<< ...ANALYSIS IN PROGRESS... >>
"The two-entry notation level grows much faster and eventually dominates the limit of the extended Buchholz’s function…"
My second level of the binary bisector notation uses curly brackets {…} for the entire sequence or the sequence ordinal, commas (,) for “hyper-level” sequences (used to diagonalize the recursion of the previous entries). In order to proceed the legacy pair sequence notation level, you need to find the particular recursion of the respective sequence ordinals.
First off, we define the earliest level of the legacy pair sequence notation where the fundamental sequence of {{0, 1}, 0} to be: A{{0, 1}, 0} = A{{0, 1}, 0}[A] where {{0, 1}, 0}[1] = {0, 0} = 1, {{0, 1}, 0}[2] = {{0, 0}, 0} = {1}, {{0, 1}, 0}[3] = {{{0, 0}, 0}, 0} = {{1}}, {{0, 1}, 0}[4] = {{{{0, 0}, 0}, 0}, 0} = {{{1}}}, etc. So, {{0, 1}, 0}[1] = {0, 0} {{0, 1}, 0}[N + 1] = {{{0, 1}, 0}[N], 0}.
The similar pattern go as follows:
{{0, 1}, 0} => 0, {0, 0} = 1, {{0, 0}, 0} = {1}, ..., {{0, 1}[N], 0} (FGH level ε0)
{{0, 1}{0, 1}, 0} => 0, {{0, 1}, 0}, {{0, 1}{{0, 1}, 0}, 0}, ..., {{0, 1}{{0, 1}{0, 1}, 0}[N], 0} (FGH level ε1)
{{0, 1}{0, 1}{0, 1}, 0} => 0, {{0, 1}{0, 1}, 0}, {{0, 1}{0, 1}{{0, 1}{0, 1}, 0}, 0}, ..., {{0, 1}{0, 1}{{0, 1}{0, 1}{0, 1}, 0}[N], 0} (FGH level ε2)
{{{{0, 1}, 0}, 1}, 0} => {{0, 1}, 0}, {{1, 1}, 0}, {{{1, 0}, 1}, 0}, ..., {{{{{0, 1}, 0}[N], 0}, 1}, 0} (FGH level εε0)
{{{0, 1}, 1}, 0} => 0, {{0, 1}, 0}, {{{{0, 1}, 0}, 1}, 0}, ..., {{{{{0, 1}, 1}, 0}[N], 1}, 0} (FGH level ζ0)
{{{0, 1}{0, 1}, 1}, 0} => 0, {{{0, 1}, 1}, 0}, {{{0, 1}{{{0, 1}{0, 1}, 1}, 0}, 1}, 0}, ..., {{{0, 1}{{{0, 1}{0, 1}, 1}, 0}[N], 1}, 0} (FGH level η0)
{{{{0, 1}, 1}, 1}, 0} => 0, {{{0, 1}, 1}, 0}, {{{{{{0, 1}, 1}, 0}, 1}, 1}, 0}, ..., {{{{{{{0, 1}, 1}, 1}, 0}[N], 1}, 1}, 0} (FGH level Γ0 (Feferman-Schütte ordinal))
{{{{{0, 1}, 1}, 1}, 1}, 0} => 0, {{{{0, 1}, 1}, 1}, 0}, {{{{{{{{0, 1}, 1}, 1}, 0}, 1}, 1}, 1}, 0}, ..., {{{{{{{{{0, 1}, 1}, 1}, 1}, 0}[N], 1}, 1}, 1}, 0} (FGH level ψ0(Ω^Ω^Ω) (Large Veblen ordinal))
{{0, 2}, 0} => 0, {{0, 1}, 0}, {{{0, 1}, 1}, 0}, ..., {{{{0, 2}, 1}[N], 1}, 0} (FGH level ψ0(Ω_2) (Bachmann-Howard ordinal))
{{0, 2}, 1} => 0, {0, 1}, {{0, 1}, 1}, ..., {{{0, 2}, 1}[N], 1} (FGH uncountable ordinal ψ1(Ω_2))
{{{0, 2}, 2}, 1} => 0, {{0, 2}, 1}, {{{{0, 2}, 1}, 2}, 1}, ..., {{{{{0, 2}, 2}, 1}[N], 2}, 1} (FGH uncountable ordinal ψ1(Ω_2^2))
{{0, 3}, 0} => 0, {{0, 2}, 0}, {{{0, 2}, 2}, 0}, ..., {{{{0, 3}, 2}[N], 2}, 0} (FGH uncountable ordinal ψ0(Ω_3))
{{0, 3}, 1} => 0, {{0, 2}, 1}, {{{0, 2}, 2}, 1}, ..., {{{0, 2}, 2}[N], 1} (FGH uncountable ordinal ψ1(Ω_3))
{{0, 3}, 2} => 0, {0, 2}, {{0, 2}, 2}, ..., {{{0, 3}, 2}[N], 1} (FGH uncountable ordinal ψ2(Ω_3))
{{0, {0, 1}}, 0} => 0, {{0, 1}, 0}, {{0, {{0, 1}, 0}}, 0}, ..., {{0, {{0, {0, 1}}, 0}[N]}, 0} (FGH level ψ0(Ω_Ω))
