User blog:Ikosarakt1/Hardy hierarchy up to psi(K)

I tried to define Hardy Hierarchy up to $$H_{\psi(\chi(\Xi(K,0)))}(n)$$, and I got the ruleset with 31 rules. It ignores the facts like $$1+\omega = \omega$$ for not making technical complexities, but it doesn't matter because ordinals can be treated just as notational symbols. Also, exponentiation isn't allowed: we must define it through $$\psi$$. So, the ruleset for which I was working for a few weeks:

The Ruleset
Rule 1:

$$H_0(n) = n$$

Rule 2:

$$H_{\alpha+1}(n) = H_\alpha(n+1)$$

Rule 3:

$$H_\alpha(n) = H_{\alpha[n]}(n)$$

Rule 4:

$$(\alpha+\beta)[n] = \alpha+(\beta[n])$$

Rule 5:

$$\Omega_0 = 1$$

Rule 6:

$$\psi_{\Omega_{\alpha+1}}(0) = \Omega_\alpha$$

Rule 7:

$$\psi_{\Omega_{\alpha+1}}(\beta+1)[0] = 0$$

Rule 8:

$$\psi_{\Omega_{\alpha+1}}(\beta+1)[n+1] = \psi_{\Omega_{\alpha+1}}(\beta)+\psi_{\Omega_{\alpha+1}}(\beta+1)[n]$$

Rule 9:

$$\psi_\alpha(\beta)[n] = \psi_\alpha(\beta[n])$$

Rule 10:

$$\psi_\alpha(\beta+1)[0] = \psi_\alpha(\beta)+1$$

Rule 11.

$$\psi_\alpha(\beta+\lambda)[0] = \psi_\alpha(\beta)$$

Rule 12.

$$\psi_\alpha(\beta+\alpha)[n+1] = \psi_\alpha(\beta+\psi_\alpha(\beta+\alpha)[n])$$

Rule 13. $$(\lambda \geq \alpha)$$

$$\psi_\alpha(\beta+\lambda) = \psi_\alpha(\psi_\lambda(\beta+\lambda))$$

Rule 14.

$$\psi_{\chi_\alpha(\beta,\lambda)(0)}[0] = 1$$

Rule 15.

$$\psi_{\chi_\alpha(\beta,\lambda+1)(0)}[0] = \chi_\alpha(\beta,\lambda)+1$$

Rule 16.

$$\psi_{\chi_\alpha(\beta+1,\lambda)(\delta)[n+1] = \chi(\beta,\psi_{\chi_\alpha(\beta+1,\lambda)(\delta)[n])$$

Rule 17.

$$\chi_\alpha(\beta,\lambda)[n] = \chi_\alpha(\beta,\lambda[n])$$

Rule 18.

$$\chi_\alpha(\beta,0)[n] = \chi_\alpha(\beta[n],0)$$

Rule 19.

$$\chi_\alpha(\beta,\lambda+1)[n] = \chi_\alpha(\beta[n],\chi_\alpha(\beta,\lambda)+1)$$

Rule 20.

$$\chi_\alpha(\beta+\alpha,0)[0] = \chi_\alpha(\beta,0)$$

Rule 21.

$$\chi_\alpha(\beta+\alpha,\lambda+1)[0] = \chi_\alpha(\beta+\alpha,\lambda)+1$$

Rule 22.

$$\chi_\alpha(\beta+\alpha,\lambda)[n+1] = \chi_\alpha(\beta+\chi_\alpha(\beta+\alpha,\lambda)[n],0)$$

Rule 23.

$$\chi_\alpha(\beta+\lambda,\delta) = \chi_\alpha(\chi_\lambda(\beta+\lambda,\delta))$$

Rule 24.

$$M_0 = 0$$

Rule 25.

$$\chi_{M_{\alpha+1}}(0,\beta) = \Omega_{M_\alpha+\beta}$$

Rule 26.

$$\chi_{Xi(\alpha+1,\beta+1)}(0,\lambda) = \Xi(\alpha,\Xi(\alpha+1,\beta)+\lambda)$$

Rule 27.

$$\chi_{\Xi(\alpha+1,0)(0,\beta)} = \Xi(\alpha,\beta)$$

Rule 28.

$$\Xi(0,\alpha) = M_\alpha$$

Rule 29.

$$\Xi(\alpha,\beta)[n] = \Xi(\alpha,\beta[n])$$

Rule 30.

$$\Xi(\alpha,0)[n] = \Xi(\alpha[n],0)$$

Rule 31.

$$\Xi(\alpha,\beta+1)[n] = \Xi(\alpha[n],\Xi(\alpha,\beta)+1)$$

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Feel free to improve it (particularly in rule-conditions, I think they must be more formal.)