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:

$\exists x. \phi x$

$\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.

insideH (rather than metatheoretically), you could take H to be any (supposedly consistent) system where it proves the consistency you want? $\endgroup$