Does ZF+AD have any unusual arithmetic consequences?

For (1) and (2), the key point here is absoluteness. For example, suppose $V$ is a model of ZFC + enough large cardinals. Then $L(\mathbb{R})^V$ satisfies ZF + DC + AD. But $L(\mathbb{R})^V$ and $V$ have the same natural numbers, hence satisfy the same arithmetic sentences. So if $(*)$ is a "big" large cardinal axiom - namely one which is strong enough to imply that determinacy holds in some inner model - then any arithmetic consequence of determinacy is also a consequence of $(*)$.

• Indeed, as far as I know all such axioms imply that determinacy holds in $L(\mathbb{R})$ - and that gives us not just the arithmetic facts, but the projective facts as well: every second-order consequence of determinacy is a consequence of any large cardinal axiom implying that determinacy holds in any model containing all the reals, in particular $L(\mathbb{R})$.

This addresses (1) and (2). The same reasoning provides a strong heuristic (in my opinion) that the answer to (3) will be "no" - any such theory would have to somehow be fundamentally unrelated to outer/inner models, to put it mildly, and we don't know any such candidate yet (ignoring things like ZFC + $\neg$ Con$(*)$). The situation, annoyingly, remains:

So far, we know of no situation where different set-theoretic axioms (ignoring "pathological" examples like inconsistency statements) yield contradictory arithmetic consequences.

I just want to record here a few things I've learned lately concerning the motivational question about the empirical linearity of "plausible" theories of arithmetic.

This empirical fact -- that all "plausible" theories of (first-order -- or even second-order) arithmetic appear to be linearly (or at least directed-ly) ordered by their direct implications -- seems to be a pretty famous fact among logicians. I recently came across some further discussion of it in Woodin's Notices article.

To see why one might expect this, one can contemplate the $\omega$-rule, i.e. the (infinitary) deduction rule which says that if you can deduce $\phi(n)$ for all numerals $n$, then you can deduce $\forall x(\phi(x))$ (where we work in the language of first-order arithmetic). It's easy to show that when first-order logic is augmented with the $\omega$-rule, the axioms of $PA$ (or even $Q$) generate a complete theory (though I found it surprisingly hard to find a source bothering to both (i) spell out what the $\omega$-rule is and (ii) explicitly state that it leads to $PA$ being complete!), and that indeed this theory coincides with true arithmetic. Therefore, any theory $T$ in the language of arithmetic extending $PA$ (or even $Q$) must either be a weakening of true arithmetic, or else fail to validate the $\omega$-rule. In particular, if $T$ is complete [1], then $T$ must be $\omega$-inconsistent, i.e. there must be a formula $\varphi(x)$ such that $T \vdash \varphi(n)$ for each numeral $n$, and yet $T \vdash \exists x (\neg \varphi(x))$. Consequently, if $T$ is any theory in the language of arithmetic extending $PA$ (or even $Q$), then either $T$ is a weakening of true arithmetic, or else in every model of $T$ there is an element which is not a numeral (i.e. not of the form $1 + 1 + \dots + 1$). This seems like a pretty good reason to think that any such $T$ is "implausible". At any rate, it means that "alternate arithmetic" can't be thought of as just an alternate way of assigning truth values to formulas with the same underlying set of numbers as we have in true arithmetic -- any alternate arithmetic can have only nonstandard models, with new elements which are not numerals.

Of course, this argument requires our metatheory to reassure us that there is a complete theory of "true arithmetic". I'm no expert, but the only way out I can see to entertain the possibility of truly "plausible" alternate theories of arithmetic to exist is if one changes one's metatheoretical assumptions. It seems to me that as long as one assumes classical logic in the metatheory, one has to concede that there is a complete theory of true arithmetic, and I don't see anywhere where choice is used in showing that the $\omega$-rule completes $PA$. (Some form of choice is required to ensure that every consistent theory $T$ of the sort considered above has a model, but if $T$ doesn't have a model, then it again looks "implausible".) So even if the metatheory is $ZF$, I think one will still be led to the conclusion that only true arithmetic is "plausible". Probably, then, one needs to weaken at least to an intuitionistic metatheory in order to contemplate the possibility of "plausible" alternatives in the theory of arithmetic.

[1] An earlier version of this post made this claim without the hypothesis that $T$ be complete. Thanks to Emil Jerabek in the comments below for pointing out the error.