The First-order Array Notation has the following form:
a[c X d X e X ... X n]b where b ≥ 1; c,d,e,...,n > 1, and X is the separators. The separators come in many forms, such as {3}, {1,4,5}, {1,1,3{2}2}, {1 / 1{1 / 3}2}, or {1 <1{1 / 1,2}2> 2}.
The number a is the base, and b is the iterator.
The {1} separator stands for comma.
In this extension, the separators are the comma, the slash (/), and {x A n} where x is an expression, A is a separator, and n is an integer.
To solve the expression, we need some basic rules as follows:
Base Rule: a[0]b = a+1, a[1]b = a+b, a[2]b = ab, a[3]b = a^b.
Tailing Rule: a[# X 0]b = a[#]b (X can be any separators).
Prime Rule: a[%]1 = a.
Recursion Rule: a[c%]b = a[c-1%](a[c%](b-1)), provided c ≥ 0 and b ≥ 1.
Comma Rule: a[c{1}d%]b = a[c,d%]b.
Followed by some previous rules by following:
If the first entries are 1, a[1,1,...,1,1,c,d%]b = a[a,a,...,a,b,c-1,d%]a.
If the first entries are greater than 1, a[c,d,e,...,n%]b = a[c-1,d,e,...,n%](a[c,d,e,...,n%](b-1)).
Reuse the rules for linear arrays to that row (after separators other than commas).
a[1{2}c]b = a[1,1,1,...,1,1,2{2}c-1]a with b 1's.
a[1{2}1{2}...{2}1{2}c]b = a[1{2}1{2}...{2}1,1,1,...,1,1,2{2}c-1]a with b string of 1's using comma as a separator.
a[1{j}1{j}...{j}1{j}c] = a[1{j}1{j}...{j}1{j-1}1{j-1}...{j-1}2{j}c-1]a with b string of 1's using {j-1} separator.
a[1{1,1,1,...,1,1,c,d%}2]b = a[1{1,1,1,...,1,b,c-1,d%}2]a, which is different from a[1,1,1,...,1,1,c,d%]b = a[a,a,a,...,a,b,c-1,d%]a.
a[1{1{2}c}2]b = a[1{1,1,1,...,1,1,2{2}c-1}2], and so on. Nested arrays reuse the rules 4 through 7.
At this part, we define the new rules by following:
Reuse the previous rules above and ignore the slash separators.
Define the simplest expression for that part as a[1{1 / 2}2]b = a[1{1{...1{1,2}2...}2}2]a with b pairs of 1,2's.
If the first entry before a slash is just 1, change the {1 / n+1#} to Sb (b is the iterator), where S1 is {1 / n#}, Sn+1 = {1{Sn}2 / n#}, then change the iterator to the base.
Now the time to extend the First-order Array Notation, using the same rule as the previous chapters.
For the array 1{1 `1 / 2` 2}2, let S1 to be 1 / 2, Sn to be 1 `1{S(n-1)}2` 2, hence a[#]b to be Sb. For example, S2 is equal to 1 `1{1 / 2}2` 2, so a[1 `1 / 2` 2]b = a[1{Sa}2]a.
For example, 3[1{1 `1 / 2` 2}2]4, we can do in that manner by:
3[1{1 `1 / 2` 2}2]2 = 3[1{1 `1{1 / 2}2` 2}2]3
3[1{1 `1 / 2` 2}2]3 = 3[1{1 `1{1 `1{1 / 2}2` 2}2` 2}2]3
Therefore, 3[1{1 `1 / 2` 2}2]4 is equal to 3[1{1 `1{1 `1{1 `1{1 / 2}2` 2}2` 2}2` 2}2]3.
For higher arrays in the XFoAN, we can do the same rules in the manner as follows:
1{1 / 2 `1 / 2` 2}2 iterates Sn = 1{S(n-1)}2 `1 / 2` 2 where S1 = 1 `1 / 2` 2.
1{1 `1 / 2` 3}2 iterates Sn = 1 `1{S(n-1)}2` 2 `1 / 2` 2 where S1 = 1 `1 / 2` 2.
1{1 `1 / 2` 1 / 2}2 iterates Sn = 1 `1 / 2` 1{S(n-1)}2 where S1 = 1 `1 / 2` 2.
1{1 `1 / 2` 2}2 iterates Sn = 1 `1 / 2` 1{S(n-1)}2 where S1 = 1 `1 / 2` 2.