{{0, {0, 2}}, 0} => 0, {{0, {0, 1}}, 0}, {{0, {{0, {0, 1}}, 1}}, 0}, ..., {{0, {{0, {0, 2}}, 1}[N]}, 0} (FGH level ψ0(Ω_Ω_2))
{{0, {0, {0, 1}}}, 0} => 0, {{0, {0, 1}}, 0}, {{0, {0, {{0, {0, 1}}, 0}}}, 0}, ..., {{0, {0, {{0, {0, {0, 1}}}, 0}[N]}}, 0} (FGH level ψ0(Ω_Ω_Ω))
It is important to remember that {{A, X}, 0}[0] must be 0 in the innermost sequence, ignoring the outermost {A, 0}. So that {{{{0, 1}, 0}, 1}, 0}[0] is not 0, but instead {{0, 1}, 0}.
You can also simplify the two-entry {X, 0} to just the one-entry {X}. This is the most effective on the outer level of the sequence.
The limit of the legacy pair sequence binary bisector notation is the countable limit ψ0(Ω_Ω_Ω_…) of the extended Buchholz’s function, where “Ω_Ω_Ω_…” denotes the least omega fixed point. This ordinal is also known as the Extended Buchholz’s ordinal (EBO). It pretty much simulates the extended Buchholz's function.
And here, we define the formal rules for the legacy pair sequence notation:
Rule 1 (with empty curly brackets or with just 0 in it, or if the base is 0 or 1):
A{} = A{0} = A{0, 0} = 2^A
0{X, Y}{Q} = 1{Q}, where X can be anything inside the first entry of the bracket, no matter the value of the second entry Y is, and {Q} can be any sequence of the function – First degeneration rule.
1{X, Y}{Q} = 2{Q}, where X can be anything inside the first entry of the bracket, no matter the value of the second entry Y is, and {Q} can be any sequence of the function – Second degeneration rule.
A{X} = A{X, 0} where X can be anything inside the first entry – Level simplification rule.
Rule 2 (with natural numbers inside {} of the first entry or being appended at the end inside {}; {Q} can be any sequence of the function):
A{X + 1, Y}{Q} = A{X, Y}{X, Y}{X, Y}…{X, Y}{X, Y}{X, Y}{Q} with A copies of {X}, and X can be any sequences inside the first entry, no matter the value of the second entry Y is.
Rule 3 (where the inner bracket contains {} or {0}, whether the inner bracket is in the first or the second entry, {S} indicates the rest of the expression inside the operating curly bracket):
A{{S}{}}{Q} = A{{S}{0}}{Q} = A{{S}{0, 0}}{Q} = A{{S} + 1}{Q}
Rule 4 (using the “dot” multiplication to copy the same brackets; plus signs are optional between two brackets):
A{{S}{X, Y}}{Q} = A{{S}+{X, Y}}{Q}
A{{S}{}.F}{Q} = A{{S}{0}.F}{Q} = A{{S}{0, 0}.F}{Q} = A{{S} + F}{Q}
A{{S}{X, Y}.F}{Q} = A{{S}{X, Y}{X, Y}{X, Y}…{X, Y}{X, Y}{X, Y}}{Q} with F copies of {X}
Rule 5 (the rightmost part of the active inner sequence, not the base sequence, is added by a natural number):
A{{S}{1}}{Q} = A{{S}{1, 0}}{Q} = A{{S} + A}{Q}
A{{S}{Y + 1, Z}}{Q} = A{{S}{Y, Z}.A}{Q}
A{{S}{{…{{{S}{Y + 1, Z}}}…}}}{Q} = A{{S}{{…{{{S}{Y, Z}.A}}…}}}{Q}
Like in the primitive sequence notation level, 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. Just like before, 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.
