strict fz分析Part1：修订间差异

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2026年7月8日 (三) 07:05的最新版本

以下使用strict fz和BMS对照分析，用 $ψ$ 简称 $ψ_{Z S}$ （strict fz的符号，S即strict）

$ψ (0) = 1$

$ψ (1) = ω$

$ψ [ω] (ω) = ω^{ω}$

$ψ [ω] (ψ [ω] (ω)) = ω^{ω^{ω}}$

$ψ (ω) = (0) (1, 1)$

$ψ (ω + 1) = (0) (1, 1) (1)$

$ψ [ω \times 2] (ω \times 2) = (0) (1, 1) (1) (2)$

$ψ [ω \times 2] (ε_{0}) = (0) (1, 1) (1) (2, 1)$

$ψ (ω \times 2) = (0) (1, 1) (1, 1)$

$ψ [ω^{2}] (ω^{2}) = (0) (1, 1) (2)$

$ψ [ω^{2}, ω^{2} + ω] (ω^{2} + ω) = (0) (1, 1) (2) (1) (2)$

$ψ [ω^{2}] (ω^{2} + ω) = (0) (1, 1) (2) (1, 1)$

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

$ψ [ω^{2}] (ω^{2} + ω) = (0) (1, 1) (2) (1, 1) (2)$

$ψ [ω^{2}] (ω^{2} \times 2) = (0) (1, 1) (2) (2)$

$ψ [ω^{2}] (ω^{3}) = (0) (1, 1) (2) (3)$

$ψ [ω^{2}] (ω^{ω}) = (0) (1, 1) (2) (3) (4)$

$ψ [ω^{2}] (ε_{0}) = (0) (1, 1) (2) (3, 1)$

$ψ [ω^{2}] (ψ [ω^{2}] (ω^{2})) = (0) (1, 1) (2) (3, 1) (4)$

$ψ [ω^{2}] (ψ [ω^{2}] (ψ [ω^{2}] (ω^{2}))) = (0) (1, 1) (2) (3, 1) (4) (5, 1) (6)$

$ψ (ω^{2}) = (0) (1, 1) (2, 1)$

$ψ (ω^{2} \times 2) = (0) (1, 1) (2, 1) (1, 1) (2, 1)$

$ψ (ω^{2} \times 3) = (0) (1, 1) (2, 1) (1, 1) (2, 1) (1, 1) (2, 1)$

$ψ [ω^{3}] (ω^{3}) = (0) (1, 1) (2, 1) (2)$

$ψ [ω^{3}] (ε_{0}) = (0) (1, 1) (2, 1) (2) (3, 1)$

$ψ (ω^{3}) = (0) (1, 1) (2, 1) (2, 1)$

$ψ (ω^{4}) = (0) (1, 1) (2, 1) (2, 1) (2, 1)$

$ψ [ω^{ω}] (ω^{ω}) = (0) (1, 1) (2, 1) (3)$

$ψ [ω^{ω}, ω^{ω + 1}] (ω^{ω + 1}) = (0) (1, 1) (2, 1) (3) (2)$

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

$ψ [ω^{ω}, ω^{ω + 1}] (ω^{ω + 1} \times 2) = (0) (1, 1) (2, 1) (3) (2) (1, 1) (2, 1) (3) (2)$

$ψ [ω^{ω}, ω^{ω + 1}] (ω^{ω + 2}) = (0) (1, 1) (2, 1) (3) (2) (2)$

$ψ [ω^{ω}, ω^{ω + 1}] (ω^{ω + 3}) = (0) (1, 1) (2, 1) (3) (2) (2) (2)$

$ψ [ω^{ω}, ω^{ω + 1}] (ω^{ω + ω}) = (0) (1, 1) (2, 1) (3) (2) (3)$

$ψ [ω^{ω}, ω^{ω + 1}] (ω^{ω + ε_{0}}) = (0) (1, 1) (2, 1) (3) (2) (3, 1) = ψ [ω^{ω}, ω^{ω + 1}] (ε_{0})$

