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This question is inspired by this previous question.

It is not too hard - although a bit subtle - to show that if $F_1, F_2$ are two theories in the language of arithmetic which

  • are recursively axiomatizable,

  • extend PA, and

  • $(\dagger)$ PA proves that $F_1, F_2$ each extend PA,

then if $F_1\vdash Con(F_2)$ and $F_2\vdash Con(F_1)$ we have $F_1, F_2$ are inconsistent.

These hypotheses can be weakened in a few ways; my question is whether the third can be done away with completely:

Are there recursively axiomatizable theories $F_1, F_2$ extending $PA$ which are consistent, but which prove each others' consistency?

A brief word on why $(\dagger)$ appears necessary: the idea is to argue that if $F_1$ is inconsistent, then $F_2$ knows this, and moreover $F_1$ knows that. To do this, however, we need to argue that $F_1$ knows that $F_2$ proves every true $\Sigma_1$ sentence. Of course, if $F_2$ contains $PA$ then that is correct, but why should $F_1$ know it without $(\dagger)$?

EDIT: Let me clarify how there can be recursively axiomatizable theories which contain $PA$, but don't satisfy $(\dagger)$. First, note that the statement "$Prov_{PA}(\varphi)\implies Prov_T(\varphi)$ (in English, "$T$ contains $PA$") is not $\Sigma_1$, but rather $\Sigma_1\vee\Pi_1$; so while $PA$ does prove every true $\Sigma_1$ sentence, that doesn't help us here.

That said, how can we construct an example of a theory containing $PA$ but not "visibly"?

Fix an effective enumeration $\{\varphi_n: n\in\mathbb{N}\}$ of $PA$, and let $$T=\{\varphi_n: \mbox{there is no contradiction in $PA+Con(PA)$ of length $<n$}\}$$ (here it doesn't matter exactly how we define the length of a proof, so long as it is effective and there are only finitely many proofs of each length). Now, clearly (:P) this contains $PA$. But consider the theory $S=PA+Con(PA)+\neg Con(PA+Con(PA))$:

  • $S$ contains $PA$, obviously.

  • $S$ is consistent, as long as we assume some basic niceness (e.g. that $PA$ is true) - apply Goedel's Second to $PA+Con(PA)$.

  • $S$ proves that $T$ is finite (since for some $n$ we find a contradiction in the theory $PA+Con(PA)$ of length $n$.

  • $S$ proves that $PA$ is consistent.

  • And, finally, $S$ proves that $PA$ is not finitely axiomatizable (this is a neat fact - if $PA$ is consistent, then it is not finitely axiomatizable, and $PA$ knows this).

So $S$ is a consistent extension of $PA$ which proves that $T$ is strictly weaker than $PA$; but then this means that $PA$ does not prove that $T$ is not strictly weaker than $PA$. So $T$ "invisibly" contains $PA$.

