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From what I have gathered online about these numbers, they say that Loader's Number is larger than TREE(3) or SSCG(3) or similar. The reasoning I have seen goes is that Loader's Number is the largest computable number and TREE(3) and SSCG(3) are ostensibly computable numbers. But how do we know that:

  • Loader's Number is larger than TREE(3) or SSCG(3) (to which we have only a weak lower bound of 187196 nested Ackermann functions) assuming they are computable?
  • TREE(3) or SSCG(3) are computable in the first place? As far as I can tell these are just the longest potential game one could play given a certain number of node colors or maximum nodes. How do we know that these are computable functions? How could we solve the numbers besides brute forcing every possible game which would make them noncomputable functions?
CPlus
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  • "Loader's number, TREE, SSCG, ..." : don't you think you could at least give some references for poor beotians like me ? – Jean Marie Apr 05 '23 at 18:47
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    @JeanMarie I added some wiki links. – CPlus Apr 05 '23 at 21:46
  • Since all games must be of length at most TREE(3), and there are finitely many three-colored trees of a given node count, there are finitely many games. This makes algorithmic brute force possible and not noncomputable – C7X Sep 11 '23 at 21:18
  • @C7X But then why is Rayo's number considered noncomputable? You can brute force every possible combination of 1 googol first order set theory symbols and see which ones give the biggest result. – CPlus Sep 11 '23 at 21:21
  • @user16217248 There is a computable predicate for checking if a TREE game is valid, but Rayo's function isn't computable because the Entscheidungsproblem has a negative answer: there is no computable predicate for determining if a given first-order formula is true. In particular there is no algorithm to determine if "φ defines the number N" is true, nor is there a way to throw out the φ that don't define any natural number because that would require deciding "φ defines a natural number at all". – C7X Sep 11 '23 at 21:24
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    There is a similar problem with the Busy Beaver function, where there is no way to brute-force all the n-state Turing machines even though there are comparatively few of them, because you'd have to throw out the non-halting ones in advance before trying to find their halting time - but that would require knowing whether a Turing machine halts in advance. – C7X Sep 11 '23 at 21:25

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More or less, the reasoning is that all both $\mathrm{TREE}(3)$ and $\mathrm{SSCG}(3)$ are about finite objects, thus you can enumerate all these objects. For $\mathrm{TREE}(3)$, you can test all possible sequences of length $n$ for the desired property, and if such a sequence exists, determine that $\mathrm{TREE}(3) \ge n$. This makes them computable. Since the calculus of constructions is quite strong, the tree theorem (hopefully, I doubt anyone has checked) holds within it, and thus the $\mathrm{TREE}$ function is definable within, say, $10^9$ symbols. Since Loader's number is defined to be the largest one definable in the calculus of constructions in so-and-so ($> 10^9$) many symbols, so is $\mathrm{TREE}(3)$ and more generally, $\mathrm{TREE}(n)$ for some very large $n$.

univalence
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