Does anything precede zeration in the hyperoperators

Question

"Hyperoperator" family of operators is defined recursively like so: $$ a\langle{0}\rangle{b} = a+b\\ a\langle{1}\rangle{b} = ab = a+a+a+a... \; \text{(w/ b a's)}\\ a\langle{2}\rangle{b} = a^{b}= aaaa... \; \text{(w/ b a's)}\\ ... $$ We can also extend the hyperoperators to "-1st operator" using the method described in this old post: $$ a\langle{-1}\rangle{b} = a@b = \{a\neq{b}:\text{max}(a,b)+1,a=b:a+2\}\\ $$ This operator (called zeration) satisfies the property that $a+b = a\langle-1\rangle{a\langle-1\rangle}a...\text{(w/ b a's)}\\$.
My question is: Can we extend hyperoperators below zeration indefinitely while preserving the property I described above?

Answer 1

Your "zeration" definition is unusual. It is usualally defined as increment by one operation:

$$a\langle0\rangle b=b+1$$ Its repeating $b\langle0\rangle b$ for $c$ times is addition $c+b$. Increment is not commutative and does not depend on the first argument.

Moreover, Since $\Gamma(1)=\Gamma(2)=1$, addition and multiplication are both commutative, but since $\Gamma(0)$ goes to infinity, this means that zeration does not depend on the first argument at all. And this is true for any hyperoperation of negative integer order as well. Because the Gamma function has poles there.