In these slides (see especially slide 26), Steel emphasizes the phenomenon that for all known "natural" extensions of ZFC, the ordering by consistency strength agrees with the ordering by containment of arithmetic consequences. That is, if $T$ and $T'$ are natural extensions of ZFC, then $T \leq T'$ with respect to consistency strength if and only if $T'$ proves all the arithmetic consequences of $T$. Indeed, Steel points out that as long as $T$ and $T'$ have high enough consistency strength, the same could be said of their analytical consequences, not just their arithmetical consequences.

But I find it remarkable that the first-order theories of the strongest consistency strength known (which are not known to be inconsistent) are most naturally formulated in ZF without the axiom of choice. Namely, if you look at the theories listed at Cantor's attic, the highest consistency strength theories are ZF + Reinhardt cardinals or ZF + Berkeley cardinals (I'm being vague about how many of these cardinals are stated to exist). Of course you can get stronger theories by taking Con(ZF + Reinhardt) etc. in some language, but these are the basis of things.

My question is, basically: does the phenomenon Steel discusses continue to hold for natural extensions of ZF, rather than ZFC? In particular, does ZF + Reinhardt cardinals imply all the arithmetical consequences of ZFC + $I_0$ (the strongest large cardinal principle not known to be inconsistent with choice)? How about ZF + Berkeley cardinals?

EDIT Steel says a little more about "naturalness" on p. 6, footnote 10 of the related paper. I'll state it using some notation:

  • If $A$ is a set of statements in the language of set theory, let $\overline{A}$ denote the deductive closure of $A$
  • If $A$ is a set of statements in the language of set theory, let $\mathrm{Con}^\mathrm{Fin}(A)$ denote the set of statements $\{\mathrm{Con}(\phi) \mid \phi \in A\}$ where $\mathrm{Con}(\phi)$ is the statement that $\phi$ is consistent.
  • If $S,T$ are extensions of ZFC, write $S \leq_{\mathrm{Con}}^{\ast} T$ if $\mathrm{Con}^\mathrm{Fin}(S) \subseteq T$. Note by compactness that $S\leq_\mathrm{Con}^\ast T$ implies $S \leq_\mathrm{Con} T$, where $\leq_\mathrm{Con}$ is the usual consistency strength preorder.
  • If $T$ is an extension of ZFC, let $(\Pi^0_1)_T$ denote the $\Pi^0_1$ consequences of $T$.

According to Steel, the reflection principle implies for any $T$ extending ZFC that $(\Pi^0_1)_T = \overline{\mathrm{Con}^\mathrm{Fin}(T)}$. This implies that

  • $(\Pi^0_1)_S \subseteq (\Pi^0_1)_T \Leftrightarrow S \leq_\mathrm{Con}^\ast T$

That is, the ordering on extensions of ZFC by $\Pi^0_1$ consequences can be recast in terms of consistency statements. Steel says that for the "natural" extensions he's talking about, it turns out that $S \leq_\mathrm{Con} T \Leftrightarrow S \leq_\mathrm{Con}^\ast T$. In some sense this is just a re-statement of another version of the phenomenon, but I think it sheds some light.

I'm not sure whether all of this should go through with ZF in place of ZFC, but if it does, it would be interesting to know whether $ZFC +I_0 \leq_\mathrm{Con}^\ast ZF + \mathrm{Reinhardt}$ for example. Of course, this is still quite a ways from talking about containment of all arithmetic statements. I would be happy to say something about the $\Sigma^0_1$ statements -- I'm particularly keen (for kind of frivolous reasons) to know in $ZF + \mathrm{Reinhardt}$ that a given Turing machine will halt so long as $ZFC + I_0$ thinks it will halt.

