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Proof step 9 from proof inv-inverts (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
nσ[n + 1] = nσ[n] + 1, leninv(nσ) = ⌜n1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[nσ[n]] = n, leninv(nσ) = ∞

Rule argument

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

Additional information

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