Is $\omega$ -consistency, soundess, or completeness expressible in the language of arithmetic?

Question

Obviously through the second incompleteness theorem consistency is expressible in La, I was wondering if $\omega$ -consistency and completeness were the same and how so. I know soundness is not but I don't know why not so if someone could explain that too :)

Answer 1

Working in some fixed appropriate theory $T$ (e.g. PA), completeness and $\omega$-consistency are expressible for the same reason consistency is: because provability is expressible. That is, there is a formula $Prov(x)$ such that $\mathbb{N}\models Prov(k)$ iff $k$ is the Godel number of a theorem of $T$. With this we have:

• $T$ is consistent iff $\exists x(\neg Prov(x))$.

• $T$ is complete iff $\forall x(Prov(x)\vee Prov(x'))$, where "$x'$" is the definable function corresponding to negation (that is, $k'$ is the Godel number of $\neg \varphi$ whenever $k$ is the Godel number of $\varphi$).

• $\omega$-consistency is messier to state precisely, but not hard: the definition is, again, directly in terms of the provability predicate ("if $T$ proves $\varphi(k)$ for each $k$, then it does not prove $\exists x(\neg\varphi(x))$").

This provability predicate itself is really the fundamental object.

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What about soundness? Well, here we run afoul of Tarski's undefinability theorem: soundness talks about truth instead of proof, and this we can't do.

(Actually there's a subtlety around what "expressibility" means here, but I'm ignoring that for now and treating it as a pretty informal notion. To see the issue, suppose $T$ actually is sound; then "$0=0$" expresses the soundness of $T$, since they're each true! This is obviously silly, but it takes some work to satisfyingly untangle this, which would get us off-track here.)

However, fragments of soundness are indeed expressible. Of particular interest are the $\Sigma_k$-soundness principles: $T$ is $\Sigma_k$-sound iff it only proves true $\Sigma_k$ sentences. The point is that while there is no "global truth predicate," for each $k$ there is a truth predicate $True_k$ for $\Sigma_k$ formulas. (Basically, $True_k$ is itself $\Sigma_k$, so we never "catch our tail").