Please note that the basic rules work anywhere whether the rule applies either in the first or the second entry. You don’t have to worry about the basic rule applies exclusively in the first entry as in the primitive sequence notation level. Many experts would conflate with those rules and make some mistakes. For instance, 4{{0, {{1}}}} is actually equal to 4{{0, {4}}}. Those previous rules are not limited to the first entry of the legacy pair sequence notation.
We can also have:
A{X, Y}^N{Q} = A{X, Y}{X, Y}{X, Y}…{X, Y}{X, Y}{X, Y}{Q} with N copies of {X}, and X can be any sequences inside it, no matter the value of the second entry Y is.
Rule 6 (earliest diagonalization rule):
A{{0, 1}, 0} = A{{0, 1}, 0}[A] = A{{{0, 1}, 0}[A - 1], 0} (where {{0, 1}, 0}[0] = 0)
A{{S}{{0, 1}, 0}} = A{{S}{{0, 1}, 0}}[A] = A{{S}{{0, 1}, 0}[A]} = A{{S}{{0, 1}, 0}[A - 1]} (where {{0, 1}, 0}[0] = 0)
Rule 7 (setting up the secondary recursion or adjust the recursion of the particular bracket by jumping to the innermost one, ignoring the higher sequence {S} and the second entry P):
A{X, Y} = A{X, Y}[A]
A{{S}{X, Y}, P} = A{{S}{X, Y}, P}[A] = A{{S}{X, Y}[A], P}
A{0, {S}{X, Y}} = A{0, {S}{X, Y}}[A] = A{{S}{X, Y}[A]}
Rule 8 (if any of the respective bracket is met with the second entry’s diagonalization, where {S} can be any preceding brackets, and there is a successor next to the second entry’s bracket Y, no matter what the value of the P is):
For {0, 1}:
{{S}{0, 1}, P}[0] = 0
{{S}{0, 1}, P}[1] = {{S}, P}
{{S}{0, 1}, P}[N + 1] = {{S}{{{S}{0, 1}, P}[N], 0}, P}
For {0, Y} where Y > 1:
{{S}{0, Y + 1}, P}[0] = 0
{{S}{0, Y + 1}, P}[1] = {{S}, P}
{{S}{0, Y + 1}, P}[N + 1] = {{S}{{{S}{0, Y + 1}, P}[N], Y}, P}
For the second entry:
{0, {S}{0, 1}}[0] = 0
{0, {S}{0, 1}}[1] = {0, {S}}
{0, {S}{0, 1}}[N + 1] = {0, {S}{{0, {S}{0, 1}}[N], 0}}
{0, {S}{0, Y + 1}}[0] = 0
{0, {S}{0, Y + 1}}[1] = {0, {S}}
{0, {S}{0, Y + 1}}[N + 1] = {0, {S}{{0, {S}{0, Y + 1}}[N], Y}}
Rule 9 (if there is just a lone legacy pair sequence bracket with the nonzero second entry (Y must be greater than 1)):
{0, 1}[0] = 0
{0, 1}[1] = {0, 0} = 1
{0, 1}[N + 1] = {{0, 1}[N], 0}
{0, Y + 1}[0] = 0
{0, Y + 1}[1] = {0, Y}
{0, Y + 1}[N + 1] = {{0, Y + 1}[N], Y}
It is important to note that if the second entry is zero, you must neglect the sequences outside of the {X, 0} and have to fixate on the specific, innermost bracket with the second entry is zero before applying the others.
