# Consistency strength: If Con($T+A$) implies Con($T+B$), can we infer anything about $A$ and $B$?

To be more specific, let $T$ be a first order theory and let $A$ and $B$ be two different first-order sentences, both in the same language as $T$ but independent of $T$. Additionally, suppose we have (working in some meta-theory) that Con($T+A$) implies Con($T+B$) (but not vice-versa).

$\textbf{Question}$: What, if anything, can we say about the sentences $A$ and $B$?

That is, can we only speak of the difference between $A$ and $B$ in terms of $T$? Are we limited to saying that adding $A$ to $T$ results in a theory which is "more likely" to derive a contradiction than adding $B$ to $T$? Or can we infer anything about $A$ and $B$ on their own? (Does one prove or refute the other? Might it be that $A$ has more first-order consequences than $B$ does?)

$\textbf{Follow-up question}$: Could it ever be the case that replacing $T$ with a different theory $S$ (also in the same language and unable to decide $A$ or $B$) leads to Con($S+B$) being equivalent to, or stronger than, Con($S+A$)?

Here is an example to better illustrate my questions:

It is known that the consistency strength of ZF together with the Axiom of Choice is exactly that of ZF:

$$\text{Con(ZF+AC)} \iff \text{Con(ZF)}$$

However, the consitency strength of ZF together with the Axiom of Determinacy is very much greater:

$$\text{Con(ZF+AD)} \iff \text{Con(ZF}+\ \psi )$$

where $\psi$ is the statement "there are infinitely many Woodin cardinals".

Do these relations of consitency strength tell us about AC and AD as sentences set apart from ZF? Is AD in some regard more powerful than AC (even though it's refutable in ZFC)? Could, for some new theory of sets $T$ in the same language as ZF, it be possible that $T$+AC is equiconsistent with $T$+AD?

The answers to these questions may turn out to be trivial, but I'm unable to settle on a satisfactory conclusion myself. Any responses offering a greater intuition about consistency strength would be very much appreciated.

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For the last part, take $T=\mathrm{ZF}+\psi$. – Asaf Karagila Jul 25 '12 at 21:38

Sadly, there isn't anything we can glean from this. For example, working in ZF, we find that the principle of dependent choice--DC--is equivalent to AC for countable sets of non-0 finite sets--CC(fin)--together with the principle of dependent multiple choice--DMC. It is also equivalent to DMC together with the assumption that Cantor cubes $\{0,1\}^X$ are Baire spaces. Thus, taking T=ZF+DMC, we have Con(T+CC(fin))$\Leftrightarrow$Con(T+Cantor cubes $\{0,1\}^X$ are Baire). Moreover, we know that if Cantor cubes $\{0,1\}^X$ are Baire spaces, then CC(fin) holds. However, the converse does not hold.

There are other, similar examples, where T+A, T+B are equiconsistent, but where A, B aren't logically related to each other outside of the context of T.

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Thanks for your answer, Cameron. I wasn't aware of the example you provide and will read further into it. Can I take from what you say that the consistency strength never has an influence on logical strength? – ajcr Jun 11 '12 at 19:59
That is a good question, to which I don't know the answer. :-( We can certainly conclude that consistency strength doesn't always have an influence on logical strength. Let me sum up. In answer to your first question, we can't necessarily infer anything about A and B outside the context of T, solely by observing the relative consistency strengths of T+A and T+B. As to your follow-up question, I suppose it may be possible in some circumstances, but I can't imagine how one might (dis)prove it for general A, B...or even specific A, B, for that matter. – Cameron Buie Jun 11 '12 at 20:23
(apologies for my month-late reply!) Thanks again for your help with the question. I've left things unaccepted (for now) in case a wise logician stumbles upon the question and can offer any other clues towards a solution... – ajcr Jul 25 '12 at 20:58
That's fine. I'm glad I could help. – Cameron Buie Jul 25 '12 at 21:10

This is really more of a philosophical overview, rather than a mathematical answer. $\newcommand{ZF}{\mathrm{ZF}}\newcommand{ZFC}{\mathrm{ZFC}}\newcommand{AC}{\mathrm{AC}}\newcommand{AD}{\mathrm{AD}}$

Let us agree to work in $\ZF$ as our theory and $\ZFC$ as the meta-theory. First we consider the seemingly strange equiconsistency of $\ZF$ and $\ZFC$ alongside with many weak choice principles in between.

What does that tell us? It means that if $\ZF$ is our meta-theory, then whenever we happen to run into something which is a model of $\ZFC$ then we are guaranteed to run into a model of $\ZF+\lnot\AC$ somewhere in the universe. Often proofs of equiconsistency also tell us how to find such model, for example by forcing arguments or inner model arguments, etc.

Philosophically this means that if $\ZFC$ is true in some structure then $\ZF+\lnot\AC$ is true in another. Since $\ZF$ is a reasonable theory to consider, it tells us that there is no "leap of faith" if we assume $\ZFC$ instead. In particular it means that we are free of the worry that $\AC$ introduced some inconsistency to the system, that of course if $\ZF$ did not have one already.

From this we can deduce that Lebesgue and the school of people leading a mathematical life that $\AC$ is false are actually rejecting $\ZF$ as a whole, since from a proof of contradiction in $\ZFC$ we know how to generate a contradiction in $\ZF$. Then again, these proofs were given quite later than the early days of $\AC$ where people raged over its unthinkable consequences.

On the other hand, we have $\ZF+\AD$ which is infinitely stronger than $\ZF$ or $\ZFC$ in the consistency strength. This tells us that in contrast to the situation above even if we believe that $\ZF$ itself is consistent we still have to believe that there are large cardinals, and that there are many of them. So many that anything less than $\omega$ Woodin cardinals is not even close to enough.

This is quite the leap of faith, specifically since we know that anything that proves the consistency of $\ZF$ is quite the strong theory, and coming from outside set theory it may seem suspicious and borderline contradictory. However there are still arguments in favour of large cardinals, and either way these are interesting.

### Coda:

It seems nowadays like a reasonable idea to believe that the axioms of $\ZFC$ are consistent. I personally believe that if no shocking contradictions are found, in several decades most people will assume $\ZFC+\varphi$ where $\varphi$ is some large cardinal axiom which is strong enough to support category theory related constructions.

Until that happens, people may be wary of large cardinals, or indifferent to them.

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An interesting read, thanks. I find justifications (both philosophical and mathematical) for accepting large cardinal axioms intriguing. I've read a little about the programs to find reasonable-sounding (even arithmetical) mathematical statements that are independent of ZFC + large cardinal axioms and look forward to learning more about the discoveries that have been made through this research. – ajcr Jul 25 '12 at 21:59
@AJCR: It's an interesting question. As a strongly agnostic person, I refuse to decide any question on the existence or consistency of mathematical theories (or any other theory which is non-mathemtical). My main justification for large cardinal axioms is that they generate a lot of interesting questions and results. The surprising thing is that they are tied with "low cardinality" mathematics like determinacy of projective sets and measurability. – Asaf Karagila Jul 25 '12 at 22:03