User blog comment:Ikosarakt1/Fast-growing hierarchy/@comment-7484840-20130620190007

wouldn't it be simpler to say this (first 3 rules must be used in order): $$(\alpha_0+\alpha_1+\alpha_2+...+\alpha_{\beta}+\alpha_{\beta+1})[n] = \alpha_0+\alpha_1+\alpha_2+...+\alpha_{\beta}+(\alpha_{\beta+1})[n]$$ $$(\alpha_0\times\alpha_1\times\alpha_2\times...\times\alpha_{\beta}\times\alpha_{\beta+1})[n] = \alpha_0\times\alpha_1\times\alpha_2\times...\times\alpha_{\beta}\times(\alpha_{\beta+1})[n]$$ $$(\alpha_0^{\alpha_1^{\alpha_2^{...^{\alpha_{\beta}^{\alpha_{\beta+1}}}}}})[n] = \alpha_0^{\alpha_1^{\alpha_2^{...^{\alpha_{\beta}^{\alpha_{\beta+1}[n]}}}}})$$ $$\alpha$$ and $$\beta$$ are limit ordinals: $$\vartheta(\alpha,\beta)[n] = \vartheta(\alpha[n],\beta[n])$$ $$\vartheta(\alpha+1,\beta)[n]  = \vartheta(\alpha+1,\beta[n])$$ $$\vartheta(\alpha,\beta+1)[n]  = \vartheta(\alpha[n],\beta+1)$$ $$\vartheta(\bullet\Omega[n+1],\alpha+1) = \vartheta(\bullet\vartheta(\bullet\Omega[n],\alpha+1),\alpha+1)$$</li> $$\vartheta(\alpha,\beta)[0] = 1$$</li> $$\Omega[0] = 1$$</li></ul> $$\bullet$$ is any ordinal structure.