User blog:Ikosarakt1/Variant of BEAF-definitions for up-arrow notation-notation level.

I found a new, quite easy way of defining structures for BEAF up to $$X \uparrow^X X$$. The definition goes as follows:

Rule 1. (n=1)

A{n}B = A^B

Rule 2. (B=0)

A{n}B = 1

Rule 3. (B doesn't contain X's)

A{n+1}(B+1) = A{n}(A{n+1}B)

Rule 4. (otherwise)

A{n+1}(B+1) = (A{n+1}B){n}X

Note that it works as strong hyper-operators when B is just a number, and weak hyper-operators when B contains X's. It means that, for

example, $$X\{2\}5 = X^{X\{2\}4}$$ while $$X\{2\}(X+1) = (X\{2\}X)^X$$. Note that "true" solution of $$X\{2\}(X

+1)$$, supposed by Bowers, is $$X^{X\{2\}X}$$, but then we don't know how to evaluate it further.

Structural comparisons with traditional ordinal notations
While "Rule 4" gives us a system which certainly weaker than the hypothetical which works as pure right-associative, they grow at the

same rate when we diagonalize over them.

It think the following comparisons are correct:

X{2}X = $$\varepsilon_0$$

(X{2}X){1}2 = $$\varepsilon_0^2 = \omega^{\varepsilon_0*2}$$

(X{2}X){1}3 = $$\varepsilon_0^3 = \omega^{\varepsilon_0*3}$$

X{2}(X+1) = $$\varepsilon_0^\omega = \omega^{\omega^{\varepsilon_0+1}}$$

X{2}(X+2) = $$\varepsilon_0^{\omega^2} = \omega^{\omega^{\varepsilon_0+2}}$$

X{2}(X*2) = $$\varepsilon_0^{\omega^\omega} = \omega^{\omega^{\varepsilon_0+\omega}}$$

X{2}(X{2}X) = $$\varepsilon_0^{\varepsilon_0} = \omega^{\omega^{\varepsilon_0*2}}$$

X{2}(X{2}(X+1)) = $$\varepsilon_0^{\varepsilon_0^\omega} = \omega^{\omega^{\omega^{\varepsilon_0+1}}}$$

X{2}(X{2}(X{2}X)) = $$\varepsilon_0^{\varepsilon_0^{\varepsilon_0}} = \omega^{\omega^{\omega^{\varepsilon_0*2}}}$$

X{3}X = $$\varepsilon_1$$

So it turned out that X^^^X by that definition is just at level of $$\varepsilon_1$$. Don't be fooled however, believing that

X^^^^X will be at level of just $$\varepsilon_2$$. Just look at further behavior:

(X{3}X){1}X = $$\varepsilon_1^\omega = \omega^{\omega^{\varepsilon_1+1}}$$

(X{3}X){1}(X{3}X) = $$\varepsilon_1^{\varepsilon_1} = \omega^{\omega^{\varepsilon_1*2}}$$

(X{3}X){1}(X{3}X){1}(X{3}X) = $$\varepsilon_1^{\varepsilon_1^{\varepsilon_1}} = \omega^{\omega^{\omega^{\varepsilon_1*2}}}$$

(X{3}X){2}X = X{3}(X+1) = $$\varepsilon_2$$

(X{3}(X+1)){2}X = X{3}(X+2) = $$\varepsilon_3$$

X{3}(X+n) = $$\varepsilon_{n+1}$$

X{3}(X*2) = $$\varepsilon_\omega$$

X{3}(X{2}X) = $$\varepsilon_{\varepsilon_0}$$

X{3}(X{3}X) = $$\varepsilon_{\varepsilon_1}$$

X{4}X = $$\zeta_0$$

Consider that if A has level $$\alpha$$, then A{2}X has level $$\varepsilon_{\alpha+1}$$:

(X{4}X){2}X = $$\varepsilon_{\zeta_0+1}$$

(X{4}X){2}(X+1) = $$\omega^{\omega^{\varepsilon_{\zeta_0+1}+1}}$$

(X{4}X){2}(X{4}X) = $$\omega^{\omega^{\varepsilon_{\zeta_0+1}*2}}$$

(X{4}X){3}X = X{4}(X+1) = $$\varepsilon_{\zeta_0+2}$$

X{4}(X+2) = $$\varepsilon_{\zeta_0+3}$$

X{4}(X*2) = $$\varepsilon_{\zeta_0+\omega}$$

X{4}(X{4}X) = $$\varepsilon_{\zeta_0*2}$$

X{4}(X{4}(X{4}X)) = $$\varepsilon_{\varepsilon_{\zeta_0*2}}$$

X{5}X = $$\zeta_1$$

We see that each adding to B in A{3}B (when B contains X's) leads to the level of new epsilon number.

(X{5}X){2}X = $$\varepsilon_{\zeta_1+1}$$

(X{5}X){3}X = $$\varepsilon_{\zeta_1+2}$$

(X{5}X){3}(X+1) = $$\varepsilon_{\zeta_1+3}$$

(X{5}X){3}(X+n) = $$\varepsilon_{\zeta_1+n+2}$$

(X{5}X){3}(X{5}X) = $$\varepsilon_{\zeta_1*2}$$

(X{5}X){3}(X{5}X){3}(X{5}X) = $$\varepsilon_{\varepsilon_{\zeta_1*2}}$$

(X{5}X){4}X = X{5}(X+1) = $$\zeta_2$$

So "hexating" A in this system will lead to the level $$\zeta_{\alpha+1}$$

X{5}(X+1) = $$\zeta_3$$

X{5}(X+2) = $$\zeta_4$$

X{5}(X{5}X) = $$\zeta_{\zeta_0}$$

X{6}X = $$\eta_0$$

Consider the sequence:

X{2}X = $$\varepsilon_0 = \varphi(1,0)$$

X{3}X = $$\varepsilon_1 = \varphi(1,1)$$

X{4}X = $$\zeta_0 = \varphi(2,0)$$

X{5}X = $$\zeta_0 = \varphi(2,1)$$

X{6}X = $$\eta_0$$

Using the pattern, we conclude that X{2n}X = $$\varphi(n,0)$$ and X{2n+1}X = $$\varphi(n,1)$$. So we indeed reach the power of $$\varphi(\omega,0)$$.

Pros and cons
Pros:


 * Easy to define and doesn't need additional notation like explained here.


 * Exhibits the same growth rate as hypothetical "normal" variant of BEAF.

Cons:


 * General avoiding the principle of BEAF: when in A & n, A composed with X's and operations, the total number of entries can be found just replacing X's to n's and solving. When we solve X{2}(X+1), we expect $$3^{3^{3^3}}$$ entries, but replacing the first expression to (X{2}X){1}X we get instead only $$3^{3^3+1}$$ of them.


 * Definition of increment to B in up-arrow notation is dependent on existence of X's. So the pattern breaks on base-recursing and polyponent-recursing. It looks awkwardly.