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Proof step 33 from proof inv-get-inverse-between (previous step)

# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦ nσ[n1 + m + 1] = n + 1

User applied rule cut formula

And obtained the following new goals

Proof step 34:
# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , max∞(λ n0. nσ[n0] = n + 1) = ∞, n1 + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[n1 + m + 1] = n + 1
Proof step 47:
# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , n1 + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[n1 + m + 1] = n + 1, max∞(λ n0. nσ[n0] = n + 1) = ∞

Rule argument

fma: max∞(λ n0. nσ[n0] = n + 1) = ∞

Additional information

Used formulas: max∞(λ n0. nσ[n0] = n + 1) = ∞