Moreover, in case of the rule 7 or rule 8 that apply on {{S}{0, Y}[N], 0} where N = 0 ({{S}{0, Y}[0], 0}), we have to clear the sequence and return 0 immediately, not retaining the {0, Y}. So:
Rule 10 (clear the sequence and return 0 if the diagonalization rule is met with {X, 0}):
{{{S}{0, 1}, P}[0], 0} = {{{S}{0, 1}, P}, 0}[0] = 0
{{{S}{0, Y + 1}, P}[0], 0} = {{{S}{0, Y + 1}, P}, 0}[0] = 0
{{0, {S}{0, Y + 1}}[0], 0} = {{0, {S}{0, Y + 1}}, 0}[0] = 0
4{{0, 1}{0, 1}}
= 4{{0, 1}{0, 1}}[4]
= 4{{0, 1}{0, 1}[4]}
= 4{{0, 1}{{0, 1}{0, 1}[3]}}
= 4{{0, 1}{{0, 1}{{0, 1}{0, 1}[2]}}}
= 4{{0, 1}{{0, 1}{{0, 1}{{0, 1}{0, 1}[1]}}}}
= 4{{0, 1}{{0, 1}{{0, 1}{{0, 1}}}}}
4{{{{0, 1}, 1}, 1}}
= 4{{{{0, 1}, 1}, 1}}[4]
= 4{{{{0, 1}, 1}, 1}[4]}
= 4{{{{{{{0, 1}, 1}, 1}[3]}, 1}, 1}}
= 4{{{{{{{{{{0, 1}, 1}, 1}[2]}, 1}, 1}}, 1}, 1}}
= 4{{{{{{{{{{{{{0, 1}, 1}, 1}[1]}, 1}, 1}}, 1}, 1}}, 1}, 1}}
= 4{{{{{{{{{{{{0, 1}, 1}}, 1}, 1}}, 1}, 1}}, 1}, 1}}
4{{0, 3}}
= 4{{0, 3}}[4]
= 4{{0, 3}[4]}
= 4{{{0, 3}[3], 2}}
= 4{{{{0, 3}[2], 2}, 2}}
= 4{{{{{0, 3}[1], 2}, 2}, 2}}
= 4{{{{{0, 2}, 2}, 2}, 2}}
4{{0, {1} + 1}{0, 2}}
= 4{{0, {1} + 1}{0, 2}}[4]
= 4{{0, {1} + 1}{0, 2}[4]}
= 4{{0, {1} + 1}{{0, {1} + 1}{0, 2}[3], 1}}
= 4{{0, {1} + 1}{{0, {1} + 1}{{0, {1} + 1}{0, 2}[2], 1}, 1}}
= 4{{0, {1} + 1}{{0, {1} + 1}{{0, {1} + 1}{{0, {1} + 1}{0, 2}[1], 1}, 1}, 1}}
= 4{{0, {1} + 1}{{0, {1} + 1}{{0, {1} + 1}{{0, {1} + 1}{0, 1}, 1}, 1}, 1}}
4{{0, {0, 2}}}
= 4{{0, {0, 2}}}[4]
= 4{{0, {0, 2}}[4]}
= 4{{0, {{0, {0, 2}}[3], 1}}}
= 4{{0, {{0, {{0, {0, 2}}[2], 1}}, 1}}}
= 4{{0, {{0, {{0, {{0, {0, 2}}[1], 1}}, 1}}, 1}}}
= 4{{0, {{0, {{0, {{0, {0, 1}}, 1}}, 1}}, 1}}}
Using the extended Buchholz’s function for analysis:
<< ...WORK IN PROGRESS... >>
The {a, b} bracket corresponds to the ψ_{b}(a) argument in the extended Buchholz's function, and the adjacent sequences correspond to the "+" operator in the extended Buchholz's function. {a} is a shorthand for {a, 0}.
First off, let's analyze the very early stages of the legacy pair sequence notation level, from {{0, 1}} to {{0, 1}{0, 1}} (FGH level ε0 to ε1).
{{0, 1}} has level ε0 (ψ0(Ω) in the extended Buchholz's function - this is straightforward.)