1{1 `1 / 3` 2}2 iterates Sn = 1 `1{S(n-1)}2 / 2` 2 where S1 = 1 `1 / 2` 2.
1{1 `1 / 1 / 2` 2}2 iterates Sn = 1 `1 / 1{S(n-1)}2` 2 where S1 = 1 `1 / 2` 2.
1{1 `1 `1 / 2` 2` 2}2 iterates Sn = 1 `1 `1{S(n-1)}2` 2` 2 where S1 = 1 `1 / 2` 2
And so on. The other rules remain unchanged.
Examples:
3[1{1 / 2 `1 / 2` 2}2]2 = 3[1{1{1 `1 / 2` 2}2 `1 / 2` 2}2]3
3[1{1 `1 / 2` 3}2]2 = 3[1{1 `1{1 `1 / 2` 2}2` 2 `1 / 2` 2}2]3
3[1{1 `1 / 2` 1 / 2}2]2 = 3[1{1 `1 / 2` 1{1 `1 / 2` 2}2}2]3
3[1{1 `1 / 3` 2}2]2 = 3[1{1 `1{1 `1 / 2` 2}2 / 2` 2}2]3
3[1{1 `1 / 1 / 2` 2}2]2 = 3[1{1 `1 / 1{1 `1 / 2` 2}2` 2}2]3
3[1{1 `1 `1 / 2` 2` 2}2]2 = 3[1{1 `1 `1{1 `1 / 2` 2}2` 2` 2}2]3
Due to the ambiguity of the punctuation mark, I will also use <>. For example: 1 <1 / 2> 2 = 1 `1 / 2` 2.
1 <2> 2 has level φ(ω,0)
1 / 2 <2> 2 has level ε(φ(ω,0)+1)
1 / 1 / 2 <2> 2 has level ζ(φ(ω,0)+1)
1 <2> 3 has level φ(ω,1)
1 <2> 1,2 has level φ(ω,ω)
1 <2> 1{1 <2> 2}2 has level φ(ω,φ(ω,0))
1 <2> 1 / 2 has level φ(ω+1,0)
1 <2> 1 / 1 / 2 has level φ(ω+2,0)
1 <2> 1 <2> 2 has level φ(ω2,0)
1 <2> 1 <2> 1 <2> 2 has level φ(ω3,0)
1 <3> 2 has level φ(ω^2,0)
1 <4> 2 has level φ(ω^3,0)
1 <1,2> 2 has level φ(ω^ω,0)
1 <1{2}2> 2 has level φ(ω^ω^ω,0)
1 <1{1 / 2}2> 2 has level φ(ε0,0)
1 <1{1 / 1 / 2}2> 2 has level φ(ζ0,0)
1 <1{1 <2> 2}2> 2 has level φ(φ(ω,0),0)
1 <1{1 <1{1 / 2}2> 2}2> 2 has level φ(φ(ε0,0),0)
1 <1{1 <1{1 <1{1 / 2}2> 2}2> 2}2> 2 has level φ(φ(φ(ε0,0),0),0)
...
