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Proof step 151 from proof inv-get-inverse-between (previous step)

# nσ = ⌜m0⌝ + ⌜1⌝ , stutf(nσ), n < nσ[m0], max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n ⊦ nσ[m0] = 0, nσ[max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1] = n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n = 0

User applied rule cut formula

And obtained the following new goals

Proof step 152:
max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n) = ∞, # nσ = ⌜m0⌝ + ⌜1⌝ , stutf(nσ), n < nσ[m0], max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n ⊦ nσ[m0] = 0, nσ[max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1] = n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n = 0
Proof step 162:
# nσ = ⌜m0⌝ + ⌜1⌝ , stutf(nσ), n < nσ[m0], max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n ⊦ max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n) = ∞, nσ[m0] = 0, nσ[max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1] = n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n = 0

Rule argument

fma: max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n) = ∞

Additional information

Used formulas: max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n) = ∞