Let $\mathrm{R}\_0,\cdots,\mathrm{R}\_8$ be the following functions:


$\mathrm{R}\_0(x,y)=\{x,y\}$


$\mathrm{R}\_1(x,y)=x-y$


$\mathrm{R}\_2(x)=\bigcup x$


$\mathrm{R}\_3(x,y)=x\times y$


$\mathrm{R}\_4(x)=\mathrm{Dom}(x)$


$\mathrm{R}\_5(x)=\{(a,b)\ |\ a\in b\ \wedge\ a\in x\ \wedge\ b\in x\}$


$\mathrm{R}\_6(x)=\{(a,b,c)\ |\ (b,a,c)\in x\}$


$\mathrm{R}\_7(x)=\{(a,b,c)\ |\ (b,c,a)\in x\}$


$\mathrm{R}\_8(x,y)=\{x[\{a\}]\ |\ a\in y\}\ \ $ where $x[\{a\}]=\{b\ |\ (a,b)\in x\}$ .


Gandy-Jensen system $\mathsf{GJ}\_0$ is the system of which the set-theoretic axioms are the Axiom of Extensionality and the following nine set-existence axioms:


$\mathrm{R}\_0(x,y)\in\mathrm{V},\ \cdots,\ \mathrm{R}\_8(x,y)\in\mathrm{V}$ .


The class of rudimentary functions is the closure of $\mathrm{R}\_0,\cdots,\mathrm{R}\_8$ under composition.


It is obvious that if $F$ is a $n$-ary rudimentary function then $\mathsf{GJ}\_0\vdash F(x\_1,\cdots,x\_n)\in\mathrm{V}$ .


My question is that whether the converse of the above proposition is true, i.e. is it true that:


For every class term $A(x\_1,\cdots,x\_n)$ whose free variables are in $x\_1,\cdots,x\_n$ , if $\mathsf{GJ}\_0\vdash A(x\_1,\cdots,x\_n)\in\mathrm{V}$ , then $(x\_1,\cdots,x\_n)\mapsto A(x\_1,\cdots,x\_n)$ is a $n$-ary rudimentary function.




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Edit 1 : I think I should make my terminologies more precise:


Let $\mathsf{T}$ be an axiom system of set theory (such as $\mathsf{GJ}\_0,\ \mathsf{ZFC},$ and so on). We say that a function F is a n-ary $\mathsf{T}$-rudimentary function if there is a n-ary rudimentary function G such that $\mathsf{T}\vdash F=G$ .


I actually asked that whether the following statement is true:


For every class term $A(x\_1,\cdots,x\_n)$ whose free variables are in $x\_1,\cdots,x\_n$ , if $\mathsf{GJ}\_0\vdash A(x\_1,\cdots,x\_n)\in\mathrm{V}$ , then $(x\_1,\cdots,x\_n)\mapsto A(x\_1,\cdots,x\_n)$ is a $n$-ary $\mathsf{ZFC}$-rudimentary function.


Where $\mathsf{ZFC}$ can be replaced by other systems such as $\mathsf{GJ}\_0$ and then we have more similar questions.




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Edit 2 : The question become trivial if I restrict "rudimentary" to "$\mathsf{T}$-rudimentary" defined in Edit 1 . So I will remove this restriction, and still ask that whether the original proposition (which is weaker than the false proposition in Edit 1 ) is true. 



A question on Gandy-Jensen system and the rudimentary functions
Rudimentary functions can only increase rank by a fixed finite amount. So $$A(x,y) = \begin{cases} x + y & \text{when $x,y \in \omega$,} \\ 0 & \text{otherwise,} \end{cases}$$ is not rudimentary even if $\mathsf{GJ}\_0 \vdash A(x,y) \in V$.




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This is true even for nonstandard models of $\mathsf{GJ}\_0$ but one has to be careful about the definition of $A(x,y)$. For simplicity, I will use $F(x) = 1+A(x,x) = 1+2x$. Then $F(x) = y$ iff $x \in \omega$ ($\omega$ being understood as a definable class, not necessarily a set) and there is exactly one sequence $s$ with domain $x+1$ such that if $s(0) = 1$ and $s(z+1) = s(z)+2$ whenever $z \lt x$ then $s(x) = y$; otherwise, $F(x) = 0$.


In ill-founded models, $\{x \in \omega : F(x) \neq 0\}$ could be a proper cut in $\omega$ but that doesn't change the fact that this provably total function $F$ is not rudimentary.


