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Are there known examples of statements which are strong from a proof-theoretic standpoint but which are indistinguishable by one set of axioms (or proof system) yet distinct according to a stronger set of axioms?

More specifically, I'm wondering if there are examples of the following kind:

Let $T_1$ and $T_2$ be theorems in some formal language $\mathcal{L}$. Let $A_1$ and $A_2$ be two distinct sets of axioms in $\mathcal{L}$ but which are not (obviously) incompatible. Are there known examples of $T_1$ and $T_2$ where $A_1$ proves "$T_1$ is equivalent to $T_2$" but $A_2$ proves "$T_1$ is strictly stronger than $T_2$"?

At first glance, this notion of "stronger" conflicts with the received notion of "stronger" as "proving the same and more theorems" so the notion of "stronger theory" I'm asking about rules out characterizations like "the collection of formulas deducible from $A_1$ is properly contained in the collection of formulas deducible from $A_2$". I'm wondering if there is a sense in which two theories can disagree on the equivalence of two propositions because the weaker theory views the propositions as the same in some sense and the stronger theory witnesses some kind of first-order (or higher?) distinction between the propositions. Is this a useful notion in general? Or would this require some kind of axiom like "indiscernible objects are identical" to make precise and/or useful?

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  • $\begingroup$ A partial motivation for this question comes from the fact that I've seen something like "Rowbottom cardinals (or maybe Jonsson cardinals) are Ramsey in $L[\mu]$" the minimal model for a measurable. Is this correct? $\endgroup$
    – Everett Piper
    Mar 11, 2012 at 1:11
  • $\begingroup$ I meant "inner model for a measurable" above. $\endgroup$
    – Everett Piper
    Mar 11, 2012 at 1:12

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When I compare the question with the example in the comment, I infer that you regard "in $L[\mu]$" as weak, presumably because there's "only" a measurable cardinal there. But that theory is strong in another direction, namely by saying that the universe is only $L[\mu]$ rather than something bigger.

In a similar vein, V=L is a strong theory. But if you want to regard it as weak because it doesn't allow very impressive large cardinals, then another example of what you asked for would be that V=L proves that the cardinal of the continuum and the first uncountable cardinal are the same.

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  • $\begingroup$ I definitely agree with you in this regard. But I'm worried that, at least in this context, the theory ZFC + V=L can't distinguish between interesting objects like the bounding number or many other cardinal characteristic numbers. Do you think there is a sense of "stronger" in this sense that is mathematically legitimate (or maybe more concretely: interesting to pursue for set theorists)? $\endgroup$
    – Everett Piper
    Mar 11, 2012 at 1:55

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