Why $\mathsf{GJ}_0\vdash A(x,y)\in\mathrm{V}$ ? There is no axiom of induction or foundation in $\mathsf{GJ}_0$ .

Mathias doesn't say that " $\mathsf{GJ}_0$ is a subsystem of $\mathsf{fReR}_0$ ". So he indeed asserts a weaker proposition ($\mathsf{fReR}$ can prove the existence of cartesian products).

Mathias [2], p.14 . He says that "$\mathsf{GJ}$ is a subsystem of $\mathsf{fReR}$ would follow from the theory of companions". But $\mathsf{GJ}$ can prove the existence of cartesian products (there is an right arrow above $v$ in the axiom schema of rudimentary replacement(see page 6)).

By a class term I mean an informal abbreviation of the form $\{x\ |\ \phi\}$ , and then $\{x\ |\ \phi\}\in\mathrm{V}$ is an abbreviation which has the same meaning with $\exists y(\forall x(x\in y\leftrightarrow\phi))$ .

Yes, You are right. My question is trivial with this restriction. In fact, for every non-$\Delta_0^{\mathsf{T}}$ sentence $\phi$ , the function $x\mapsto \{x\}\cap\{x\ |\ \phi\}$ is not $\mathsf{T}$-rudimentary, where $\mathsf{GJ}_0\subseteq\mathsf{T}$ .

If $A$ is $x\cap B$ and $B$ is $\Delta_0$ , then it is sufficient to show that these functions produce Godel operators.

See my new remark above. In this sense, how can we prove $x\mapsto x\cap\{x\ |\ \mathrm{CH}\}$ is $\mathsf{ZFC}$-rudimentary ? Let this function be $F(x)$ , neither $\mathsf{ZFC}\vdash F(x)=x$ nor $\mathsf{ZFC}\vdash F(x)=\varnothing$ is true, since $\mathrm{CH}$ can't be decided in $\mathsf{ZFC}$ .

I think the same with you. But how to prove that "definably piecewise rudimentary functions need not be rudimentary" ? Consider an example such as $x\mapsto x\cap\{x\ |\ \mathrm{CH}\}$ , where $\mathrm{CH}$ means the Continuum Hypothesis. How can we prove this function is not rudimentary(assume our meta-system is ZFC)?