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Let $T$ be a consistent theory, and let $A$ be a statement in the same language. Consider the three theories

  • $T$
  • $T+A$
  • $T+\neg A$

Is it possible for them to be pairwise distinct in consistency strength?

As a follow-up, is it possible for $T+A$ and $T+\neg A$ to be incomparable in consistency strength? (Clearly they are both stronger than $T$, so a fortiori this would mean that the three consistency strengths are distinct.)

I'm primarily interested in theories in classical, finitary first-order logic, but if it makes a difference to consider other logics, I'd find that interesting.

If $T+A$ is inconsistent, then $\neg A$ is provable in $T$, so $T+\neg A$ and $T$ are certainly equiconsistent; similarly if $T+\neg A$ is inconsistent, then $T$ and $T+A$ are equiconsistent. So the question is only interesting if all theories involved are consistent.

This question is motivated by the fact that typically if $T$ is ZFC and $A$ is a large cardinal hypothesis, then $T+\neg A$ is modeled by a substructure of any model of $T$ (by "chopping the universe off" at an inaccessible), so that $T$ and $T+\neg A$ are equiconsistent.

tcamps
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    According to the Stanford Encyclopedia of Philosophy, there is no known natural such example. – Hanul Jeon May 09 '15 at 10:21
  • @AsafKaragila Could you explain why the proof-theory tag is inappropriate? Consistency is ultimately a proof-theoretic notion. For example, the consistency strength of the theories I'm asking about could be distinguished by ordinal analysis. – tcamps May 09 '15 at 13:28
  • You might be right, and it might be a proof theoretic question. But it doesn't feel like it to me, it felt a lot more like a set theoretic/straight up logic type of question. – Asaf Karagila May 09 '15 at 13:34
  • @tetori The SEP article talks about the interpretability hierarchy, so let me see if I have this straight. If $T$, $T+A$ have different consistency strengths, then $T$ is interpretable into $T+A$ in the obvious way, but $T+A$ can't be interpretable into $T$ because then a model of $T$ would give a model of $T+A$, so by the completeness theorem (and assuming $T$ is consistent), consistency of $T$ would imply consistency of $T+A$. So a proper interpretability jump implies a proper consistency strength jump (continued)... – tcamps May 09 '15 at 13:45
  • The SEP says there are no natural examples where $T+A$ and $T+\neg A$ both have a proper interpretability jump over $T$, hence no natural examples where they both have a proper consistency strength jump over $T$, hence no natural examples where all the consistency strengths are distinct. Right? – tcamps May 09 '15 at 13:45

2 Answers2

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You can construct such sentences via fixed point theorem; viz. there exists $\phi$ such that both $Con(ZFC + \phi)$, $Con(ZFC + \neg \phi)$ are unprovable in $ZFC + Con(ZFC)$. Such a $\phi$ is what is called a double jump sentence. A reference for this is P. Lindstrom, Aspects of Incompleteness, 2003.

Guest
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  • That's pretty nice. Although it doesn't quite show that $\sf ZFC+\phi$ and $\sf ZFC+\lnot\phi$ have different consistency strengths. – Asaf Karagila May 10 '15 at 22:04
  • @Guest Thanks for the hint, and the reference! Which fixed point theorem is this? Also, is there a specific part of Lindstrom's book that talks about this? – tcamps May 11 '15 at 02:00
  • You suggest to take $\phi+\exists\kappa\text{ inaccessible}$. But then the negation is $\lnot\phi\lor\lnot\exists\kappa\text{ inaccessible}$ and $\sf ZFC+\operatorname{Con}(ZFC)$ can in fact prove $\operatorname{Con}\sf (ZFC+\lnot\exists\kappa\textrm{ inaccessible})$. So your example fails if we add the inaccessible part. – Asaf Karagila May 11 '15 at 12:00
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Here is something that might be worth noting.

It is impossible that both $A$ and $\lnot A$ can prove something like $\operatorname{Con}\sf (ZFC)$. Suppose that they were, given any model of $\sf ZFC$ either $A$ holds there or $\lnot A$ holds there. In either case it must be that $\operatorname{Con}\sf (ZFC)$ must hold in that model; so by the completeness theorem $\sf ZFC$ must prove its own consistency.

So it raises the question, what does it mean to have different consistency strengths? Of course that $\sf ZFC$ must have the weakest consistency strength of the three; so we might expect both $\mathsf{ZFC}+A$ and $\mathsf{ZFC}+\lnot A$ to prove $\operatorname{Con}\sf (ZFC)$, since they should be strictly stronger.

But as the above shows, it's impossible for both of them to prove that. So while this is not exactly an argument while such $A$ cannot be found in the case of $\sf ZFC$, or some theory strong enough to be subjected to the second incompleteness theorem, it does suggest that the answer will not come from large cardinal and the likes of it. And that the consistency "jumps" must go in a non-obvious direction here.

Asaf Karagila
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  • Interesting! This can be stated non-model-theoretically, too: if $T$ is consistent, subject to the second incompleteness theorem, and validates the law of excluded middle, then $T+A$ and $T+\neg A$ can't both prove $\mathrm{Con}(T)$, because then $T$ proves $\mathrm{Con}(T)$. And $T+\mathrm{Con}(T)$ is usually considered to be quite close in consistency strength to $T$. – tcamps May 10 '15 at 20:51