LPrSS VS veblen函数

本条目展示LPrSS和veblen函数的列表分析。

$1,3$之前等价于PrSS，这里不再枚举，详情可以参见PrSS VS 康托范式。

$1,3=\phi (1,0)$

$1,3,1=\phi (1,0)+1$

$1,3,1,3=\phi (1,0)\times 2$

$1,3,2={\omega }^{\phi (1,0)+1}$

$1,3,2,1,3,2={\omega }^{\phi (1,0)+1}\times 2$

$1,3,2,2={\omega }^{\phi (1,0)+2}$

$1,3,2,3={\omega }^{\phi (1,0)+\omega }$

$1,3,2,3,2,3={\omega }^{\phi (1,0)+\omega \times 2}$

$1,3,2,3,3={\omega }^{\phi (1,0)+{\omega }^2}$

$1,3,2,3,4={\omega }^{\phi (1,0)+{\omega }^{\omega }}$

$1,3,2,4={\omega }^{\phi (1,0)\times 2}$

$1,3,2,4,2,4={\omega }^{\phi (1,0)\times 3}$

$1,3,2,4,3={\omega }^{{\omega }^{\phi (1,0)+1}}$

$1,3,2,4,3,2,4,3={\omega }^{{\omega }^{\phi (1,0)+1}\times 2}$

$1,3,2,4,3,3={\omega }^{{\omega }^{\phi (1,0)+2}}$

$1,3,2,4,3,4={\omega }^{{\omega }^{\phi (1,0)+\omega }}$

$1,3,2,4,3,5={\omega }^{{\omega }^{\phi (1,0)\times 2}}$

$1,3,2,4,3,5,4={\omega }^{{\omega }^{{\omega }^{\phi (1,0)+1}}}$

$1,3,3=\phi (1,1)$

$1,3,3,2={\omega }^{\phi (1,1)+1}$

$1,3,3,2,4={\omega }^{\phi (1,1)+\phi (1,0)}$

$1,3,3,2,4,4={\omega }^{\phi (1,1)\times 2}$

$1,3,3,3=\phi (1,2)$

$1,3,4=\phi (1,\omega )$

$1,3,4,3=\phi (1,\omega +1)$

$1,3,4,3,4=\phi (1,\omega \times 2)$

$1,3,4,4=\phi (1,{\omega }^2)$

$1,3,4,5=\phi (1,{\omega }^{\omega })$

$1,3,5=\phi (1,\phi (1,0))$

$1,3,5,3=\phi (1,\phi (1,0)+1)$

$1,3,5,3,5=\phi (1,\phi (1,0)\times 2)$

$1,3,5,4,6=\phi (1,{\omega }^{\phi (1,0)\times 2})$

$1,3,5,5=\phi (1,\phi (1,1))$

$1,3,5,6=\phi (1,\phi (1,\omega ))$

$1,3,5,7=\phi (1,\phi (1,\phi (1,0)))$

$1,4=\phi (2,0)$

$1,4,3=\phi (1,\phi (2,0)+1)$

$1,4,3,4=\phi (1,\phi (2,0)+\omega )$

$1,4,3,5=\phi (1,\phi (2,0)+\phi (1,0))$

$1,4,3,6=\phi (1,\phi (2,0)\times 2)$

$1,4,3,6,4=\phi (1,{\omega }^{\phi (2,0)+1})$

$1,4,3,6,4,7=\phi (1,{\omega }^{\phi (2,0)\times 2})$

$1,4,3,6,5=\phi (1,\phi (1,\phi (2,0)+1))$

$1,4,3,6,5,8=\phi (1,\phi (1,\phi (2,0)\times 2))$

$1,4,4=\phi (2,1)$

$1,4,5=\phi (2,\omega ))$

$1,4,6=\phi (2,\phi (1,0))$

$1,4,7=\phi (2,\phi (2,0))$

$1,5=\phi (3,0)$

$1,5,2={\omega }^{\phi (3,0)+1}$

$1,5,3=\phi (1,\phi (3,0)+1)$

$1,5,4=\phi (2,\phi (3,0)+1)$

$1,5,5=\phi (3,1)$

$1,6=\phi (4,0)$

$Limit=\phi (\omega ,0)$