{{0, 1} + 1} has level ε0 + 1
{{0, 1} + 2} has level ε0 + 2
{{0, 1} + 3} has level ε0 + 3
{{0, 1} + A} has level ε0 + α
{{0, 1}{1}} has level ε0 + ω
{{0, 1}{1} + 1} has level ε0 + ω + 1
{{0, 1}{1}{1}} has level ε0 + ω2
{{0, 1}{1}{1}{1}} has level ε0 + ω3
{{0, 1}{2}} has level ε0 + ω^2
{{0, 1}{3}} has level ε0 + ω^3
{{0, 1}{{1}}} has level ε0 + ω^ω
{{0, 1}{{2}}} has level ε0 + ω^ω
{{0, 1}{{{1}}}} has level ε0 + ω^ω^ω
{{0, 1}{{{{1}}}}} has level ε0 + ω^ω^ω^ω
{{0, 1}{{{{{1}}}}}} has level ε0 + ω^ω^ω^ω^ω
{{0, 1}{{0, 1}}} has level ε0·2
So why does the {{0, 1}{{0, 1}}} sequence have the FGH ordinal level of ε0·2 instead of ε0^2? The solution is rather straightforward. We establish the fundamental sequence of the earliest part of the legacy pair sequence system for {{0, 1}}, using all the previous rules from the primitive sequence system. Not only that, it is odd that the equalities of ε0^2 = ψ0(Ω + ψ0(Ω)) versus ε0·2 = ψ0(Ω) + ψ0(Ω) in the extended Buchholz's function rather uses the properties of the Veblen hierarchy, that is ε0^2 = ω^(ε0·2). So the ordinal level ε0^2 must be used for the the sequence {{0, 1}{{0, 1}{{0, 1}}}} with three {0, 1}'s instead of just two.
The same rationale also holds for sequences from {{0, 1}{{0, 1}}} to {{0, 1}{0, 1}} as well. Yes, we can have:
{{0, 1}{{0, 1}} + 1} has level ε0·2 + 1
{{0, 1}{{0, 1}}{1}} has level ε0·2 + ω
{{0, 1}{{0, 1}}{{0, 1}}} has level ε0·3
{{0, 1}{{0, 1}}{{0, 1}}{{0, 1}}} has level ε0·4
{{0, 1}{{0, 1} + 1}} has level ε0·ω = ω^(ε0 + 1)
{{0, 1}{{0, 1} + 1} + 1} has level ε0·ω + 1 = ω^(ε0 + 1) + 1
{{0, 1}{{0, 1} + 1}{1}} has level ε0·ω + ω = ω^(ε0 + 1) + ω
{{0, 1}{{0, 1} + 1}{{0, 1}}} has level ε0·ω + ε0 = ω^(ε0 + 1) + ε0
{{0, 1}{{0, 1} + 1}{{0, 1}}{{0, 1}}} has level ε0·ω + ε0·2 = ω^(ε0 + 1) + ε0·2
{{0, 1}{{0, 1} + 1}{{0, 1} + 1}} has level ε0·ω2 = ω^(ε0 + 1)·2
{{0, 1}{{0, 1} + 1}{{0, 1} + 1}{{0, 1} + 1}} has level ε0·ω3 = ω^(ε0 + 1)·3
{{0, 1}{{0, 1} + 2}} has level ε0·ω^2 = ω^(ε0 + 2)
{{0, 1}{{0, 1} + 2}{0, 1} + 2}} has level ε0·ω^2·2 = ω^(ε0 + 2)·2
{{0, 1}{{0, 1} + 3}} has level ε0·ω^3 = ω^(ε0 + 3)
{{0, 1}{{0, 1} + 4}} has level ε0·ω^4 = ω^(ε0 + 4)
{{0, 1}{{0, 1}{1}}} has level ε0·ω^ω = ω^(ε0 + ω)
{{0, 1}{{0, 1}{1} + 1}} has level ε0·ω^(ω + 1) = ω^(ε0 + ω + 1)
{{0, 1}{{0, 1}{1}{1}}} has level ε0·ω^ω2 = ω^(ε0 + ω2)
{{0, 1}{{0, 1}{2}}} has level ε0·ω^ω^2 = ω^(ε0 + ω^2)
{{0, 1}{{0, 1}{{1}}}} has level ε0·ω^ω^ω = ω^(ε0 + ω^ω)