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    $\begingroup$ Steel asserts the phenomenon only for "natural" theories extending ZFC, and I wonder how strictly one needs to interpret that. For example, if $T$ is a natural theory, then I would find $\text{Con}(T)$ to be natural, and strictly higher in consistency strength. But do we know in general that $\text{ZFC}+\text{Con}(T)$ proves all the arithmetic consequences of $T$? For example, do we know that the consistency of a supercompact cardinal implies over ZFC all the arithmetic consequences of an actual supercompact cardinal? $\endgroup$ Apr 16, 2016 at 18:45
  • $\begingroup$ @JoelDavidHamkins I would love to know the answer to questions of that form! My sense is that Steel has a more restrictive meaning of "natural" in mind, but I really have no idea. $\endgroup$
    – Tim Campion
    Apr 16, 2016 at 18:59
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    $\begingroup$ To answer my question, my updated answer shows that ZFC+Con(T) does not prove all the arithmetic consequences of T when T is any of the usual large cardinal axioms. So Steel cannot view ZFC+Con(T) as a natural theory, even when T is. $\endgroup$ Apr 17, 2016 at 18:36
  • $\begingroup$ Are your frivolous reasons to do with naming a bigger number than one that uses $I_0$, by any chance? ;-) $\endgroup$
    – David Roberts
    Apr 17, 2016 at 23:12
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    $\begingroup$ That $(\Pi^0_1)_S\subseteq(\Pi^0_1)_T$ iff $S\le^*_{\mathrm{Con}}T$ iff $S$ is interpretable in $T$ is the Orey-Hájek characterization of interpretability in reflexive theories. $\endgroup$ Apr 17, 2016 at 23:48

1 Answer 1


Steel makes his assertions only for "natural" theories extending $\newcommand\ZFC{\text{ZFC}}\ZFC$, and so in order to focus attention on the extent to which his phenomenon relies on this idea of naturality, let me describe a few counterexamples to the phenomenon, even in the $\ZFC$ case. The fact of the matter is that there are some natural-seeming theories for which the phenomenon can fail.

First example. Imagine a large cardinal set theorist who asserts that inconsistency is no more likely at the level of Mahlo cardinals than it is at the level of inaccessible cardinals. I believe that many large cardinal set theorists actually hold this view, and I have heard some prominent set theorists assert similar claims explicitly. Indeed, Woodin has asserted that if $\ZFC$ is found inconsistent, then he would look to $\text{PA}$ to fall next. So in this sense, the view is not unreasonable.

Let $T$ be the theory that expresses the view explicitly, so that $T$ asserts $\newcommand\Con{\mathop{Con}}\ZFC$ plus the implication $\Con(\ZFC+I)\to\Con(\ZFC+M)$, where $I$ asserts that there is an inaccessible cardinals and $M$ asserts that there is a Mahlo cardinal. So $T$ asserts over $\ZFC$ that if an inaccessible cardinal is consistent, then so is a Mahlo cardinal.

This theory is actually equiconsistent with $\ZFC$, because if $\ZFC$ is consistent, then there is a model of $\ZFC$ where $\ZFC$ is inconsistent, which would make $T$ vacuously true, by denying the antecedent. Thus, $T\leq_{\Con}\ZFC$ in the consistency strength hierarchy. But meanwhile, $T$ has additional arithmetic consequences, such as the implication $\Con(\ZFC+I)\to\Con(\ZFC+M)$ itself, which is not provable in $\ZFC$ unless inaccessible cardinals are inconsistent, since $\Con(\ZFC+M)$ implies $\Con(\ZFC+\Con(\ZFC+I))$, which would violate the incompleteness theorem for $\ZFC+\Con(\ZFC+I)$ if this theory were consistent. So this is a counterexample to Steel's phenomenon.

One can construct many similar counterexamples to the phenomenon, by using theories of the form $\ZFC+\Con(\ZFC+LC_1)\to\Con(\ZFC+LC_2)$, whenever $LC_1$ is a weaker large cardinal than $LC_2$. Steel will have to claim that these theories are not natural in order to preserve his phenomenon. But are they so unnatural? Most large cardinal set theorists would find them to be true.

