4
$\begingroup$

It is trivial to show that Peano artithmetic ($\mathsf{PA}$) supplemented with the $\omega$-rule is complete. Joseph Shoenfield (`On a Restricted $\omega$-Rule', Bull. Acad. Polon. Sci. 7 (1959): 405–7) showed that this is true even if we replace the $\omega$-rule with the recursive $\omega$-rule; i.e., we admit

$$\frac{\phi(\bar{0}), \phi(\bar{1}),\ldots}{\forall x \ \phi(x)}$$

only if there exists a recursive function enumerating the proofs of $\phi(\bar{0}), \phi(\bar{1}),\ldots$.

Is it known whether this result can be strengthened? E.g., is $\mathsf{PA}$ $+$ the primitive recursive $\omega$-rule complete? $\mathsf{PA}$ $+$ the Kalmár-elementary $\omega$-rule?

$\endgroup$
2
$\begingroup$

In short the answer is yes.

Let us consider Schütte-style proof of completeness of $\omega$-logic. This proof works as follows. For any sequent $\Gamma$ we define it's canonical (cut-free) pre-proof (i.e. possibly ill-founded proof tree that locally obeys the rules of $\omega$-logic). The general idea here is to define pre-proof, whose conclusion is $\Gamma$, so that each possible rule is applied at some point. Next we show that if there is an infinite path in the cannonical pre-proof of $\Gamma$ there is an infinite path, then $\Gamma$ is false. Henceforth a sequent $\Gamma$ is true in $\mathbb{N}$ iff the canonical pre-proof of $\Gamma$ is well-founded.

The usual construction of the canonical pre-proof is in fact Kalmar elementary. Moreover, it should be even polynomial.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.