1 <1 / 2> 2 has level Γ0 (Feferman-Schütte ordinal) = φ(1,0,0)
1 / 2 <1 / 2> 2 has level ε(Γ0+1)
1 / 1 / 2 <1 / 2> 2 has level ζ(Γ0+1)
1 <2> 2 <1 / 2> 2 has level φ(ω,Γ0+1)
1 <1{1 / 2}2> 2 <1 / 2> 2 has level φ(ε0,Γ0+1)
1 <1{1 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(Γ0,1)
1 <1{1 <1 / 2> 2}2> 3 <1 / 2> 2 has level φ(Γ0,2)
1 <1{1 <1 / 2> 2}2> 1,2 <1 / 2> 2 has level φ(Γ0,ω)
1 <1{1 <1 / 2> 2}2> 1{1 <1 / 2> 2}2 <1 / 2> 2 has level φ(Γ0,Γ0)
1 <1{1 <1 / 2> 2}2> 1 / 2 <1 / 2> 2 has level φ(Γ0+1,0)
1 <1{1 <1 / 2> 2}2> 1 / 1 / 2 <1 / 2> 2 has level φ(Γ0+2,0)
1 <1{1 <1 / 2> 2}2> 1 <2> 2 <1 / 2> 2 has level φ(Γ0+ω,0)
1 <1{1 <1 / 2> 2}2> 1 <1{1 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(Γ0*2,0)
1 <2{1 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(Γ0*ω,0)
1 <3{1 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(Γ0*ω^2,0)
1 <1,2{1 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(Γ0*ω^ω,0)
1 <1{1 / 2}2{1 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(Γ0*ε0,0)
1 <1{1 <1 / 2> 2}3> 2 <1 / 2> 2 has level φ(Γ0^2,0)
1 <1{1 <1 / 2> 2}1,2> 2 <1 / 2> 2 has level φ(Γ0^ω,0)
1 <1{1 <1 / 2> 2}1{1 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(Γ0^Γ0,0)
1 <1{2 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(Γ0^Γ0^ω,0)
1 <1{1 / 2 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(ε(Γ0+1),0)
1 <1{1 <1{1 <1 / 2> 2}2> 2 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(φ(Γ0,1),0)
1 <1 / 2> 3 has level Γ1 = φ(1,0,1)
1 <1{1 <1 / 2> 2}2> 2 <1 / 2> 3 has level φ(Γ0,Γ1+1)
1 <1{1 <1 / 2> 3}2> 2 <1 / 2> 3 has level φ(Γ1,1)
1 <1 / 2> 4 has level Γ2 = φ(1,0,2)
1 <1 / 2> 1,2 has level Γω = φ(1,0,ω)
1 <1 / 2> 1{1 <1 / 2> 2}2 has level Γ(Γ0) = φ(1,0,Γ0)
1 <1 / 2> 1 / 2 has level φ(1,1,0)
1 <1 / 2> 2 / 2 has level Γ(φ(1,1,0)+1)
1 <1 / 2> 1 / 3 has level φ(1,1,1)
1 <1 / 2> 1 / 1 / 2 has level φ(1,2,0)
1 <1 / 2> 1 / 1 / 1 / 2 has level φ(1,3,0)
1 <1 / 2> 1 <2> 2 has level φ(1,ω,0)
1 <1 / 2> 1 <1{1 / 2}2> 2 has level φ(1,Γ0,0)
1 <1 / 2> 1 <1 / 2> 2 has level φ(2,0,0)
1 <1 / 2> 2 <1 / 2> 2 has level Γ(φ(2,0,0)+1)
1 <1 / 2> 1 <1{1 <1 / 2> 1 <1 / 2> 2}2> 2 <1 / 2> 2 has level φ(1,φ(2,0,0),1)
1 <1 / 2> 1 <1 / 2> 3 has level φ(2,0,1)
1 <1 / 2> 1 <1 / 2> 1 / 2 has level φ(2,1,0)
1 <1 / 2> 1 <1 / 2> 1 <2> 2 has level φ(2,ω,0)
1 <1 / 2> 1 <1 / 2> 1 <1 / 2> 2 has level φ(3,0,0)
1 <1 / 2> 1 <1 / 2> 1 <1 / 2> 1 <1 / 2> 2 has level φ(4,0,0)
1 <2 / 2> 2 has level φ(ω,0,0)
1 <2 / 2> 3 has level φ(ω,0,1)
1 <2 / 2> 1 / 2 has level φ(ω,1,0)
1 <2 / 2> 1 <1 / 2> 2 has level φ(ω+1,0,0)
1 <2 / 2> 1 <2 / 2> 2 has level φ(ω2,0,0)
1 <3 / 2> 2 has level φ(ω^2,0,0)
1 <1,2 / 2> 2 has level φ(ω^ω,0,0)
1 <1{1 / 2}2 / 2> 2 has level φ(ε0,0,0)
1 <1{1 <1 / 2> 2}2 / 2> 2 has level φ(Γ0,0,0)
1 <1 / 3> 2 has level φ(1,0,0,0)
1 <1 / 3> 3 has level φ(1,0,0,1)
1 <1 / 3> 1 / 2 has level φ(1,0,1,0)
1 <1 / 3> 1 <2> 2 has level φ(1,0,ω,0)
1 <1 / 3> 1 <1 / 2> 2 has level φ(1,1,0,0)
1 <1 / 3> 1 <2 / 2> 2 has level φ(1,ω,0,0)
1 <1 / 3> 1 <1 / 3> 2 has level φ(2,0,0,0)
1 <2 / 3> 2 has level φ(ω,0,0,0)
1 <1{1 <1 / 3> 2}2 / 3> 2 has level φ(φ(1,0,0,0),0,0,0)
1 <1 / 4> 2 has level φ(1,0,0,0,0)
1 <1 / 5> 2 has level φ(1,0,0,0,0,0)
1 <1 / 1,2> 2 has level ψ(Ω^Ω^ω) (small Veblen ordinal)
1 <2 / 1,2> 2 has level ψ(Ω^(Ω^ω*ω))
1 <1 / 2,2> 2 has level ψ(Ω^Ω^(ω+1))
1 <1 / 1,3> 2 has level ψ(Ω^Ω^(ω2))
1 <1 / 1,1,2> 2 has level ψ(Ω^Ω^(ω^2))
1 <1 / 1{2}2> 2 has level ψ(Ω^Ω^(ω^ω))
1 <1 / 1{1 / 2}2> 2 has level ψ(Ω^Ω^ψ(Ω))
1 <1 / 1{1 <1 / 2> 2}2> 2 has level ψ(Ω^Ω^ψ(Ω^Ω))
1 <1 / 1{1 <1 / 1,2> 2}2> 2 has level ψ(Ω^Ω^ψ(Ω^Ω^ω))
...
