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Proof step 11 from proof inv-get-inverse-between (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 12:
max∞(λ n0. nσ[n0] = n) = ∞, ∀ n. ∃ m. nσ[m] = n, # nσ = ∞, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), max∞(λ n0. nσ[n0] = n).n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ max∞(λ n0. nσ[n0] = n + 1).n = 0, nσ[max∞(λ n0. nσ[n0] = n).n + m + 1] = n + 1
Proof step 30:
∀ n. ∃ m. nσ[m] = n, # nσ = ∞, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), max∞(λ n0. nσ[n0] = n).n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ max∞(λ n0. nσ[n0] = n) = ∞, max∞(λ n0. nσ[n0] = n + 1).n = 0, nσ[max∞(λ n0. nσ[n0] = n).n + m + 1] = n + 1

Rule argument

fma: max∞(λ n0. nσ[n0] = n) = ∞

Additional information

Used formulas: max∞(λ n0. nσ[n0] = n) = ∞