$ψ [ω^{ω}] (ω^{ω + 1}) = (0) (1, 1) (2, 1) (3) (2, 1)$

$ψ [ω^{ω}, ω^{ω + 2}] (ω^{ω + 2}) = (0) (1, 1) (2, 1) (3) (2, 1) (2)$

$ψ [ω^{ω}] (ω^{ω + 2}) = (0) (1, 1) (2, 1) (3) (2, 1) (2, 1)$

$ψ [ω^{ω}] (ω^{ω \times 2}) = (0) (1, 1) (2, 1) (3) (2, 1) (3)$

$ψ [ω^{ω}] (ω^{ω^{2}}) = (0) (1, 1) (2, 1) (3) (3)$

$ψ [ω^{ω}] (ε_{0}) = (0) (1, 1) (2, 1) (3) (4, 1)$

@@ 第42行： / 第42行： @@
 <math>\psi(\omega^2)=(0)(1,1)(2,1)</math>
+<math>\psi(\omega^2\times2)=(0)(1,1)(2,1)(1,1)(2,1)</math>
+<math>\psi(\omega^2\times3)=(0)(1,1)(2,1)(1,1)(2,1)(1,1)(2,1)</math>
+<math>\psi[\omega^3](\omega^3)=(0)(1,1)(2,1)(2)</math>
+<math>\psi[\omega^3](\varepsilon_0)=(0)(1,1)(2,1)(2)(3,1)</math>
+<math>\psi(\omega^3)=(0)(1,1)(2,1)(2,1)</math>
+<math>\psi(\omega^4)=(0)(1,1)(2,1)(2,1)(2,1)</math>
+<math>\psi[\omega^\omega](\omega^\omega)=(0)(1,1)(2,1)(3)</math>
+<math>\psi[\omega^\omega,\omega^{\omega+1}](\omega^{\omega+1})=(0)(1,1)(2,1)(3)(2)</math>
+<math>\color{red}{\psi[\omega^\omega,\omega^{\omega+1}](\omega^{\omega+1}+\omega^\omega)=(0)(1,1)(2,1)(3)(2)(1,1)(2,1)(3)}</math>（P.S.这里按照规则没有<math>[\omega^\omega,\omega^{\omega+1},\omega^{\omega+1}+\omega^\omega]</math>提升，因为这个伪链不存在）
+<math>\psi[\omega^\omega,\omega^{\omega+1}](\omega^{\omega+1}\times2)=(0)(1,1)(2,1)(3)(2)(1,1)(2,1)(3)(2)</math>
+<math>\psi[\omega^\omega,\omega^{\omega+1}](\omega^{\omega+2})=(0)(1,1)(2,1)(3)(2)(2)</math>
+<math>\psi[\omega^\omega,\omega^{\omega+1}](\omega^{\omega+3})=(0)(1,1)(2,1)(3)(2)(2)(2)</math>
+<math>\psi[\omega^\omega,\omega^{\omega+1}](\omega^{\omega+\color{orange}{\omega}})=(0)(1,1)(2,1)(3)\color{orange}{(2)(3)}</math>
+<math>\psi[\omega^\omega,\omega^{\omega+1}](\omega^{\omega+\color{orange}{\varepsilon_0}})=(0)(1,1)(2,1)(3){\color{orange}{(2)(3,1)}}=\psi[\omega^\omega,\omega^{\omega+1}](\varepsilon_0)</math>
+<math>\psi[\omega^\omega](\omega^{\omega+1})=(0)(1,1)(2,1)(3)(2,1)</math>
+<math>\psi[\omega^\omega](\omega^{\omega+1})=(0)(1,1)(2,1)(3)(2,1)</math>
+<math>\psi[\omega^\omega,\omega^{\omega+2}](\omega^{\omega+2})=(0)(1,1)(2,1)(3)(2,1)(2)</math>
+<math>\psi[\omega^\omega](\omega^{\omega+2})=(0)(1,1)(2,1)(3)(2,1)(2,1)</math>
+<math>\psi[\omega^\omega](\omega^{\omega\times2})=(0)(1,1)(2,1)(3)(2,1)(3)</math>
+<math>\psi[\omega^\omega](\omega^{\omega^2})=(0)(1,1)(2,1)(3)(3)</math>
+<math>\psi[\omega^\omega](\varepsilon_0)=(0)(1,1)(2,1)(3)(4,1)</math>
+<math>\psi[\omega^\omega](\varepsilon_0)=(0)(1,1)(2,1)(3)(4,1)</math>