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Proof step 77 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, n ≤ nσ[m], stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ max∞(λ n0. nσ[n0] = n) = ∞, nσ[max∞(λ n0. nσ[n0] = n).n + 1] = n + 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 78:
max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , # nσ = ∞, n ≤ nσ[m], stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ max∞(λ n0. nσ[n0] = n) = ∞, nσ[n1 + 1] = n + 1

Rule argument

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

Additional information

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