Zeration

[GFR]

This operation has no inverse, you can not even cancel it:
aob = cob as long as b>a,c

Indeed, you cannot cancel, under ... any situatiom.
aob = cob = b+1, if b>a,c
aob = cob --> a = c, if b < a,c
Moreover, let us see these examples:
3#2 = 27, 2#3 = 16; non-commutable;
3^2 = 9, 2^3 = 8; non-commutable;
3*2 = 6, 2*3 = 6 ; commutable;
3+2 = 5, 2+3 = 5 ; commutable;
3o2 = 4, 2o3 = 4 ; commutable.

See also the inverse operations:
slog(/3)27 = 2 ; srt(\2)27 = 3; slog(/2)16 = 3 ; srt(\3)16 = 2;
log(/3)9 = 2 ; rt(\2)9 = 3 ; log(/2)8 = 3 ; rt(\3)8 = 2;
6/3 = 2 ; 6/2 = 3;
- but ... attention, now ...
4ò3 = 2 ; 4ò2 = 3 (with "ò": inverse operator of "o"), and indeed also:
4ò2 = 2 (because 2o2 = 4 and 2o3 = 3o2 = 4).

In conclusion, we could say that:
4ò2 = {2,3}. Strange, but not unique, in fact:
rt(\2)4 = sqrt 4 = {-2,+2}.

It happens, in the best families!

To be consistent with the other hyperoperations we would surely demand that
aoa=a+2
ao(aoa)=a+3
ao(ao(aoa))=a+4

or generally ao(a+n)=a+n+1 in analogy to \( a^{^n a} = {^{n+1}}a \) and \( a\times a^n = a^{n+1} \) and \( a+a\times n = a\times (n+1) \)

But this leads to a contradiction for n=0:
ao(a+0)=a+1 as opposed to aoa=a+2

Well, as a matter of fact, for n=0, we have:
ao(a+0) = aoa = a+2 , but:
ao(a+n) = (a+n)+1 , for n>0, which gives:

ao(a+1) = (a+1)+1 = a+2 (again, but as ... successor of a+1))
ao(a+2) = (a+2)+1 = a+3
ao(a+3) = (a+3)+1 = a+4
..............................

And, of course:
ao(ao(ao(aoa))) = a+5

The definition of zeration has a discontinuous and multi-valued character, which we should be prepared to accept in the execution of such operation. In fact:

ao(a+0) = ao(a+1) = a+2

The result of zeration is given by the successor of the greatest of the operands, if they are different, or by the second successor of any of them, if they are equal.

Così è, .... se vi pare (Pirandello, ... I presume).

GFR