A Better example. Let me now explain a much better counterexample to the phenomenon (thanks to Emil for explaining how to make it work). If $T$ is a natural theory extending $\ZFC$, then most people would find $\ZFC+\Con(T)$ also to be a natural theory. Indeed, the literature is full of theorems proved under the assumption $\Con(T)$, where $T$ is any of the usual large cardinal theories. The theory is strictly stronger than $T$ in consistency strength. But meanwhile, I claim that if $T$ is $\ZFC$ plus any of the usual large cardinal axioms, then $T$ will have arithmetic consequences not provable in $\ZFC+\Con(T)$, and so this is a counterexample to the phenomenon.

The reason is that all of the usual large cardinal axioms prove that there is a transitive model of $\ZFC$, and this is enough to establish the implication $$\newcommand\Prov{\text{Prov}}\Prov_{\ZFC}(\varphi)\to\varphi$$ for arithmetic assertions $\varphi$. Note that this implication is itself an arithmetic assertion. For example, if $\kappa$ is inaccessible, then since $V_\kappa\models\ZFC$, it follows that $\Prov_{\ZFC}(\varphi)$ implies that $V_\kappa$ thinks the arithmetic assertion $\varphi$ is true, and this is absolute to $V$, since $V_\kappa$ has the same arithmetic as $V$. So $\varphi$ is true.

But meanwhile, $\ZFC+\Con(T)$, if consistent, will not be able to prove all instances of this implication. The reason is that by the incompleteness theorem, there is a model of $\ZFC+\Con(T)+\neg\Con(\ZFC+\Con(T))$. Because of the negated consistency assertion, the model will satisfy $\Prov_{\ZFC}(\neg\Con(T))$, while also satisfying $\Con(T)$, and this violates the desired implication $\Prov_{\ZFC}(\varphi)\to\varphi$ in the instance where $\varphi$ is $\neg\Con(T)$.

So the theory $\ZFC+\Con(T)$ is a counterexample to the phenomenon discussed by Steel, where $T$ is any of the usual large cardinal axioms, because $T$ proves $\Prov_{\ZFC}(\varphi)\to\varphi$ for arithmetic assertions $\varphi$, but these implications are not all provable in $\ZFC+\Con(T)$, if this theory is consistent.

  • $\begingroup$ I agree that in order to prove a general metatheorem, one will have to be precise about what "natural" means. But I don't think this is necessary in order to show the phenomenon holds for the comparison between ZF + Reinhardt and ZFC + I0. For example it would follow if we knew that ZFC + I0 had the same arithmetic consequences as ZF + I0 [since ZF + Reinhardt extends ZF + I0, right?], which would follow from a version of Shoenfield absoluteness(but I guess that doesn't work,given that Shoenfield absoluteness works by comparing things to L, where these large-cardinal hypotheses can't hold...). $\endgroup$
    – Tim Campion
    Apr 17, 2016 at 2:13
  • $\begingroup$ Joel David Hamkins, I don't know about ZFC, but for theories for which we have ordinal analyses (or, say, functional interpretation), we can typically add any set of true $\Pi^0_1$-sentences without changing the provably recursive functions ($\Pi^0_2$-sentences). $\endgroup$ Apr 17, 2016 at 2:52
  • $\begingroup$ @UlrikBuchholtz That is what I would have expected. So can you provide "natural" counterexamples to Steel's phenomenon, if we work over PA, say, instead of ZFC? $\endgroup$ Apr 17, 2016 at 3:02
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    $\begingroup$ ZFC (and in fact, PA + Goodstein's theorem) proves the uniform $\Sigma_1$-reflection principle for PA, which is not provable in any consistent extension of PA by $\Pi_1$ sentences (as otherwise its finite fragment would imply its own consistency). (Nothing special about PA in this argument.) $\endgroup$ Apr 17, 2016 at 11:26
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    $\begingroup$ So, in the situation in the answer: if T proves uniform $\Sigma_1$-reflection for ZFC (any large cardinal axiom does), then this arithmetic consequence of T is not provable in ZFC + Con(T). $\endgroup$ Apr 17, 2016 at 11:30

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