strict fz分析Part1

以下使用strict fz和BMS对照分析，用$\psi$简称${\psi }_{ZS}$（strict fz的符号，S即strict）

$\psi (0)=1$

$\psi (1)=\omega$

$\psi [\omega ](\omega )={\omega }^{\omega }$

$\psi [\omega ](\psi [\omega ](\omega ))={\omega }^{{\omega }^{\omega }}$

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

$\psi (\omega +1)=(0)(1,1)(1)$

$\psi [\omega \times 2](\omega \times 2)=(0)(1,1)(1)(2)$

$\psi [\omega \times 2]({\epsilon }_0)=(0)(1,1)(1)(2,1)$

$\psi (\omega \times 2)=(0)(1,1)(1,1)$

$\psi [{\omega }^2]({\omega }^2)=(0)(1,1)(2)$

$\psi [{\omega }^2,{\omega }^2+\omega ]({\omega }^2+\omega )=(0)(1,1)(2)(1)(2)$

$\psi [{\omega }^2]({\omega }^2+\omega )=(0)(1,1)(2)(1,1)$

$\psi [{\omega }^2]({\omega }^2+\omega \times 2)=(0)(1,1)(2)(1,1)(1,1)$

$\psi [{\omega }^2]({\omega }^2+\omega )=(0)(1,1)(2)(1,1)(2)$

$\psi [{\omega }^2]({\omega }^2\times 2)=(0)(1,1)(2)(2)$

$\psi [{\omega }^2]({\omega }^3)=(0)(1,1)(2)(3)$

$\psi [{\omega }^2]({\omega }^{\omega })=(0)(1,1)(2)(3)(4)$

$\psi [{\omega }^2]({\epsilon }_0)=(0)(1,1)(2)(3,1)$

$\psi [{\omega }^2](\psi [{\omega }^2]({\omega }^2))=(0)(1,1)(2)(3,1)(4)$

$\psi [{\omega }^2](\psi [{\omega }^2](\psi [{\omega }^2]({\omega }^2)))=(0)(1,1)(2)(3,1)(4)(5,1)(6)$

$\psi ({\omega }^2)=(0)(1,1)(2,1)$

$\psi ({\omega }^2\times 2)=(0)(1,1)(2,1)(1,1)(2,1)$

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

$\psi [{\omega }^3]({\omega }^3)=(0)(1,1)(2,1)(2)$

$\psi [{\omega }^3]({\epsilon }_0)=(0)(1,1)(2,1)(2)(3,1)$

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

$\psi ({\omega }^4)=(0)(1,1)(2,1)(2,1)(2,1)$

$\psi [{\omega }^{\omega }]({\omega }^{\omega })=(0)(1,1)(2,1)(3)$

$\psi [{\omega }^{\omega },{\omega }^{\omega +1}]({\omega }^{\omega +1})=(0)(1,1)(2,1)(3)(2)$

$\psi [{\omega }^{\omega },{\omega }^{\omega +1}]({\omega }^{\omega +1}+{\omega }^{\omega })=(0)(1,1)(2,1)(3)(2)(1,1)(2,1)(3)$（P.S.这里按照规则没有$[{\omega }^{\omega },{\omega }^{\omega +1},{\omega }^{\omega +1}+{\omega }^{\omega }]$提升，因为这个伪链不存在