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Proof step 61 from proof inv-get-inverse-last (previous step)

nσ[(# nσ).n - 1] = nσ[n1], ⌜n1⌝ < # nσ, ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ # nσ = ∞, nσ[n1 + m + 1] = nσ[n1] + 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 62:
# nσ = ⌜n⌝ , nσ[n - 1] = nσ[n1], ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, ⌜n1⌝ < # nσ, stutf(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ # nσ = ∞, nσ[n1 + m + 1] = nσ[n1] + 1

Rule argument

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

Additional information

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