# What are the known conditions for a restriction on naive comprehension that enables a generalization of a property all so constructed sets meet?

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?

• 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? Sep 25 '19 at 20:21
• @NoahSchweber, I've rephrased it. it should be clear by now! Sep 25 '19 at 20:44

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.

• No this is wrong because T is consistent Sep 26 '19 at 6:53
• The theory T with no non-logical axioms is consistent. What is wrong? Sep 26 '19 at 7:06
• 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? Sep 26 '19 at 11:36
• Yes, because that changes the question. Sep 26 '19 at 14:00
• 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.) Sep 26 '19 at 15:14