A question on Gandy-Jensen system and the rudimentary functions

Question

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.

Answer 1

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.

Answer 2

I'm not well versed in fine structure, so this may well be wrong:

I think it depends what you mean by a "class term." Since the language of set theory contains no function symbols, there are no terms as usually defined; so you would need to fix a set of function symbols to be used as the language out of which terms can be built. Note that, in this case, your function symbols must be complicated for the answer to be nontrivial, since if your function symbols are expressible in terms of rudimentary functions (e.g., "$x\mapsto \{x\}$" is just $R_0(x, x)$), then the answer to your question is trivially yes.

If on the other hand by "class term" you mean a definable function - that is, a formula $p(\overline{x}, y)$ such that $GJ_0$ (or similar) proves that for every $\overline{x}$ there is exactly one $y$ such that $p(\overline{x}, y)$ - then I think the answer is no, since we can define functions by cases (e.g., "$x\mapsto \{x\}$ if $x=\emptyset$ and $x\mapsto \emptyset$ otherwise"). It seems that in general such "definably piecewise rudimentary" functions need not be rudimentary, but of course $GJ_0$ will prove that each one is total.

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Edit: the new phrasing of the question - roughly "if $GJ_0$ proves that F is total, then $ZFC$ proves that $F=G$ for some rudimentary $G$" - yields a very simple negative answer, for the reasons you say. In fact, we can replace $ZFC$ with any reasonable theory $T$: pick some sentence $p$ undecidable in $T$, and let $F: x\mapsto\emptyset$ if $p$ and $\{x\}$ if $\neg p$. It's easy to see that $GJ_0$ proves that $F$ is total, but that $F$ is not $T$-rudimentary.

The question is much more interesting without this restriction, I feel.

(Also, you should still define what you mean by "class term.")