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The primitive fast fixation notation has X rules of operation (default to base 10):
Rule 1 (with empty curly brackets or with just 0 in it, or if the base is 0):
E[n]{}A = E[n]{0}A = n^A
E{}A = E{0}A = E[10]{}A = E[10]{0}A = 10^A
E[m]{Q}E[n]{X}0 = E[n]{Q}1, where X can be anything inside the curly brackets, and {Q} can be any sequence of the function.
Rule 2 (with natural numbers inside {} or being appended at the end inside {}; {Q} can be any sequence of the function):
E[m]{Q}E[n]{X + 1}A = E[m]{Q}E[n]{X}{X}{X}…{X}{X}{X}n with A copies of {X}, and X can be any sequences nested inside it
Rule 3 (where the inner bracket contains {} or {0}, {S} indicates the rest of the expression inside the operating curly bracket):
E[m]{Q}E[n]{{S}{}}A = E[m]{Q}E[n]{{S}{0}}A = E[m]{Q}E[n]{{S} + 1}n
Rule 4 (using the “asterisk” multiplication to copy the same brackets; plus signs are optional between two brackets):
E[m]{Q}E[n]{{S}{X}}A = E[m]{Q}E[n]{{S}+{X}}n
E[m]{Q}{{S}{}*F}A = E[m]{Q}E[n]{{S}{0}*F}A = E[m]{Q}E[n]{{S} + F}n
E[m]{Q}E[n]{{S}{X}*F}A = E[m]{Q}E[n]{{S}{X}{X}{X}…{X}{X}{X}}A with F copies of {X}
We can also use the “dot” (.) operator as an alternative to the “asterisk” (*) operator for bracket multiplication as well.
Rule 5 (the rightmost part of the sequence is not “added” by a natural number):
E[m]{Q}E[n]{{S}{1}}A = E[m]{Q}E[n]{{S} + A}n
E[m]{Q}E[n]{{S}{Y + 1}}A = E[m]{Q}E[n]{{S}{Y}*A}n
E[m]{Q}E[n]{{S}{{…{{{S}{Y + 1}}}…}}}A = E[m]{Q}E[n]{{S}{{…{{{S}{Y}*A}}…}}}n
In particular, you must consult the rules for the innermost bracket level first until the bracket sequence is added by a natural number at the end before applying the rules for the particular sequence one bracket level lower.
It is important to note that the brackets are operated from left to right, and each variable in the rules must be a non-negative integer.
We can also have:
E[m]{Q}EN[n]{X}A = E[m]{Q}E[n]{X}{X}{X}…{X}{X}{X}A with N copies of {X}.
As well as the specific shorthands like:
E[n]A = E[n]{0}A
F[n]A = E[n]{1}A
G[n]A = E[n]{2}A
H[n]A = E[n]{3}A
I[n]A = E[n]{{1}}n = I[n]{n}n