This question is motivated by wondering to what extent "natural" theories are linearly ordered (or at least ordered in a directed manner) by their (first-order) arithmetic consequences, in analogy to the phenomenon that "natural" theories seem to be linearly ordered by consistency strength.

In an answer to this question, Joel David Hamkins gave several examples illustrating how these two hierarchies -- arithmetic consequence versus consistency strength - may differ, and indeed his examples also illustrate that the ordering by arithmetic consequence is not linear, even for arguably "natural" theories.

But it so happens that these examples largely involved playing games with consistency statements, so the departure from linearity feels "small". One way of making this a bit more precise is that it seems, as far as I can tell, to still be possible that the partial ordering of theories is still directed, at least for "natural" theories. That is, what I know about the matter is consistent with the idea that there really is one "true arithmetic" with which all of our "natural" or "serious" theories (apart from things like $T + \neg Con(T)$ for various $T$) are consistent.

In order to have a hope of challenging this view, it seems one needs an alternate "coherent" picture of the mathematical universe. The only example that comes to mind for me is $ZF + AD$, which is inconsistent with ZFC, but nonetheless its consistency strength is well-calibrated (and nontrivial), and seems to have some sort of "inner logic" to it which could potentially yield a different picture of the world even at the level of arithmetic. AD paints a different picture of the universe at least if we include non-arithmetic statements in our scope. But I'm not sure it does anything unusual at the level of arithmetic.


  1. Are the arithmetic consequences of $ZF+AD$ consistent with standard theories like $ZFC + L$ for various large cardinal axioms $L$?

  2. Are the arithmetic consequences of $ZF+AD$ implied by large cardinals?

  3. More broadly, is there any good candidate out there for a theory $T$ whose arithmetic consequences are inconsistent with (or at least not implied by) ZFC + large cardinals, such that $T$ has some kind of "inner coherence" (in analogy to how large cardinals are said to have "inner coherence" by virtue of inner model theory -- as opposed to theories of the form $T + \neg Con(T)$ and the like)?

  • $\begingroup$ At least AD kills all idempotent proofs in Ramsey theory, since non-principal ultrafilters on a countable domain seize to exist... $\endgroup$ Sep 7, 2019 at 19:46

2 Answers 2


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.

  • $\begingroup$ Thanks! I suppose this line of argument would be circumvented by something like ZF+ Reinhardt or ZF+Berkeley. I gather from the lack of discussion on my earlier linked question that not much is known about these theories -- so perhaps they might yield "alternative arithmetic facts"... (Of course, because so little is known it's certainly questionable whether such theories have any kind of "inner coherence" to them.) $\endgroup$
    – Tim Campion
    Jun 21, 2019 at 19:29
  • 2
    $\begingroup$ @TimCampion Nope, those aren't any different either: they also prove that determinacy holds in $L(\mathbb{R})$. The mystery about them is whether they're consistent, but it is known that even if they are consistent they're extremely strong. Again, no set-theoretic axioms are (assuming their consistency) known to provide contradictory arithmetic facts, ignoring ones which are themselves arithmetic statements specifically chosen to do so (e.g. inconsistency statements). $\endgroup$ Jun 21, 2019 at 19:31
  • $\begingroup$ I'm confused -- granted that ZF+Reinhardt proves determinacy holds in $L(\mathbb R)$, we see that ZF+Reinhardt is stronger than ZF+AD in arithmetic consequences. But it's the other way around that we'd like -- we'd like to know that, say, the arithmetic consequences of ZF+Reinhardt and ZFC+I0 are consistent with each other, and I don't see how the argument gets us there... $\endgroup$
    – Tim Campion
    Jun 21, 2019 at 19:34
  • 1
    $\begingroup$ Sorry, I thought you were still comparing with AD. For Reinhardt vs. I0 for example, you'd want to show that any model of ZF + Reinhardt has an inner model of ZFC + I0. I believe this is known to be true but I'm not sure. (Certainly i think almost everyone believes that if they're each consistent then they're arithmetically compatible.) $\endgroup$ Jun 21, 2019 at 19:44
  • 1
    $\begingroup$ @NoahSchweber : Depending on one's taste, one might regard your final highlighted statement as comforting rather than annoying. It could be interpreted as reassuring people who really only care about arithmetic consequences that they need not worry unduly that those crazy mathematicians who assume wild axioms about sets will go off on the wrong track. $\endgroup$ Jun 22, 2019 at 18:27

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.

  • 2
    $\begingroup$ The $\omega$-rule generates all of true arithmetic, yes, but you need repeated applications of the $\omega$-rule. Therefore your conclusion that any theory of arithmetic that is not sound (weakening of true arithmetic, as you call it) must prove $\exists x\,\neg\phi(x)$ and all instances $\phi(n)$ , $n\in\omega$, for some formula $\phi(x)$, is incorrect. Theories such that no such $\phi$ exists are known as $\omega$-consistent, and there exist unsound $\omega$-consistent theories. $\endgroup$ Apr 6, 2020 at 15:23
  • $\begingroup$ See e.g. en.wikipedia.org/wiki/ω-consistent_theory for more background. $\endgroup$ Apr 6, 2020 at 15:30
  • $\begingroup$ @EmilJeřábek Thanks so much! I suppose the terminology "$\omega$-consistent" makes some sense -- it looks like a theory is $\omega$-consistent iff it is consistent in $\omega$-logic. But the terminology "sound" for "weakening of true arithmetic" strikes me as strange -- I'm used to "soundness" being a property of a logic rather than a property of a theory. $\endgroup$
    – Tim Campion
    Apr 6, 2020 at 15:52
  • 4
    $\begingroup$ No, being consistent in $\omega$-logic is a much stronger property than $\omega$-consistency. (The terminology is fairly confusing.) “Sound” is simply overloaded; in the context of theories of arithmetic (and perhaps other theories with a clearly distinguished “standard” model), it means “true in the standard model”, this is unrelated to the notion of a derivation system being sound with respect to a semantics. $\endgroup$ Apr 6, 2020 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.