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The total recursive functions are exactly those number-theoretic functions that can be represented by a $\Sigma_1$ formula of first-order arithmetic.

Is there a similar characterization of the primitive recursive functions? I'm looking for something like for example

(wild conjecture) The primitive recursive functions are those that can be represented by a $\Sigma^0_1$ formula which can be proved total and single-valued by $\Delta^0_0$ induction.

hmakholm left over Monica
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  • Possibly helpful: http://math.stackexchange.com/questions/364494/primitive-recursive-function-which-isnt-delta-0 – Asaf Karagila Feb 24 '14 at 22:30
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    It is a classical theorem in proof theory that the provably total functions in $\mathrm{I}\Sigma_1$ are exactly the primitive recursive functions. I remember this is in the Aczel–Simmons–Wainer volume, but I don't have a copy of this book at hand now. I am sure you can also find it in some other books in the subject. The theory $\mathrm{I}\Delta_0$ in the usual language of arithmetic $\{0,1,{+},{\times},{<}\}$ is not enough. For example, it does not prove the totality of exponentiation. – Lawrence Wong Feb 25 '14 at 09:07

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The provably total functions in $\mathrm I\Sigma_1$ are exactly the primitive recursive functions. This is a theorem of Parsons from the 1970s; see, for instance, Theorem 3 on page 34 of the book Proof Theory edited by Aczel, Simmons and Wainer for the standard proof.

The theory $\mathrm I\Delta_0$ in the usual language of arithmetic $\{0,1,{+},{\times},{<}\}$ is not sufficient to prove the totality of all primitive recursive functions. For example, it was shown by Parikh in his paper “Existence and feasibility in arithmetic” that the totality of exponentiation is not provable in $\mathrm I\Delta_0$.

Lawrence Wong
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  • By "provably total", does it mean that there's no requirement that the formula representing the function can be shown to be single-valued? – hmakholm left over Monica Feb 25 '14 at 11:59
  • @Henning Makholm: no, it has to be provably a function as well. The theorem can be stated formally as: a total function $f$ is primitive recursive if and only if there is an $e$ such that $(\forall x)[f(x) = U((\mu s)T( e,s,x))]$, and $I\Sigma^0_1$ proves $(\forall x)(\exists s)T(\underline e,s,x)$. Here $U$ and $T$ are the standard predicates of Kleene to represent a universal model of computation in arithmetic. So the functionality of $f$ is implicit in the statement of the theorem, in a certain way. – Carl Mummert Feb 25 '14 at 12:30
  • In that statement, we could replace $U$ and $T$ with a universal $\Sigma^0_1$ formula, which would give an interchangeable result with a slightly different syntactic appearance. – Carl Mummert Feb 25 '14 at 12:39