Recall the definition of [cardinal definable][1] sets, to re-iterate:

$Define: X \text { is cardinal definable} \iff  \\\exists \text { cardinal } \kappa \, \exists \text { cardinals } \lambda_1,.., \lambda_n <^\rho \kappa \ \exists \phi : \\ X=\{ y \in V_{\rho(\kappa)} \mid  \phi^{V_{\rho (\kappa)}} (y,\lambda_1,..,\lambda_n)\}$

Where: $\lambda_i <^\rho \kappa \iff \rho(\lambda_i) < \rho(\kappa)$, and $\rho$ is the rank function; and "*cardinal*" is defined after Scott's as an equivalence class under *bijection* of sets of the lowest possible rank.

That every set is cardinal definable is proved consistent with the failure of choice (see [here][2]), on the other hand, it is also proved consistent with choice (see [here][3]).

>Now, working in $\sf ZF$, if we say that every cardinal definable set admits a choice function, would that entail full $\sf AC$?

> If we work in $\sf ZF-Reg.$, would cardinal definable choice imply $\sf AC$?


  [1]: https://mathoverflow.net/questions/412541/is-every-set-being-cardinal-definable-consistent-with-zf-negation-of-choice
  [2]: https://mathoverflow.net/a/412591/95347
  [3]: https://mathoverflow.net/a/412536/95347