TITLE: Does the consistency strength hierarchy coincide with the "arithmetic consequence" hierarchy at ZF + Reinhardt?
QUESTION [7 upvotes]: 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.

REPLY [11 votes]: 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.