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Proof step 41 from proof inv-inverts-last (previous step)

leninv(nσ) = ⌜n1⌝ , # nσ = ⌜n⌝ + ⌜1⌝ , stutf(nσ), nσ[n] < n1 ⊦ max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]) = ∞, max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]).n = n

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 42:
max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]) = ⌜n2⌝ , leninv(nσ) = ⌜n1⌝ , # nσ = ⌜n⌝ + ⌜1⌝ , stutf(nσ), nσ[n] < n1 ⊦ max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = nσ[n]) = ∞, n2 = n

Rule argument

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

Additional information

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