User blog:Leftunknown/EUSI equivalences (OLD)

'' NOTE: The EUSI function used within this blog is outdated, as it has been replaced with a much more efficient, documented, accurately described and simpler Veblen hierarchy. The equivalents listed below also have many mistakes beginning from the epsilon part, as it becomes increasingly more difficult to decode. ''

In this blog, I will be comparing EUSI functions to their ordinal equivelances. Whenever I use "verse X" or simply "X", I am talking about the verse to which the EUSI belongs to.

1 element arrays $$\psi(\alpha)$$

 * $$\exists$$= there exists (at least one)
 * $$\in$$= contains

2 element arrays $$\psi(\alpha,\beta)$$

 * $$\in^\infty$$= "infinite nestation"
 * $$N^0$$= the lowest nested verse of verse N
 * $$N_\psi$$= the EUSI of verse N

In depth
Rules:

Succession rule: $$ \psi(\alpha) = \alpha + 1$$

Nestation rule: $$ \psi(\alpha,\beta,\gamma \cdots \delta) = \psi(\alpha-1,\psi(\alpha,\beta,\gamma \cdots \delta),\gamma \cdots \delta)$$

Zero rule: $$ \psi(0,\beta,\gamma \cdots \delta) = \psi(\beta,\gamma \cdots \delta)$$

Double zero rule: $$ \psi(0,0,\gamma \cdots \delta) = \psi(1,\gamma \cdots \delta)$$ The (1,0) decomposition:

$$\psi(1,0)\rightarrow\psi(0,\psi(1,0))\rightarrow\psi(\psi(1,0))\rightarrow\psi(1,0)+1$$

Full decomposition notation:

$$\psi(1,0)\rightarrow^\infty\omega$$ $$\psi(1,\alpha)\rightarrow^\infty\omega+\alpha$$

$$\psi(2,0)\rightarrow\psi(1,\psi(2,0))\rightarrow\omega+\psi(2,0)\rightarrow^\infty\omega^2$$

$$\psi(2,\alpha)\rightarrow\psi(1,\psi(2,\alpha))\rightarrow\omega+\psi(2,\alpha)\rightarrow^\infty\omega^2+\alpha$$

$$\psi(3,0)\rightarrow\psi(2,\psi(3,0))\rightarrow\omega^2+\psi(3,0)\rightarrow^\infty\omega^3$$

$$\psi(4,0)\rightarrow\psi(3,\psi(4,0))\rightarrow\omega^3+\psi(4,0)\rightarrow^\infty\omega^4$$

We can thus see:

$$\psi(\alpha,0)=\omega^\alpha$$

$$\psi(\alpha,\beta)=\omega^\alpha+\beta$$ $$\psi(1,0,0)\rightarrow\psi(0,\psi(1,0,0),0)\rightarrow\psi(\psi(1,0,0),0)$$

From the rule above we can show that:

$$\psi(\psi(1,0,0),0)=\omega^{\psi(1,0,0)}\rightarrow^\infty\varepsilon_0$$

$$\psi(1,0,0)=\varepsilon_0$$ $$\psi(1,0,\alpha)\rightarrow\psi(0,\psi(1,0,\alpha),\alpha)\rightarrow\psi(\psi(1,0,\alpha),\alpha)=\omega^{\psi(1,0,\alpha)}+\alpha$$

$$\psi(1,0,\alpha)\rightarrow^\infty\omega^{\omega^{\omega^{\cdots}+\alpha}+\alpha}+\alpha>\omega^{\varepsilon_0+\alpha}$$

$$\psi(1,\alpha,0)\rightarrow^\infty\varepsilon_0^{\psi(\alpha,0)}\rightarrow\varepsilon_0^{\omega^\alpha}=\varepsilon_0^\alpha=\omega^{\varepsilon_0\times\alpha}$$

$$\psi(1,\alpha,\beta)\rightarrow^\infty\varepsilon_0^{\psi(\alpha,\beta)}\rightarrow\varepsilon_0^{\omega^\alpha+\beta}>\omega^{\varepsilon_0\times\alpha+\beta}$$

