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?

  • $\begingroup$ 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, 2019 at 20:21
  • $\begingroup$ @NoahSchweber, I've rephrased it. it should be clear by now! $\endgroup$ Sep 25, 2019 at 20:44

1 Answer 1


The answer to the first question is no. Suppose no formula meets qualification Q. Let πœ“(π‘₯) be xβ‰ x.

The answer when there is at least one formula πœ™ 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 βˆƒu(tr(u)βˆ§βˆ€s(s∈y-->s∈u)βˆ§βˆƒs(s∈u∧empty(s))) where tr(u) is βˆ€wβˆ€v(w∈u∧v∈w-->v∈u) and empty(w) is βˆ€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 πœ“(π‘₯) be βˆƒt(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), βˆ€π‘₯[βˆ€π‘¦(π‘¦βˆˆπ‘₯β†”πœ™(𝑦))β†’πœ“(π‘₯)] is provable from T for all πœ™ meeting qualification Q, and 𝑇+βˆ€π‘₯(πœ“(π‘₯)) is not consistent.

  • $\begingroup$ No this is wrong because T is consistent $\endgroup$ Sep 26, 2019 at 6:53
  • $\begingroup$ The theory T with no non-logical axioms is consistent. What is wrong? $\endgroup$ Sep 26, 2019 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, 2019 at 11:36
  • $\begingroup$ Yes, because that changes the question. $\endgroup$ Sep 26, 2019 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, 2019 at 15:14

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.