Laws and Orders

[GFR]

Concerning the DL, I prefer my "implication", i.e.:
IF a[s]x = x, THEN x = a[s+1]oo

Gianfranco, that is not true!
Take \( a=\sqrt{2} \) and \( x=4 \) then surely \( a[3]x=x \). But \( 4\neq 2 = a[4]\infty \). You can not bend mathematics to suit your wishes.

I am not bending anything and I don't have any particular wish to satisfy, in this context. Could you please explain to me why, from:

a[3]x = x, we should not have, by simple substitutions:
x = a[3](a[3](a[3](a[3]x))), and therefore:
x = a[4]oo ???

Then, why I cannot say that:
a[3]x = x ----> x = a[4]oo ??

Concerning your classical example, by putting: a = sqrt(2), we indeed have:
(sqrt(2))^x = x ----> x = (sqrt(2))#oo = h = plog(-ln x) / (-ln x).

We know that, for 1 < a < e^(1/e), h has two real solutions, one for plog(z) = W/-1 (z) and one for plog(z) = W/0 (z). These two solutions can be easily calculated by Mathematica and they are:
h/inf = 2 and h/sup = 4. In my non-orthodox notation, I would put:
h = {2,4}.

Value h/inf = 2 is an asymptotical value of x = a#n, "reached" when n->oo. Value h/sup = 4 is an old mystery that has nothing to do with my mathematical weaknesses. It is not "reachable", but you must accept that the two values satisfy the selfroot condition:

selfrt(4) = selfrt(2) = sqrt(2).

You see, Henryk, you are certainly right to agitate the flag of mathematical orthodoxy. But,alway remaining calm and cool !!!

GFR