{{0, 1}{{0, 1}{{{1}}}}} has level ε0·ω^ω^ω^ω = ω^(ε0 + ω^ω^ω)
{{0, 1}{{0, 1}{{0, 1}}}} has level ε0^2 = ω^(ε0·2)
{{0, 1}{{0, 1}{{0, 1}}} + 1} has level ε0^2 + 1 = ω^(ε0·2) + 1
{{0, 1}{{0, 1}{{0, 1}}}{{0, 1}}} has level ε0^2 + ε0 = ω^(ε0·2) + ε0
{{0, 1}{{0, 1}{{0, 1}}}{{0, 1}{{0, 1}}}} has level ε0^2·2 = ω^(ε0·2)·2
{{0, 1}{{0, 1}{{0, 1}} + 1}} has level ε0^2·ω = ω^(ε0·2 + 1)
{{0, 1}{{0, 1}{{0, 1}}{1}}} has level ε0^2·ω^ω = ω^(ε0·2 + ω)
{{0, 1}{{0, 1}{{0, 1}}{{0, 1}}}} has level ε0^3 = ω^(ε0·3)
{{0, 1}{{0, 1}{{0, 1}}{{0, 1}}{{0, 1}}}} has level ε0^4 = ω^(ε0·4)
{{0, 1}{{0, 1}{{0, 1} + 1}}} has level ε0^ω = ω^ω^(ε0 + 1)
{{0, 1}{{0, 1}{{0, 1} + 1}{{0, 1} + 1}}} has level ε0^ω2 = ω^(ω^(ε0 + 1)·2)
{{0, 1}{{0, 1}{{0, 1} + 2}}} has level ε0^ω^2 = ω^ω^(ε0 + 2)
{{0, 1}{{0, 1}{{0, 1} + 3}}} has level ε0^ω^3 = ω^ω^(ε0 + 3)
{{0, 1}{{0, 1}{{0, 1}{1}}}} has level ε0^ω^ω = ω^ω^(ε0 + ω)
{{0, 1}{{0, 1}{{0, 1}{{1}}}}} has level ε0^ω^ω^ω = ω^ω^(ε0 + ω^ω)
{{0, 1}{{0, 1}{{0, 1}{{0, 1}}}}} has level ε0^ε0 = ω^ω^(ε0·2)
{{0, 1}{{0, 1}{{0, 1}{{0, 1}}{{0, 1}}}}} has level ε0^ε0^2 = ω^ω^(ε0·3)
{{0, 1}{{0, 1}{{0, 1}{{0, 1} + 1}}}} has level ε0^ε0^ω = ω^ω^ω^(ε0 + 1)
{{0, 1}{{0, 1}{{0, 1}{{0, 1}{1}}}}} has level ε0^ε0^ω^ω = ω^ω^ω^(ε0 + ω)
{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}}}}}} has level ε0^ε0^ε0 = ω^ω^ω^(ε0·2)
{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1} + 1}}}}} has level ε0^ε0^ε0^ω = ω^ω^ω^ω^(ε0 + 1)
{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}}}}}}} has level ε0^ε0^ε0^ε0 = ω^ω^ω^ω^(ε0·2)
{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}}}}}}}} has level ε0^ε0^ε0^ε0^ε0 = ω^ω^ω^ω^ω^(ε0·2)
{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}{{0, 1}}}}}}}}} has level ε0^ε0^ε0^ε0^ε0^ε0 = ω^ω^ω^ω^ω^ω^(ε0·2)
So, the limit of {{0, 1}{{0, 1}{{0, 1}{{0, 1}{...}}}}} is {{0, 1}{0, 1}}, which has the FGH ordinal level of ε1.
{{0, 1}{0, 1}}[n] is expanded as follows:
{{0, 1}{0, 1}}[1] = {{0, 1}}
{{0, 1}{0, 1}}[2] = {{0, 1}{{0, 1}}}
{{0, 1}{0, 1}}[3] = {{0, 1}{{0, 1}{{0, 1}}}}
{{0, 1}{0, 1}}[4] = {{0, 1}{{0, 1}{{0, 1}{{0, 1}}}}}
...
Furthermore, when I replace all of the {0, 1}'s with an arbitrary number of copies, where {0, 1} has the FGH ordinal level of ε0, then {0, 1}.2 = {0, 1}{0, 1} = ε1, {0, 1}.3 = {0, 1}{0, 1}{0, 1} = ε2, {0, 1}.4 = {0, 1}{0, 1}{0, 1}{0, 1} = ε3, etc. This method is simple, just increase the epsilon numbers, and that's {0, 1}.n = ε(n-1).