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First of all i am not looking for very accurate lower bound. It would be enough to me the following :

Prove that $TREE(3)>f_{\omega +2}(3)$ (here that function $f$ comes from fast growing hierarchy)

Generally i don't know how to solve it. I have a little idea. It is to show that there is a injective function between two sets. One of them would be the set of elements of the longest sequence with 3. labels second set would have cardinality equal or greater to the $f_{\omega +2}(3)$. But i don't know how to find that function

mkultra
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  • Closely related : https://math.stackexchange.com/questions/3449265/comparing-big-numbers – Peter Nov 26 '19 at 00:56
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    At https://mathoverflow.net/a/95588/20307 there is a proof that $$\text{TREE(3)} > H_{\vartheta (\Omega^{\omega}, 0)}(n(4))$$ where $H_{\alpha}$ is a version of the Hardy hierarchy and $n()$ is Friedman's block-subsequence function, so the RHS is clearly greater than $f_{\omega +2}(3)$. (For all $n$, it can be proved that $H_{\omega^{\alpha}}(n) > f_{\alpha}(n)$, where $f_\alpha$ is the Fast Growing Hierarchy.) – r.e.s. Nov 27 '19 at 21:04

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