# New Foundations and weaker forms of choice

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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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. – Ed Dean Nov 22 '11 at 7:00
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). – Carlo Von Schnitzel Nov 22 '11 at 7:20
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. – Emil Jeřábek Nov 22 '11 at 10:37
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? – Carlo Von Schnitzel Nov 22 '11 at 16:32
Make it $A\lor(B\land \neg B)$ then. – Emil Jeřábek Nov 28 '11 at 14:49