It is known that the Continuum Hypothesis is independent of ZFC. The formulation of the Collatz conjecture looks somehow more simple than that of the Continuum Hypothesis. Is it possible that the Collatz Conjecture is also independent of ZFC? Or does the form of this conjecture prevent that from happening?

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    $\begingroup$ Undecidability is always dependent on a specific axiomatic system since at minimum the statement would be decidable in an axiomatic system that had the statement you care about as an additional axiom. We cannot at this time rule out the Collatz conjecture being undecidable. And we know from a result of Conway that there are problems very similar to Collatz which are undecidable, since one can use similar functions to model Turing machines. But this seems more like a Math Stack Exchange rather than a Mathoverflow level question. $\endgroup$
    – JoshuaZ
    Commented Jun 2 at 14:04
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    $\begingroup$ @JoshuaZ I agree with you, but not your first sentence: the OP does specify an axiomatic system, namely ZFC. $\endgroup$ Commented Jun 2 at 14:10
  • $\begingroup$ Does this answer your question? Knuth's intuition that Goldbach might be unprovable $\endgroup$ Commented Jun 2 at 16:40
  • $\begingroup$ @AlexKruckman My comment about axiomatic independence was in reference to their question about it being the same as being undecidable. $\endgroup$
    – JoshuaZ
    Commented Jun 2 at 17:14
  • $\begingroup$ @TimothyChow my question is specifically about Collatz. This statement looks a bit different than Goldbach. $\endgroup$
    – Riemann
    Commented Jun 2 at 19:03