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Proof step 92 from proof inv-get-inverse-succ (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 93:
# nσ = ⌜n1⌝ , ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n⌝ + ⌜1⌝ < # nσ, stutf(nσ), n ≤ nσ[n1 - 1] ⊦ # nσ = ∞, nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + 1] = n + 1

Rule argument

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

Additional information

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