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It is well understood that there can exist undecidable statements for which it is completely impossible to prove these statements are undecidable; this raises an interesting question about if there are any famous open problems which may or may not have this "undecidable undecidable" property.

Question: If the yes/no question is P=NP true? is undecidable; and if there is no proof of its undecidability; then how can mainstream mathematics research ever make progress in such a situation?

My understanding is that if you simply add a new axiom "P=NP is true" and if we cannot ever prove that P=NP is either false or undecidable, then we will have no way to verify that this new axiom does not fundamentally break the internal consistency of logic; I can not see any way out of this "paradox" in such a situation.

Motivation: My understanding is that much work has already been done on trying to show P=NP is undecidable, it seems to me extremely plausible that given the nature of P=NP itself, that a positive resolution ("P=NP is true") would fundamentally change the questions which are decidable and undecidable (please let me know if this does not seem plausible to you).
This forms the basis for my suspicious that P=NP may be one of those undecidable undecidable statements.

References: Are there statements that are undecidable but not provably undecidable, Is there a statement whose undecidability is undecidable?, and Would undecidability of $P = NP$ imply its truth/falsity? and is motivated by my old question What is the efficiency of the algorithm which solves this word problem?

Prem
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Matt Calhoun
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    In practice, most mathematicians and computer scientists adopt a pragmatic stance toward undecidable questions. Even if a statement like P=NP were proven to be undecidable, meaningful research would not cease. Rather, progress would shift toward examining the consequences of adopting different axioms or working within distinct models. The key is that useful and non-trivial results can often be derived conditionally: for instance, we can investigate what follows if we assume P=NP and separately explore the consequences under the assumption P $\neq$ NP. – Antony Theo. Oct 20 '24 at 16:22
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    Moreover , what leads you to believe that P=NP is a more plausible statement than P$\neq$ NP? From my experience (outside of mathematics research) it would seem infinitely more likely that P$\neq$NP . Take for instance the fundamental dependance of a (cryptographic) system like RSA on that assumption. In my eyes it would be foolish for someone to base a whole economic system on a "50-50 statement" or even a statement's whos undecidability might not be provable inside ZFC. I assume you've thought of this problem more than I have so I would really want to know your answer to that. :) – Antony Theo. Oct 20 '24 at 16:39
  • @AntonyTheo. You asked " what leads you to believe that P=NP is a more plausible statement than P≠NP?", great question. With high confidence I believe I have already written the algorithm which solves BSP in N steps[1] as linked in the references section of my question. After consulting with many computer scientists and professional mathematicians, I concluded the "self replication axiom" is logically equivalent to the assertion "P=NP is true" (cont...) [1] https://math.stackexchange.com/questions/3605352/what-is-the-efficiency-of-the-algorithm-which-solves-this-word-problem – Matt Calhoun Oct 20 '24 at 17:15
  • (...cont) I also think its clear that algorithm cannot be implemented on a Turing Machine, and the "self replication axiom" I mentioned is the idea of adding an axiom to the theory of computation that "Turing Machines can self replicate at every step". I am stuck on figuring out if its a logical contradiction to add this axiom given what I described in my question. This work is the result of my attempts to use computer virus methods for a good cause (I am 100% white hat) and the work is strongly motivated by cell self replication in the human nervous system. – Matt Calhoun Oct 20 '24 at 17:17
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    Proving that your self-replication axiom implies $P=NP$ seems easy enough. It changes changes the fundamental concept of a computation into a massively parallel model. I think many would call this "cheating" because $P=NP$ was constructed with Turing's sequential model in mind, and it bears mentioning that a parallel system still performs the same amount of "work" as it merely shifts some of the burden of time complexity into resource complexity. On the other hand, proving that $P=NP$ necessarily implies your self-replication axiom seems like it would be much harder to do... and impressive. – RyRy the Fly Guy Oct 20 '24 at 18:50
  • @RyRytheFlyGuy your comment is extremely intriguing. The "self-replication axiom" is apparently logically consistent and based on a real world model. If it was possible to prove that "P=NP is true" IMPLIES "self-replication axiom" and vice versa, then this would mean that, by adding the self-replication axiom to the theory of computation we have arrived at a positive resolution of the P=NP question, resolving the problem in my OP. This seems to me to be a rather novel and promising new strategy to solve this open problem. Have I correctly understood your comment? – Matt Calhoun Oct 21 '24 at 13:37
  • (quick remark: its not possible to gain exponential computational power with anything less than doubling the number of Turing machines at every step, since you can't "cut a Turing Machine in half", and furthermore, there is no need for tripling or quadrupling at each step since doubling is sufficient to achieve exponential computational power over time. It intuitively "makes sense" to me that to solve an NP-complete problem in P time, the only way is with exponentially increasing parallel computation. Seems imho like there is something missing in the argument to make this completely rigorous – Matt Calhoun Oct 21 '24 at 13:40
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    Yes, i think you get it. you asserted that $P=NP$ and your self-replication axiom were logically equivalent. To prove this, you would need to show one implies the other (and vice versa). Such a proof would effectively solve the $P=NP$ problem because logical equivalence would imply $P \neq NP$ without self-replication. However, proving $P=NP \Rightarrow$ self-replication axiom seems like it would be challenging. You would have to couple your intuition with a rigorous proof that a polynomial time algorithm beyond self-replication does not exist. that lies at the very crux of the $P=NP$ problem. – RyRy the Fly Guy Oct 21 '24 at 15:58
  • @RyRytheFlyGuy if you would like to write up your comment as an answer I will accept it. To be clear when I asserted equivalence, it was a guess and I have no proof. After carefully considering things, I have come to the conclusion this is above my "mathematical pay grade". Best I have is use the constraint of maximum efficiency to try and prove uniqueness of the self replication algorithm and hence show no algorithm can solve BSP in less steps and prove the implication via uniqueness. I am at a loss of how to proceed and hope others will be interested in this strategy. Any advice? Thx – Matt Calhoun Oct 22 '24 at 18:41

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