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Proof step 15 from proof inv-get-inverse-last (previous step)

⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ[m1] = n ∧ (∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1);nσ[(# nσ).n - 1] = n) ⊦ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1

User applied rule cut formula

And obtained the following new goals

Proof step 16:
max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞, ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ[m1] = n ∧ (∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1);nσ[(# nσ).n - 1] = n) ⊦ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1
Proof step 41:
⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ[m1] = n ∧ (∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1);nσ[(# nσ).n - 1] = n) ⊦ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞, nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1

Rule argument

fma: max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞

Additional information

Used formulas: max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