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Proof step 105 from proof inv-get-inverse-between (previous step)

# nσ = ∞, stutf(nσ), n < nσ[m1], max∞(λ n0. nσ[n0] = n).n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ max∞(λ n0. nσ[n0] = n) = ∞, nσ[m1] = 0, nσ[max∞(λ n0. nσ[n0] = n).n + m + 1] = n + 1, max∞(λ n0. nσ[n0] = n + 1).n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 106:
max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , # nσ = ∞, stutf(nσ), n < nσ[m1], n1 + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ max∞(λ n0. nσ[n0] = n) = ∞, nσ[m1] = 0, nσ[n1 + m + 1] = n + 1, max∞(λ n0. nσ[n0] = n + 1).n = 0

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), m1] flags:

Additional information

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