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 $\sf T$.



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> 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? $$\sf Con(\forall x. \phi x) \land Con( T) \\\to Con(T+ \forall x. \phi x)$$.
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My guess is to the negative, but I don't know of a counter-example.



Is there a known counter-example to this rule?
Take $\psi(y)$ to be $(y=y)\land \exists u\, \forall v\, \neg(v\in u)$, so that $\psi$ expresses the statement that there is some $\in$-minimal object. Thus, $\pi(x)$ is just false if there are no $\in$-minimal objects, but if there are $\in$-minimal objects then it says $x$ has all objects as elements.


Let $\phi(x)$ be the statement $\pi(x)\lor \forall u\, \exists v\, (v\in u)$. So, this says either (1) $x$ has every object as a member and there is an $\in$-minimal object, or (2) there are no $\in$-minimal objects.


Clearly $\forall x\, \phi(x)$ is consistent. FOL proves $\pi(x)\rightarrow \phi(x)$ (and really all you need is propositional logic). Also $T$ is consistent. But $T+\forall x\, \phi(x)$ is not consistent.


