# Can the axiom of choice or its weaker versions be (dis)proved using reflection principles?

In On the Question of Absolute Undecidability, Peter Koellner investigates whether it is possible to prove or disprove $$V = L$$ using (EDIT: both first and second-order) reflection principles, ie. statements of the form*

$$V \vDash \varphi(A) \to \exists \alpha \ V_\alpha \vDash \varphi^\alpha(A^\alpha)$$

and shows that it cannot be done.

Starting from $$ZF$$, can these principles be used to prove or disprove the Axiom of choice or some of its weaker variants (eg. Dependent, Countable choice)?

* definition on page 13.

• Although this is unrelated to the question on AC, perhaps the Robert's paper "A strong reflection principle" would be relevant. In this paper, there is a formulation of a reflection principle, of similar form, that implies the existence of 1-extendible cardinals (so it's incompatible with V=L). Jan 12 '20 at 14:49
• That seems to be an interesting paper. Bookmarked and saved in the Wayback Machine, thanks. Jan 12 '20 at 15:07

The question seems to me asking if sufficiently large cardinals defined by indescribability properties will prove the axiom of choice or its weak variants hold below such cardinals. (Disproving is moot since these are consistent with $$V=L$$.)
• This doesn't make sense to me. If reflection principles were a consequence of ZF, why would anybody bother trying to prove $V = L$ with them? Also, on page 14, Koellner says "This [reflection] principle yields inaccessible cardinals, Mahlo cardinals, weakly compact cardinals and more". Maybe, in my question, I mischaracterized the reflection principles? (But I've taken the definition from the paper.) Or maybe you only considered first-order reflection principles? Is that I also ask about second-order reflection something that I should have included in my question? Jan 12 '20 at 12:54