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## Overview

EBOCF(Extended Buchholz's OCF) is 2 variables function. So, I decided to extend it like ψ_A(B) => ψ_0(A,B). Please be careful that this 3 variables ψ is neither OCF nor Ordinal notation. I named this notation "Kumakuma 3 variables ψ". "Kuma" means "bear" in Japanese🐻. In order to make the definiton, I referred to the following pages. I appreciate p進大好きbot corrected my definitons.

## Definition

### Notation

Here, I define the character string used for the notation.

I define sets T and PT of formal strings consisting of 0, ψ_, (, ), +, and commas in the following recursive way:

1. 0∈T.
2. For any X_1,X_2,X_3∈T, ψ_{X_1}(X_2,X_3)∈T and ψ_{X_1}(X_2,X_3)∈PT.
3. For any X_1,...,X_m∈PT (2≦m<∞), X_1+...+X_m∈T.

0 is abbreviated as $0, ψ_0(0,0) is abbreviated as$1, $1+...+$1 (n $1's) is abbreviated as$n for each $$n \in \mathbb{N}$$ greater than 1, and ψ_0(0,$1) is abbreviated as$ω.

### Ordering

Here, I define the magnitude relation between the notation.

For an X,Y∈T, I define the recursive 2-ary relation X<Y in the following recursive way:

1. If X=0, then X<Y is equivalent to Y≠0.
2. Suppose X=ψ_{X_1}(X_2,X_3) for some X_1,X_2,X_3∈T.
2-1. If Y=0, then X<Y is false.
2-2. Suppose Y=ψ_{Y_1}(Y_2,Y_3) for some Y_1,Y_2,Y_3∈T.
2-2-1. If X_1=Y_1 and X_2=Y_2, then X<Y is equivalent to X_3<Y_3.
2-2-2. If X_1=Y_1 and X_2≠Y_2, then X<Y is equivalent to X_2<Y_2.
2-2-3. If X_1≠Y_1, then X<Y is equivalent to X_1<Y_1.
2-3. If Y=Y_1+...+Y_{m'} for some Y_1,...,Y_{m'}∈PT (2≦m'<∞), then
X<Y is equivalent to X=Y_1 or X<Y_1.
3. Suppose X=X_1+...+X_m for some X_1,...,X_m∈PT (2≦m<∞).
3-1. If Y=0, then X<Y is false.
3-2. If Y=ψ_{Y_1}(Y_2,Y_3) for some Y_1,Y_2,Y_3∈T, X<Y is equivalent to X_1<Y.
3-3. Suppose Y=Y_1+...+Y_{m'} for some Y_1,...,Y_{m'}∈PT (2≦m'<∞).
3-3-1. Suppose X_1=Y_1.
3-3-1-1. If m=2 and m'=2, then X<Y is equivalent to X_2<Y_2.
3-3-1-2. If m=2 and m'>2, then X<Y is equivalent to X_2<Y_2+...+Y_{m'}.
3-3-1-3. If m>2 and m'=2, then X<Y is equivalent to X_2+...+X_m<Y_2.
3-3-1-4. If m>2 and m'>2, then X<Y is equivalent to X_2+...+X_m<Y_2+...+Y_{m'}.
3-3-2. If X_1≠Y_1, then X<Y is equivalent to X_1<Y_1.

### Cofinality

Here, I define the cofinality.

I define total recursive maps \begin{eqnarray*} \textrm{dom} \colon T & \to & T \\ X & \mapsto & \textrm{dom}(X) \\ \end{eqnarray*} in the following recursive way:

