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Proof step 39 from proof inv-get-inverse (previous step)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[max∞(λ n0. nσ[n0] = n).n] = n, max∞(λ n0. nσ[n0] = n) = ∞

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 40:
# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[n1] = n, max∞(λ n0. nσ[n0] = n) = ∞

Rule argument

spec: nat-inf name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [max∞(λ n0. nσ[n0] = n), m1]

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. nσ[n0] = n), m1