Functional super-iteration and hierarchy of functional hyper-iterations

[Base-Acid Tetration]

\( \operatorname{It}_k: \lbrace f: \mathbb{R} \rightarrow \mathbb{R} \rbrace \times \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace \rightarrow \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace, \) where It_0 is function composition (g It_0 f = g(f)). All of the operators has the property \( f \operatorname{It}_n x+1 = f \operatorname{It}_{n-1} (f \operatorname{It}_n x). \)

It is a bit sloppy and misunderstandable to write \( x+1 \) or \( x \) in the second argument. As one would associate the function \( x\mapsto x+1 \), but actually you mean the constant function:
\( f \operatorname{It}_n (x\mapsto m+1) = f \operatorname{It}_{n-1} (f \operatorname{It}_n (x\mapsto m)). \)
To make it a definition that makes at least on the natural numbers sense, we need to add the induction start:
\( f \operatorname{It}_n (x\mapsto 1) = f \)

The question then however is how do we know what \( \operatorname{It}_n \) for a non-constant function in the second argument is?

I think to properly define the hierarchy we need an additional operator
\( I_n: (\mathbb{R}\to\mathbb{R})\times \mathbb{R}\to(\mathbb{R}\to\mathbb{R}) \), \( (f,m)\mapsto I_n^m f \)

such that \( (f \operatorname{It}_n g)(x) = (I_n^{f(x)} g)(x) \).
with
\( I_0^m f = f \)
\( I_n^1 f =f \) and \( I_{n}^{m+1} f = f \operatorname{It}_{n-1} (I_n^m f). \)

My brain is flickering, I hope I got everything right.

For the function x -> x+1, you simply hyper-n-iterate the function f x0+1 times at x=x0, (eg. at x=2 you iterate 3 times, and at x=3.5 you iterate 4.5 times etc.) So I don't think that a non-constant function in the second argument would behave differently from a constant function in this respect.

Also \( I_0^m f = f \) would not be true, because at n=0, \( I^m_0 f = f \operatorname{It}_0 m = f(m), \mbox{ not }f. \) Use of the \( I \) operator would make the notation more compact, however