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Proof step 36 from proof inv-get-inverse-last (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 37:
# nσ = ⌜n⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n - 1]) = ∞, ⌜∞.n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), ∀ n0. ⌜n0⌝ < # nσ → nσ[n0] < nσ[n - 1] ⊦ nσ[∞.n + m + 1] = nσ[n - 1] + 1, # nσ = ∞

Rule argument

spec: nat-inf-basic inst: name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [# nσ, m0] flags:

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # nσ, m0