TITLE: Is there a turing machine for which halting is equivalent to the Axiom of Choice or its negation?
QUESTION [6 upvotes]: As seen in "A Turing machine for which halting is outside ZFC", Gödel's incompletness theorem can that there a turing machines for which halting can not be decided. My question is, is there a turing machine $T$, such that the statement "$T$ halts" is equivalent to the axiom of choice or its negation in ZF set theory?

REPLY [7 votes]: No, this is impossible.
While $\operatorname{Con}\sf (ZFC)$ is an arithmetic statement which can therefore be used to construct a Turing machine which will not provably half in $\sf ZFC$, the axiom of choice is not even remotely a statement about the integers. So there is no chance that this is going to happen.
Slightly more formally, though, recall that if $\varphi$ is a first-order statement about the integers, then $V\models``\omega\models\varphi"$ if and only if $L\models``\omega\models\varphi"$. Since the axiom of choice is true in $L$, if there was such Turing machine, it would have to halt in $L$ and therefore halt in $V$. So the axiom of choice would be true in $V$, and therefore provable from $\sf ZF$.
When consistency is involved, you can go to non-standard models where $\operatorname{Con}\sf (ZFC)$ is false. But with the axiom of choice this is not going to work, since whenever $M$ is a model of $\sf ZF$, then $M$ and $L^M$ have the same ordinals, so the same $\omega$.