So I checked up on it, and no, it is not the case. Also, I have some bad news. If we assume the Schroder functions (accordingly, the Abel functions) are iterated exponentials, then that means that the associated hyperoperation hierarchy would have the recursion rule:
\( HE_n(x, y) := \exp(HE_{n-1}(\log(x), y)) \)
which (if n=2 is multiplication) requires that (n=1) is \( HE_1(x, y) = x + \log(y) \), which is not addition, so this scheme does not form a hyperoperation hierarchy. However, if I might draw your attention to the Tetration Ref. section 2.3.4, the recursion rule:
\( HR_n(x, y) := \exp(HR_{n-1}(\log(x), \exp^{n-3}(y))) \)
has many of the same characteristics, and satisfies addition (n=1), mul (n=2), and powers (n=3). The only difference is in the heighest exponent.
\( HR_n(x, y) = HE_n(x, \exp^{\circ\left({n-2 \atop 2}\right)}(y)) \)
\( HE_n(x, y) := \exp(HE_{n-1}(\log(x), y)) \)
which (if n=2 is multiplication) requires that (n=1) is \( HE_1(x, y) = x + \log(y) \), which is not addition, so this scheme does not form a hyperoperation hierarchy. However, if I might draw your attention to the Tetration Ref. section 2.3.4, the recursion rule:
\( HR_n(x, y) := \exp(HR_{n-1}(\log(x), \exp^{n-3}(y))) \)
has many of the same characteristics, and satisfies addition (n=1), mul (n=2), and powers (n=3). The only difference is in the heighest exponent.
\( HR_n(x, y) = HE_n(x, \exp^{\circ\left({n-2 \atop 2}\right)}(y)) \)
