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Proof step 43 from proof inv-get-inverse-last (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 44:
max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ⌜n1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = 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 + m + 1] = n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞

Rule argument

spec: nat-inf name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n), m2]

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n), m2