If $C$ is the nonexistence of non-trivial cycles in the Collaz conjecture, and $CC$ is the Collatz conjecture itself.
If it were proven that $C$ were unproveable, would this mean that $\lnot C$ would be unproveable too?
If so, this would prove that one can never identify a counterexample.
This would prove $C$ - a contradiction. Therefore $C$ cannot be unproveable.
Therefore any proof claiming that $CC$ is unproveable, which does not first prove $C$, is incorrect.
Is my logic correct?
I've highlighted the question to avoid misuse of "Unclear what the question is".
Am I right in thinking this rule does not apply to sequences ascending to infinity, because although any sequence might exist ascending to infinity, it may not be provable that it ascends to infinity - since one cannot follow it all the way. Therefore a proof that one can never find a counterexample wouldn't necessarily imply that no counterexample exists. So step $2$ would be incorrect in respect of $CC$.
But we can deduce that any proof that $CC$ is unproveable, must prove $C$ must it not?
As a corollorary to this question, I'm interested in to what extent the overall power of any logical framework can be augmented by this principle.
For example, it's well known that the strengthened finite Ramsey theorem (which is true) implies the consistency of Peano, meaning it is unproveable within Peano. Knowing this, can we deduce that Peano can generate no counterexample to the strengthened finite ramsey theorem and therefore the strengthened finite ramsey theorem is true?
Finally, how complicated an exercise would it be to attempt to show that Collatz implies the consistency of Peano?