Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 120 / 2

We prove the claim by structural induction for primitive recursive functions. A formula is bounded when every quantifier is a bounded quantifier; this is a Delta-0 formula. The graphs of the initial functions have quantifier-free definitions:

Suppose $f(\mathbf{x})=g(h_1(\mathbf{x}),\dots ,h_m(\mathbf{x}))$ and the graphs of $g,h_1,\dots ,h_m$ have existential bounded definitions. Introduce variables $u_1,\dots ,u_m$ for the intermediate values and conjoin the graph formulasPulling all unrestricted existential witnesses to the front leaves a bounded matrix, so the required form is preserved by composition.

For primitive recursion, use Gödel beta-function sequence coding. The relationis bounded in the language of ordered rings. Given any finite sequence $a_0,\dots ,a_n$, choose $c$ larger than all its entries and divisible by $1,\dots ,n+1$. The moduli $1+(i+1)c$ are pairwise coprime, so the Chinese remainder theorem supplies $b$ with $\beta (b,c,i)=a_i$ for every $i\le n$.

Supposeand the induction hypothesis gives existential bounded graph formulas for $g$ and $h$. Then $y=f(\mathbf{x},n)$ exactly when there are $b,c$ coding values $a_0,\dots ,a_n$, together with a common witness bound $W$, such that:Write each $a_i$ using $\beta (b,c,i)$, bound the variables representing $a_i$ by $1+(n+1)c$, and bound every witness used by the graph formulas for $g$ and $h$ by $W$. All quantifiers checking the displayed finite recursion are then bounded; only $b,c,W$ and the finitely many outer coding witnesses are unrestricted existential variables. Conversely, any code passing these bounded checks satisfies the recursion equations, and induction on $i\le n$ forces its $i$th entry to be $f(\mathbf{x},i)$. This proves the existential bounded representation of a primitive recursive function.