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I was wondering how Σ1-sound theories in the language of first-order arithmetic can go wrong. As far as I can tell, they cannot prove false claims about consistency, since such claims (e.g., that there is a proof of 0=1" from the axioms) are equivalent (in weak theories of arithmetic) to Σ1 sentences. Can they prove that a sequence is finite when it is really infinite? It would be helpful to hear about some concrete Σ1-sound theories that prove blatantly false things. Thanks!

davidp
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  • I think you mean $\Sigma_1$ soundness rather than $\Pi_1$. Generally, interesting theories (e.g. anything stronger than Robinson arithmetic) are $\Sigma_1$-complete, so $\Pi_1$ soundness is just equivalent to consistency. For instance, if I prove Con(XYZ) in my favorite system and it winds up that XYZ is actually inconsistent, that will generally mean that my favorite system was inconsistent. On the other hand it's routine that a consistent system incorrectly proves that some other consistent system is inconsistent and is thereby $\Sigma_1$-unsound (e.g. PA + not Con(PA)). – spaceisdarkgreen Jan 11 '20 at 10:25
  • I guess I do mean that, since I have in mind theories that do not imply any false claims of the form "there is an x which codes a proof of 0=1". I hadn't noticed that a theory could be Π1 sound but Σ1 unsound. So, yes, please understand my question as being about Σ1 sound theories. Thanks! – davidp Jan 11 '20 at 22:13
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    I don't know why this got a -1. – Noah Schweber Jan 11 '20 at 23:49

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Basically, soundness at one level of the arithmetical hierarchy doesn't prevent errors higher up.

For example, let $PA_n$ be PA + all true $\Pi_n$ sentences (incidentally, note that every true $\Sigma_{n+1}$ sentence is a theorem of $PA_n$). Then we can form the Godel-Rosser sentence for $PA_n$:

$(*)_n$: "For every proof of me from $PA_n$, there is a shorter disproof of me from $PA_n$."

The usual arguments show that $(*)_n$ is independent of $PA_n$ and true (in particular, $PA_n$ is fully sound and $PA_n$ can decide whether something is a valid $PA_n$ proof); so $PA_n+(*)_n$ is $\Pi_n$-complete and hence $\Sigma_n$-sound, but not fully sound.

As a cautionary tale, note that Godel's second incompleteness theorem breaks down when we try to push it beyond computably axiomatizable theories: there is an arithmetically definable theory which proves its own consistency, in fact it's just an unusual arithmetic description of PA itself! I believe this is folklore, but see e.g. here.

For a more non-constructive but simpler argument, note that $Thm(PA_n)$ has Turing degree ${\bf 0^{(n+1)}}$ but $Th(\mathbb{N})$ is not arithmetic, so we must have $Thm(PA_n)\subsetneq Th(\mathbb{N})$ for all $n$. Pick $\sigma\in Th(\mathbb{N})\setminus Thm(PA_n)$, and consider $PA_n+\neg\sigma$.

(I'm writing "$Thm(S)$" for the deductive closure of $S$ here.)

Noah Schweber
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  • Thanks very much! One more question: requiring arithmetic soundness of a theory (in the language of first-order arithmetic) is enough to avoid all such consequences, correct? This despite the fact that, e.g., there are arithmetically sound but omega-inconsistent theories. – davidp Jan 12 '20 at 01:27
  • @davidp In the language of arithmetic, there are not any arithmetically sound but $\omega$-inconsistent theories - since in the language of arithmetic, "arithmetically sound" is just "sound." In order to get non-$\omega$-consistent theories which are arithmetically sound, you need to be working in a richer setting (e.g. set theory). – Noah Schweber Jan 12 '20 at 02:47
  • Thank you! What syntactic requirement on arbitrary theories interpreting the language of arithmetic suffices to ensure that they prove no false arithmetic sentences? – davidp Jan 12 '20 at 03:27
  • @davidp I'm not sure what you're looking for - "proves no false arithmetic sentences" (= arithmetic soundness) is a pretty good requirement already. – Noah Schweber Jan 13 '20 at 05:17
  • OK, thank you! :) – davidp Jan 13 '20 at 07:21