always satisfy $\phi$ in just the background language of $\sf T$.

My question is that given the above conditions, is there a known system that is thought to be a consistent extension of first order logic in which the following is provable? $$\sf Con(\forall x. \phi x) \land Con( T) \\\to Con(T+ \forall x. \phi x)$$.

My guess is to the negative, but I don't know of a counter-example.

Working in first order logic with equality and membership $``\sf FOL(=,\in)"$

Let $\phi x$ be a formula in which only $x$ occur free, and never bound.

Let $\psi_i x\vec{z}$ be a formula in which only symbols $``x,z_1,..,z_n"$ occur free, and never bound, such that:

$\sf FOL(=,\in)$ $ \vdash \forall \vec{z} \forall x: \psi_i x\vec{z} \to \phi x $

Let $\sf T$ is a theory that extends $\sf FOL(=, \in)$, with only the following axioms:

1. $\exists x. \phi x$

2. $\forall \vec{z} (\exists x. \psi_{i \leq n} x\vec{z}) $

The idea is that T only says that there exists a set that fulfills $\phi$, and stipulate a comprehension axioms assuring the existence of sets satisfying particular finite set of formulas $\psi_1,..,\psi_n$ which are in turn provable to always satisfy $\phi$ in just the background language of T.

My question is that given the above conditions, is there a known system that is thought to be a consistent extension of first order logic in which the following is provable? $$\sf [\exists H: (H \vdash \forall x. \phi x) \land Con(H)] \land Con( T) \\\to Con(T+ \forall x. \phi x)$$.

My guess is to the negative, but I don't know of a counter-example.

Working in first order logic with equality and membership $``\sf FOL(=,\in)"$

Let $\phi x$ be a formula in which only $x$ occur free, and never bound.

Let $\psi_i x\vec{z}$ be a formula in which only symbols $``x,z_1,..,z_n"$ occur free, and never bound, such that:

$\sf FOL(=,\in)$ $ \vdash \forall \vec{z} \forall x: \psi_i x\vec{z} \to \phi x $

Let $\sf T$ is a theory that extends $\sf FOL(=, \in)$, with only the following axioms:

1. $\exists x. \phi x$

2. $\forall \vec{z} (\exists x. \psi_{i \leq n} x\vec{z}) $

The idea is that T only says that there exists a set that fulfills $\phi$, and stipulate a comprehension axioms assuring the existence of sets satisfying particular finite set of formulas $\psi_1,..,\psi_n$ which are in turn provable to always satisfy $\phi$ in just the background language of T.

My question is that given the above conditions, is there a known system that is thought to be a consistent extension of first order logic in which the following is provable? $$\sf Con(\forall x. \phi x) \land Con( T) \\\to Con(T+ \forall x. \phi x)$$.

My guess is to the negative, but I don't know of a counter-example.