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Jun 29 at 14:37 comment added Hanul Jeon @C7X Yes. It should be plausible that every recursive ordinal is isomorphic to an $\mathsf{EA}$-provably-linear well-order, but I haven't checked it in detail. I believe the standard proof for the Kleene normal form theorem works over $\mathsf{RCA}_0$, and the resulting linear order might be $\mathsf{EA}$-provably linear order. However, I am unfamiliar with $\mathsf{EA}$ so I am unsure if it really works.
Jun 29 at 0:17 comment added C7X @HanulJeon Page 7 of "Reflection Ranks and Ordinal Analysis" states that the approach based on fixing a $\Delta_0$, well-ordered, $\mathsf{EA}$-provable linear order will also give an isomorphic ordinal notation of the $\langle\phi(x),\phi(x,y),p,n\rangle$ kind, provably in $\mathsf{EA}^+$. Was part of the difficulty in showing that every recursive ordinal is isomorphic to the former kind of ordinal notation?
Jun 24 at 7:45 comment added Hanul Jeon @Noah Do you mean Kleene's $\mathcal{O}$? I know that, but what is unsure for me is that it fits with the definition of an ordinal notation that Pakhomov-Walsh used. Their ordinal notation is a tuple $\langle \phi(x), \psi(x,y),p,n\rangle$ of two bounded formulas, an $\mathsf{EA}$-proof $p$ for $(\phi,\psi)$ gives a linear order, and $n$ denoting a member in the field of the linear order. It sounds believable that every recursive ordinal is isomorphic to an ordinal notation, but I don't immediately see how to prove it.
Jun 24 at 6:59 comment added Noah Schweber To be clear: yes, every computable ordinal is isomorphic to one given by a notation. See Sacks' book Higher recursion theory.
Jun 24 at 6:49 history answered Hanul Jeon CC BY-SA 4.0