$$\psi(2,0,0)\rightarrow\psi(1,\psi(2,0,0),0)\rightarrow\varepsilon_0^{\psi(2,0,0)}\rightarrow^\infty\varepsilon_1$$

$$\psi(2,\alpha,0)\rightarrow\psi(1,\psi(2,\alpha,0),0)\rightarrow\varepsilon_0^{\psi(2,\alpha,0)}\rightarrow^\infty\varepsilon_1^{\psi(\alpha,0)}\rightarrow^\infty\varepsilon_1^\alpha=\omega^{\varepsilon_1\times\alpha}$$

$$\psi(3,0,0)\rightarrow\psi(2,\psi(3,0,0),0)\rightarrow\varepsilon_1^{\psi(3,0,0)}\rightarrow^\infty\varepsilon_2$$

And in general:

$$\psi(\alpha,0,0)=\varepsilon_{\alpha-1}$$

$$\psi(\alpha,0,\gamma)>\omega^{\varepsilon_{\alpha-1}+\gamma}$$

$$\psi(\alpha,\beta,0)=\omega^{\varepsilon_{\alpha-1}\times\beta}$$

$$\psi(\alpha,\beta,\gamma)>\omega^{\varepsilon_{\alpha-1}\times\beta+\gamma}$$ $$\psi(1,0,0,0)\rightarrow\psi(0,\psi(1,0,0,0),0,0)\rightarrow\psi(\psi(1,0,0,0),0,0)\rightarrow\zeta_0$$

$$\psi(1,0,0,\epsilon)=\psi(\psi(1,0,0,\epsilon),0,\epsilon)\rightarrow^\infty\geq\omega^{\varepsilon_\omega^{\varepsilon_\omega^{\varepsilon_\cdots}}}\simeq\zeta_0$$

$$\psi(1,0,\gamma,0)\geq\psi(1,0,0,\epsilon)\simeq\zeta_0$$

$$\psi(1,\beta,0,0)=\varphi_1^\infty(\varphi_1(\alpha))=\zeta_0$$

$$\psi(2,0,0,0)=\varphi_1^\infty(\psi(2,0,0,0))\begin{cases} \zeta_1, & \text{if we follow previous patterns} \\ \zeta_\omega, & \text{if this case is the oddity} \end{cases}$$

If we go with the first, more satisfactory option, we quickly get:

$$\psi(\alpha,0,0,0)=\zeta_{\alpha-1}$$

and

$$\psi(1,0,0,0,0)=\eta_0$$

and then can fairly confidently assume, that:

$$\psi(\Omega^\alpha)=\varphi_{\alpha-1}(0)$$

in the Veblen hierarchy.

This chain of patterns would be very nice indeed, however, I'm unable to find much about zeta numbers in general to confirm whether $$\varphi_1^{\omega,\omega}(0)=\varphi_1^{\omega^2}(0)=\zeta_1\text{ or }\zeta_\omega$$. The only reason as to why I'm entertaining the possibility of $$\zeta_\omega$$is because the only tool I have in possesion which deals with transfinite ordinals up to gamma nought seems to suggest so. I can't be confident in this either, though. The first transfinite ordinal: $$\omega=\psi(1,0)$$

The first additively indecomposable ordinal: $$\gamma_0=\psi(2,0)$$

The first multiplicatively indecomposable ordinal: $$\delta_0=\psi(\psi(2,0),0)$$

Strength of Peano arithmetic / the first exponentially indecomposable ordinal: $$\varepsilon_0=\psi(1,0,0)$$

Epsilon fixed point base exponentiation of $$1<\gamma<\varepsilon_\alpha$$: $$\psi(\psi(1,0),0,0)$$

Cantor's ordinal: $$\zeta_0=\psi(1,0,0,0)$$

Feferman-Schütte ordinal: $$\Gamma_0=\psi(\Omega^\Omega)$$