interpolating the hyper operators

[MphLee]

I've found an approach to showing that it satisfies the recursive identity, which is: \( x,y,z \in \mathbb{N}\,\,s_0 \in \mathbb{R} \)

\( x [s_0 + 1] (y-1) = z \)
\( x [s_0] z = x [s_0+1] y \)

The approach is a little tricky and I don't know how to explain it so clearly.

Isn't this condition equivalent to you old attempt to limit the recursion to a subset of the real?
In fact your condition for the recursion

\( a)\,x,y,z \in \mathbb{N}\,\,s_0 \in \mathbb{R} \)
\( b)\,x [s_0 + 1] (y-1) = z \)
\( c)\,x [s_0] z = x [s_0+1] y \)

translatable in

\( i)\, s_0 + 1 \in \mathbb{I}_{x,y-1} \)
\( ii)\,x [s_0] z = x [s_0+1] y \)

That bring me to ask you why you abandoned the study on the \( \mathbb{I} \) sets.

Anyways, even if you limit the recursion to ALL the \( \mathbb{I} \) sets you can work only on a countable subset of reals:
in other words most of the reals (\( 2^{\aleph_0} \) ) are out of the recursion.

PS: If was not clear, what I mean is:
if for your interpolation you can show that:

if \( a) \) and \( b) \) hold than \( c) \) holds (recursion)

then is the same that you proved for you interpolation this:

if \( i) \) holds than \( ii) \) holds

as you can see, if \( i) \) holds then your proof is valid only for a subset of reals.