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Proof step 20 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 = ∞ ∧ (∀ n1. ¬ n1 < n0 → nσ[n1] = n) ∨ (∀ n0. ⌜n0⌝ < # nσ → nσ[n0] < n) ⊦ nσ[∞.n + m + 1] = n + 1

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 5

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