1 <1 / 1 / 2> 2 has level ψ(Ω^Ω^Ω) (large Veblen ordinal)
1 / 2 <1 / 1 / 2> 2 has level ψ(Ω^Ω^Ω+Ω)
1 <1 / 2> 2 <1 / 1 / 2> 2 has level ψ(Ω^Ω^Ω+Ω^Ω)
1 <1 / 1 / 2> 3 has level ψ(Ω^Ω^Ω*2)
1 <1 / 1 / 2> 1,2 has level ψ(Ω^Ω^Ω*ω)
1 <1 / 1 / 2> 1 / 2 has level ψ(Ω^(Ω^Ω+1))
1 <1 / 1 / 2> 1 <2> 2 has level ψ(Ω^(Ω^Ω+ω))
1 <1 / 1 / 2> 1 <1 / 2> 2 has level ψ(Ω^(Ω^Ω+Ω))
1 <1 / 1 / 2> 1 <1 / 1,2> 2 has level ψ(Ω^(Ω^Ω+Ω^ω))
1 <1 / 1 / 2> 1 <1 / 1 / 2> 2 has level ψ(Ω^(Ω^Ω*2))
1 <2 / 1 / 2> 2 has level ψ(Ω^(Ω^Ω*ω))
1 <1 / 2 / 2> 2 has level ψ(Ω^Ω^(Ω+1))
1 <1 / 1,2 / 2> 2 has level ψ(Ω^Ω^(Ω+ω))
1 <1 / 1 / 3> 2 has level ψ(Ω^Ω^(Ω2))
1 <1 / 1 / 1,2> 2 has level ψ(Ω^Ω^(Ωω))
1 <1 / 1 / 1 / 2> 2 has level ψ(Ω^Ω^Ω^2)
1 <1 / 1 / 1 / 1 / 2> 2 has level ψ(Ω^Ω^Ω^3)
1 <1 <2> 2> 2 has level ψ(Ω^Ω^Ω^ω)
1 <1 <1{1 <1 / 2> 2}2> 2> 2 has level ψ(Ω^Ω^Ω^ψ(Ω^Ω))
1 <1 <1 / 2> 2> 2 has level ψ(Ω^Ω^Ω^Ω)
1 <1 <1 / 2> 1 / 2> 2 has level ψ(Ω^Ω^Ω^(Ω+1))
1 <1 <1 / 2> 1 <1 / 2> 2> 2 has level ψ(Ω^Ω^Ω^(Ω2))
1 <1 <2 / 2> 2> 2 has level ψ(Ω^Ω^Ω^(Ωω))
1 <1 <1 / 3> 2> 2 has level ψ(Ω^Ω^Ω^Ω^2)
1 <1 <1 / 1 / 2> 2> 2 has level ψ(Ω^Ω^Ω^Ω^Ω)
1 <1 <1 <1 / 2> 2> 2> 2 has level ψ(Ω^Ω^Ω^Ω^Ω^Ω)
As you can see each layer of separators adds two Ω's into the ordinal power of the first uncountable cardinal, and hence the limit of the level is ψ(ε(Ω+1)) = ψ(Ω_2) (which is the Bachmann-Howard ordinal).