1. If X=0, then dom(X)=0.
2. Suppose X=ψ_{X_1}(X_2,X_3) for some X_1,X_2,X_3∈T.
2-1. Suppose dom(X_3)=0.
2-1-1. Suppose dom(X_2)=0.
2-1-1-1. If dom(X_1)=0 or dom(X_1)=$1, then dom(X)=X. 2-1-1-2. If dom(X_1)≠0,$1, then dom(X)=dom(X_1).
2-1-2. If dom(X_2)=$1, then dom(X)=X. 2-1-3. Suppose dom(X_2)≠0,$1.
2-1-3-1. If dom(X_2)<X, then dom(X)=dom(X_2).
2-1-3-2. Otherwise, then dom(X)=$ω. 2-2. If dom(X_3)=$1 or dom(X_3)=$ω, then dom(X)=$ω.
2-3. Suppose dom(X_3)≠0,$1,$ω.
2-3-1. If dom(X_3)<X, then dom(X)=dom(X_3).
2-3-2. Otherwise, then dom(X)=$ω. 3. If X=X_1+...+X_m for some X_1,...,X_m∈PT (2≦m<∞), then dom(X)=dom(X_m). ### Fundamental Sequences Here, I define the fundamental sequences using cofinality. I define total recursive maps \begin{eqnarray*} [ \ ] \colon T \times T & \to & T \\ (X,Y) & \mapsto & X[Y] \end{eqnarray*} in the following recursive way: 1. If X=0, then X[Y]=0. 2. Suppose X=ψ_{X_1}(X_2,X_3) for some X_1,X_2,X_3∈T. 2-1. Suppose dom(X_3)=0. 2-1-1. Suppose dom(X_2)=0. 2-1-1-1. If dom(X_1)=0, then X[Y]=0. 2-1-1-2. If dom(X_1)=$1, then X[Y]=Y.
2-1-1-3. If dom(X_1)≠0,$1, then X[Y]=ψ_{X_1[Y]}(X_2,X_3). 2-1-2. If dom(X_2)=$1, then X[Y]=Y.
2-1-3. Suppose dom(X_2)≠0,$1. 2-1-3-1. If dom(X_2)<X, then X[Y]=ψ_{X_1}(X_2[Y],X_3). 2-1-3-2. Otherwise, then take a unique P,Q∈T such that dom(X_3)=ψ_{P}(Q,0). 2-1-3-2-1. Suppose Q=0. 2-1-3-2-1-1. If Y=$h (1≦h<∞) and X[Y[0]]=ψ_{X_1}(Γ,X_3) for unique Γ∈T, then
X[Y]=ψ_{X_1}(X_2[ψ_{P[0]}(Γ,0)],X_3).
2-1-3-2-1-2. Otherwise, then X[Y]=ψ_{X_1}(X_2[ψ_{P[0]}(Q,0)],X_3).
2-1-3-2-2. Suppose Q≠0.
2-1-3-2-2-1. If Y=$h (1≦h<∞) and X[Y[0]]=ψ_{X_1}(Γ,X_3) for unique Γ∈T, then X[Y]=ψ_{X_1}(X_2[ψ_{P}(Q[0],Γ)],X_3). 2-1-3-2-2-2. Otherwise, then X[Y]=ψ_{X_1}(X_2[ψ_{P}(Q[0],0)],X_3). 2-2. Suppose dom(X_3)=$1.
2-2-1. If Y=$1, then X[Y]=ψ_{X_1}(X_2,X_3[0]). 2-2-2. If Y=$k (2≦k<∞), then
X[Y]=ψ_{X_1}(X_2,X_3[0])+...+ψ_{X_1}(X_2,X_3[0]) (k's ψ_{X_1}(X_2,X_3[0])).
2-2-3. If neither of them, then X[Y]=0.
2-3. If dom(X_3)=$ω, X[Y]=ψ_{X_1}(X_2,X_3[Y]). 2-4. Suppose dom(X_3)≠0,$1,$ω. 2-4-1. If dom(X_3)<X, X[Y]=ψ_{X_1}(X_2,X_3[Y]). 2-4-2. Otherwise, then take a unique P,Q∈T such that dom(X_3)=ψ_{P}(Q,0). 2-4-2-1. Suppose Q=0. 2-4-2-1-1. If Y=$h (1≦h<∞) and X[Y[0]]=ψ_{X_1}(X_2,Γ) for unique Γ∈T, then
X[Y]=ψ_{X_1}(X_2,X_3[ψ_{P[0]}(Γ,0)]).
2-4-2-1-2. Otherwise, then X[Y]=ψ_{X_1}(X_2,X_3[ψ_{P[0]}(Q,0)]).
2-4-2-2. Suppose Q≠0.
2-4-2-2-1. If Y=$h (1≦h<∞) and X[Y[0]]=ψ_{X_1}(X_2,Γ) for unique Γ∈T, then X[Y]=ψ_{X_1}(X_2,X_3[ψ_{P}(Q[0],Γ)]). 2-4-2-2-2. Otherwise, then X[Y]=ψ_{X_1}(X_2,X_3[ψ_{P}(Q[0],0)]). 3. Suppose X=X_1+...+X_m for some X_1,...,X_m∈PT (2≦m<∞). 3-1. If X_m[Y]=0 and m=2, then X[Y]=X_1. 3-2. If X_m[Y]=0 and m>2, then X[Y]=X_1+...+X_{m-1}. 3-3. If X_m[Y]≠0, then X[Y]=X_1+...+X_{m-1}+X_m[Y]. ### FGH Here, I define the FGH. I define total recursive maps \begin{eqnarray*} f \colon T \times \mathbb{N} & \to & \mathbb{N} \\ (X,n) & \mapsto & f_X(n) \end{eqnarray*} in the following recursive way: 1. If X=0, then $$f_X(n) = n+1$$. 2. If X=$1 or X=Y+\$1 for some Y∈T, then $$f_X(n) = f_{X[0]}^n(n)$$.
3. If neither of them, then $$f_X(n) = f_{X[n]}(n)$$.

### Large Function and Large Number

Here, I define large function and large number.

I define total recursive maps \begin{eqnarray*} g \colon \mathbb{N} & \to & PT \\ n & \mapsto & g(n) \end{eqnarray*} in the following recursive way:

1. If n=0, then $$g(n) = ψ_0(0,0)$$.
2. Otherwise, then $$g(n) = ψ_{g(n-1)}(0,0)$$.

I define total recursive maps \begin{eqnarray*} F \colon \mathbb{N} & \to & \mathbb{N} \\ n & \mapsto & F(n) \end{eqnarray*} as $$F(n) = f_{ψ_0(0,g(n))}(n)$$.

I name $$F^{10^{100}}(10^{100})$$ "Kumakuma 3 variables ψ number".

### Naming

Here, I give names for ordinals.

I name an ordinal which correspond to $$ψ_0(0,ψ_{2}(0,0))$$ "KBHO" (Kuma-Bachmann–Howard Ordinal).

I name an ordinal which correspond to $$ψ_0(0,ψ_{ω}(0,0))$$ "KBO" (Kuma-Buchholz Ordinal).

Since ψ_0(ψ_{ψ_{ψ_{ψ_...}(0)}(0)}(0)) in EBOCF is called EBO (Extended Buchholz Ordinal) among Japanese googologists, I name an ordinal which correspond to the limitation of Kumakuma 3-ary ψ "EKBO" ( Extended Kuma- Buchholz Ordinal).

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