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Proof step 17 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[∞.n + m + 1] = n + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 18:
⊦ stutf(nσ) → (max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m) = ∞ ↔ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m) ∨ (∀ n. ⌜n⌝ < # nσ → nσ[n] < m))
Proof step 19:
stutf(nσ) → (max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞ ↔ # nσ = ∞ ∧ (∃ n0. ∀ n1. ¬ n1 < n0 → nσ[n1] = n) ∨ (∀ 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[∞.n + m + 1] = n + 1

Rule argument

spec: inst: name: stutf-maxinf-is-fin seq: ⊦ stutf(nσ) → ( max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m) = ∞ ↔ # nσ = ∞ ∧ (∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = m) ∨ (∀ n. ⌜n⌝ < # nσ → nσ[n] < m)) subst: [nσ, m] → [nσ, n] flags: current, []

Additional information

Used terms: nσ, n