Question: is the conjecture below true?
Consider decision problems in which the instance is (the PR index, definition, or LOOP program of) a primitive recursive function. Denote the PR function (with PR index $i$) by $\varphi_i$. Examples of PR problems (input $i$):
Problem P1 (decidable): let $n=\varphi_i(0)$. Are all integers $\varphi_i(1)$,..., $\varphi_i(n)$ prime?
Problem P2: (undecidable, see http://www.dcc.fc.up.pt/~acm/pr3.pdf): $\exists n:\varphi_i(n)=0$?
Conjecture. Looking to the program (index) is not more powerful than evaluating the function. In more detail:
The computational model $\langle$Turing machine M($i$) with the PR index $i$ of a PR function $\varphi_i$ as input$\rangle$ can not decide more properties than the (more restricted) model $\langle$Turing machine M$^f$ whose "input" is an oracle for computing $f(n)$ given $n$$\rangle$.
Definition (text from user aws). A property $P$ is decidable if there is a recursive function [computable, halting for all inputs] that, given the PR index $i$ of a PR function $\varphi_i$, returns 1 if the function $\varphi_i$ is in $P$, and returns 0 otherwise.
The instance is a PR index ($\varphi_i$ is always PR), not a TM index. Say it represents a LOOP program, not a set of quadruples.
There is some information that we can get by looking to the index of $f$. The maximum loop nesting of the LOOP program that defines $f$ allows us, in some cases, to establish a positive answer to to a question (property) like "$\exists n_0\forall n\geq n_0:f(n)\leq g(n)$?" (where $g$ is fixed and depends on that maximum nesting) is positive. But this is not sufficient to decide the property. Apparently this "semi-answer" cannot be obtained by the oracle machine $M^f$.
In "Rice (like) Theorem" for primitive recursive functions? I posted a similar but more vague question (no conjectures). Excuse me if this problem is not appropriate to mathoverflow...