Is there a decidable problem which has a proof that it cannot be proved to have a particular deciding Turing machine? I came across the question Is there an algorithm that provably exists although we don&#39;t know what it is? I was able to follow the example "Given an integer $n\ge0$ is there a run of $n$ or more consecutive 7s in the decimal expansion of $\pi?$" given in one of the answers. The answer shows that the problem in example is decidable although we don't know which is the correct algorithm. Let $P$ be a problem which is decidable, we know that for some for some finite integer $k &gt; 1$ one of the Turing machines $M_1,M_2,..,M_k$ decides the problem, but we don't know which of them exactly is the decider. Is it possible that $P$ has a finite proof which tells us that it is not possible to prove exactly which of the machines is the decider ? Or is it always, that the proof exists proving that exactly one of the machines is the decider but we don't know it ?

I am not even sure if my question makes sense. I could make changes or remove the question if it does not not. Sorry for the vague definition of a proof.
Let us examine the following language:

$x\in L \iff ZFC \hspace{1mm} \text{is consistent}$.

Clearly, $L$ is decidable, and is decided by either the constant 1 or 0 machines.

Proofs from ZFC for the statements "$M_0$ decides $L$", "$M_1$ decides $L$" cannot exist  (if ZFC is consistent), since any of those would imply a proof from ZFC about its own consistency (see Godels's second incompleteness).

Note that i can however, provide a proof that the following machine $M$ decides $L$:

Given input $x$, check if there exists a proof for the consistency of ZFC, of length$\le |x|$. If such proof is found, return 0, otherwise return 1. 

If ZFC is consistent, the above machine outputs 1 for all $x$. If it isn't, we also have a proof that the above machine decides $L$ (An inconsistent system can prove any statement). The problem is that i cant prove whether or not $M$ computes the constant 1 function.