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Let $\mathcal Q$ be some qualification on formulas in the first order language of set theory (FOL($\in$)), that is met by at least one formula; Let $T$ be the first order set theory whose extra-logical axioms are the following sole axiom schema:

$\mathcal Q$-Comprehension schema: if $\phi(y)$ is a formula that meets qualification $\mathcal Q$, in which $x$ doesn't occur, and in which the symbol $y$ occurs free, and only free; then all closures of: $$\exists x \forall y (y \in x \leftrightarrow \phi(y))$$, are axioms.

Now suppose that $T$ is consistent, and that $\psi(x)$ is some formula in one free variable $x$, and $x$ only occur free in it, and suppose that theory $T$ proves that per the same conditions written above for $\mathcal Q$-Comprehension, all closures of following: $$\forall x [\forall y (y \in x \leftrightarrow \phi(y)) \to \psi(x)]$$, are theorems.

Would it always follow that: $T + \forall x (\psi(x))$ is consistent?

The other question is:

If not, then: are there known conditions that if qualifcation $\mathcal Q$ meets then $T + \forall x (\psi(x))$ would be consistent?

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  • $\begingroup$ I'm confused, "$T$" seems to be doing double-duty here - as both an arbitrary theory (per the first half of the first sentence) and as the specific theory "$\mathcal{Q}$-comprehension." Can you clarify? $\endgroup$ Sep 25 '19 at 20:21
  • $\begingroup$ @NoahSchweber, I've rephrased it. it should be clear by now! $\endgroup$ Sep 25 '19 at 20:44
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The answer to the first question is no. Suppose no formula meets qualification Q. Let 𝜓(𝑥) be x≠x.

The answer when there is at least one formula 𝜙 that meets qualification Q and the language does not have = as a primitive symbol, is still no. Suppose that the only formulas which meet qualification Q are (y∈y or not(y∈y)), and ∃u(tr(u)∧∀s(s∈y-->s∈u)∧∃s(s∈u∧empty(s))) where tr(u) is ∀w∀v(w∈u∧v∈w-->v∈u) and empty(w) is ∀x(not(x∈w)(that is y is contained in a transitive set which has an empty set as an element). By Q-Comprehension, there is an empty set. A universal set(guaranteed to exist by Q-Comprehension) is a transitive set which has an empty set as an element. Let 𝜓(𝑥) be ∃t(t∈x). Then for this Q, T is consistent(it holds in the 2 element set {a,b} with the binary relation E, where E is defined by xEy iff y=b), ∀𝑥[∀𝑦(𝑦∈𝑥↔𝜙(𝑦))→𝜓(𝑥)] is provable from T for all 𝜙 meeting qualification Q, and 𝑇+∀𝑥(𝜓(𝑥)) is not consistent.

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  • $\begingroup$ No this is wrong because T is consistent $\endgroup$ Sep 26 '19 at 6:53
  • $\begingroup$ The theory T with no non-logical axioms is consistent. What is wrong? $\endgroup$ Sep 26 '19 at 7:06
  • $\begingroup$ when I said that $\mathcal Q$ is a qualification on the formulas of the language of theory $T$, this means that some formulas of the language must meet that qualification and others must not, otherwise why should one stipulate an empty schema, that doesn't make sense. Should I stipulate that there is at least one formula $\phi$ that meets qualification $\mathcal Q$ to make that clear? $\endgroup$ Sep 26 '19 at 11:36
  • $\begingroup$ Yes, because that changes the question. $\endgroup$ Sep 26 '19 at 14:00
  • $\begingroup$ changes made I'll edit it further, also notice that $=$ is not a primitive of the language here, and there are no identity axioms in the underlying logic. (although this of course doesn't affect your argument in case $\mathcal Q$ is not met.) $\endgroup$ Sep 26 '19 at 15:14

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