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Proof step 24 from proof inv-get-inverse-last (previous step)

# nσ = ∞, max∞(λ n. nσ[n] = nσ[m1]) = ∞, stutf(nσ), m1 < n0 → nσ[n0] = nσ[m1] + 1, ∀ n1. ¬ n1 < n0 → nσ[n1] = nσ[m1], ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ nσ[∞.n + m + 1] = nσ[m1] + 1

User applied rule cut formula

And obtained the following new goals

Proof step 25:
# nσ = ∞, max∞(λ n. nσ[n] = nσ[m1]) = ∞, m1 < n0, stutf(nσ), m1 < n0 → nσ[n0] = nσ[m1] + 1, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ n1. ¬ n1 < n0 → nσ[n1] = nσ[m1] ⊦ nσ[∞.n + m + 1] = nσ[m1] + 1
Proof step 29:
# nσ = ∞, max∞(λ n. nσ[n] = nσ[m1]) = ∞, stutf(nσ), m1 < n0 → nσ[n0] = nσ[m1] + 1, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ n1. ¬ n1 < n0 → nσ[n1] = nσ[m1] ⊦ nσ[∞.n + m + 1] = nσ[m1] + 1, m1 < n0

Rule argument

fma: m1 < n0

Additional information

Used formulas: m1 < n0