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Noah Schweber
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  • To head off a couple attempts at a trivial solution: there are weak theories which do prove their own consistency, as well as strong arithmetically definable theories which prove their own consistency. Being strong, and recursively axiomatizable, makes things tricky. – Noah Schweber Jan 02 '17 at 17:44
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    Just when I think I am out, they pull me back in. – Rene Schipperus Jan 02 '17 at 18:19
  • @ReneSchipperus Hehehe. I don't know how to construct a counterexample here - both theories would have to contain PA in a cruddy way, so that kills off every natural example I know. And I don't know how to beef up e.g. Willard's theories to get them above PA without tripping over Goedel . . . – Noah Schweber Jan 02 '17 at 18:21
  • It seems to me that if $PA\subseteq S\subseteq T$ recursively axiomatizable then $PA\vdash Pr_S(\sigma)\rightarrow Pr_T(\sigma)$. Is this not what you are asking ? – Rene Schipperus Jan 02 '17 at 18:26
  • Just a thought but doesn't PA prove that PA is $\Sigma_n$ consistent ? So PA will know that each $\Sigma_n$ part is in $S$. Don't know yet if this gives enough to finish the argument. – Rene Schipperus Jan 02 '17 at 18:44
  • @ReneSchipperus Yes, PA proves the consistency of $I\Sigma_n$ (I think this is what you mean by "$\Sigma_n$ consistent") for each $n$. The problem, of course, is that PA does not prove "$\forall n[Con(I\Sigma_n)]$", so we can have weird extensions of PA which prove that there is some natural number $c$ such that $I\Sigma_c$ is inconsistent. – Noah Schweber Jan 02 '17 at 18:46
  • @ReneSchipperus Interestingly, the example I gave of a theory invisibly containing PA doesn't obviously work - and in fact I think nothing of the kind will (so I've deleted that comment). We need to work a bit harder to get a theory invisibly containing PA. I've edited my answer to include such an example. – Noah Schweber Jan 02 '17 at 18:57
  • I am just trying to get a hold on things. I reviewed our comments on the other question, it seem that the essential point is that $PA\vdash \sigma \rightarrow Pr_S(\sigma)$ for $\Sigma_1$ sentences $\sigma$. For this implies $T\vdash \sigma \rightarrow Pr_S(\sigma)$. Then $T\vdash \neg con(T) \rightarrow Pr_S( \neg con(T) )$ and thus $T\vdash Pr_S( \neg con(T) ) \rightarrow \neg con(S) $ and so $T\vdash con(T) $. So $PA\vdash \sigma \rightarrow Pr_S(\sigma)$ is what one needs to contradict. – Rene Schipperus Jan 02 '17 at 19:13
  • @ReneSchipperus That's true, but note that in that answer we assumed that $S$ and $T$ both visibly contained $PA$. That's essential to $PA\vdash \sigma\rightarrow Pr_S(\sigma)$ (or similarly with $T$), so the lack of this assumption here is a serious obstacle. (Note that "$PA\vdash \sigma\rightarrow Pr_S(\sigma)$" doesn't follow just from $PA$ proving all true $\Sigma_1$ sentences, since "$\sigma\rightarrow Pr_S(\sigma)$" is $\Sigma_1\vee\Pi_1$ and not $\Sigma_1$, even if $\sigma$ is $\Sigma_1$.) – Noah Schweber Jan 02 '17 at 19:15
  • Wait a minute isn't $PA\vdash \sigma \rightarrow Pr_S(\sigma)$ for $Sigma_1$ formulas $\sigma$ just a property of the provability predicate $Pr_S$, valid for all recursive $S$ ? – Rene Schipperus Jan 02 '17 at 19:20
  • @ReneSchipperus Nope. It's certainly not valid for all recursive $S$ - what if $S$ is the empty theory? – Noah Schweber Jan 02 '17 at 19:23
  • Or do you think there is the implicit assumption that $S$ extends $PA$? Course of you don't have this relation you might have some trouble proving the second Godel theorem for $S$. – Rene Schipperus Jan 02 '17 at 19:23
  • @ReneSchipperus But you're right if we make a very weak assumption on $S$ - it has to contain the seriously-super-weak theory $R$. Since $R$ is finitely axiomatizable, any theory containing $R$ visibly contains $R$; so this seems to be a problem. Hmm . . . – Noah Schweber Jan 02 '17 at 19:26
  • At this point I would look into the proof of this relation. It could be that $S$ only has to have a small part of $PA$. – Rene Schipperus Jan 02 '17 at 19:27
  • @ReneSchipperus Hehe, overlapping comments. – Noah Schweber Jan 02 '17 at 19:27
  • Let us continue this discussion in chat. – Rene Schipperus Jan 02 '17 at 19:30

1 Answers1

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Aha! Rene Schipperus and I figured it out (see the comment thread above).

The answer is no: the requirement $(\dagger)$ that $F_1, F_2$ "visibly" contain $PA$ is unnecessary.

Here is why: the only time $(\dagger)$ was used in this answer was to show

$F_i$ proves that $F_j$ proves every true $\Sigma_1$ sentence

for $i, j\in\{1, 2\}$. But for this, it's enough for $F_i$ to prove that $F_j$ contains Robinson's Q, a much weaker theory than $PA$, since $PA$ (which $F_i$ contains) proves that $Q$ proves every true $\Sigma_1$ sentence.

So the real question is, "Can a theory invisibly contain $Q$?"

The answer to this is no, for a simple reason: $Q$ is finitely axiomatizable! This means that the statement "$F$ contains $Q$" is $\Sigma_1$ - by contrast, note that if we replace "$Q$" with a non-finitely-axioatizable theory $R$ (like, say, the weak arithmetic $R$ :P), the statement "$F$ contains $R$" is $\Pi_2$.

Why does this difference matter? Well, since $F_i$ contains $PA$, $F_i$ proves "any theory containing $Q$ proves any true $\Sigma_1$ sentence." And since $F_j$ contains $Q$, $F_i$ proves "$F_j$ contains $Q$" by the above paragraph. So we don't need any visibility assumption at all.

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