Laws and Orders

[bo198214]

In other words, I (& al.) started asking to ourselves:
"Do we have hyperoperation hierarchy definable by: a[s]<r>a = a[s+1](r+1) ?"

And the answer is "No!" as I already showed (if we allow s=0). We are a group of mathematicians, it does not matter if someone has this opinion or that opinion if the facts are different. So again:

We start with the law
a[s]<r>a = a[s+1](r+1)

then I put r=0, as a[s]<r> is the r times iteration of f(x)=a[s]x, it follows that a[s]<0> is the identity, a[s]<0>x=x, so:

a = a[s]<0>a = a[s+1]1

now we set s=0:

a = a[1]1

as we agree that in our operation sequence x[1]y=x+y, the above is a contradiction.