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New Foundations (introduced by Quine) proves that $AC$ is false. Out of curiosity, is $NF$ consistent with countable choice or dependent choice? What's the strongest consequence of choice still consistent with $NF$ and has it been localized at what point consequences of $AC$ become inconsistent with $NF$? Hope my question is clear. Thx.

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    $\begingroup$ NF is not known to disprove countable choice or dependent choice, but I'm pretty sure there are no extant consistency proofs for these relative to NF either. That seems to be backed up by this article of Tom Forster's (and he should know): plato.stanford.edu/entries/quine-nf/#4 Of course, the variant NFU is consistent with full choice. $\endgroup$
    – Ed Dean
    Nov 22, 2011 at 7:00
  • $\begingroup$ I should have checked this article before asking. Unrelated to $NF$, as a general question, is there a general technique when we want to localize, for some arbitrary statement $A$, the strongest statement implying (or implied by) statement $A$ but still consistent with that theory? I don't if I am being clear. Maybe my question is completely silly and this is just the technique of forcing in ZF (measuring consistency strength). $\endgroup$ Nov 22, 2011 at 7:20
  • $\begingroup$ The strongest statement implying $A$ (over $T$, I suppose) and consistent with $T$ exists if and only if $T+A$ is a consistent complete theory, in which case it is $A$ itself. This cannot happen for any recursively axiomatized theory of at least arithmetical strength (such as set theory) by Gödel’s theorem. The strongest statement implied by $A$ over $T$ and consistent with $T$ exists if and only if $T+A$ is consistent (in which case it is $A$) or $T$ is complete and consistent (in which case it is any theorem of $T$). Again, the latter cannot happen if $T$ obeys Gödel’s theorem. $\endgroup$ Nov 22, 2011 at 10:37
  • $\begingroup$ Haha, slick answer Emil. What if we want to add the condition "other than A itself", which is what I had in mind first? Is there a technique for this? $\endgroup$ Nov 22, 2011 at 16:32
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    $\begingroup$ Make it $A\lor(B\land \neg B)$ then. $\endgroup$ Nov 28, 2011 at 14:49

2 Answers 2

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Thanks to Tim Chow for telling me about this thread. The general idea is that weak versions of choice that concern only sets that are in some sense small all seem to be consistent wrt NF. No proofs yet of course. In contrast, versions of choice that make claims about all sets can go wrong. Thus DC_\alpha for big alpha is refutable. But some quite strong choice principles seem to be open (for large sets). We don't know whether or not the prime ideal theorem is consistent with NF. Nor do we know whether or not the cardinal principle alpha = 2.alpha for alpha infinite is consistent. It's not known whether or not the partition principle is consistent wrt NF. I have spent some time on these last three questions and got nowhere!

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Perhaps I am not qualified to answer this question, but there are some experts claiming in email newsletters that NF + PP[1] is consistent. (no proof)

In a general way, the NF's choiceless set is very "close" to V.

Not all weaker forms of choice axiom are consistent with NF. For example, Thomas Forster claimed[2] a proof that $NF + DC_α$[3] is inconsistent, where α is a non-Cantorian ordinal. No one has studied the specific upper / lower boundaries, But $\Omega$: "The order type of all the ordinals" satisfies the requirement.

Consider the family of all wellorderings under the binary relation of end-extension and consider a maximal chain, using (some form of) DC. The union of this chain will be a wellordering that all wellorderings embed into. But this is impossible.


[1] Partition Principle: if A is nonempty and surjects onto B, B injects into A

[2] http://www.dspace.cam.ac.uk/handle/1810/223940

[3] $DC_α$: Let $S$ be a nonempty set and let $R$ be a binary relation such that for every $α$-sequence $s = [ x_γ : γ < α ]$ of elements of $S$ there exists $y ∈ S$ such that $sRy$.Then there is a function f such that for every $γ < α, (f | γ) R f(γ)$.

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  • $\begingroup$ This is very hearsay at this point. $\endgroup$
    – Asaf Karagila
    Jan 25, 2023 at 20:01
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    $\begingroup$ @AsafKaragila This is true but it's also true that an unfortunate amount of the consistency results surrounding NFU only exist in email exchanges apparently. $\endgroup$ Jan 26, 2023 at 0:46
  • $\begingroup$ fills in the details of the second claim. I can't help with the first claim, I haven't even seen how to forcing when ground model is NF. $\endgroup$ Jan 26, 2023 at 16:29

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