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Proof step 95 from proof inv-get-inverse-succ (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 96:
max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = n) = ∞, # nσ = ⌜m⌝ + ⌜1⌝ , stutf(nσ), n ≤ nσ[m], max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = n).n < m ⊦ nσ[max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = n).n + 1] = n + 1, m = 0
Proof step 106:
# nσ = ⌜m⌝ + ⌜1⌝ , stutf(nσ), n ≤ nσ[m], max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = n).n < m ⊦ max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = n) = ∞, nσ[max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = n).n + 1] = n + 1, m = 0

Rule argument

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

Additional information

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