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Proof step 46 from proof inv-get-inverse-between (previous step)

# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , n1 + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, stutf(nσ), ⌜n⌝ + ⌜1⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦ max∞(λ n0. nσ[n0] = n + 1) = ∞, nσ[n1 + 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 47:
max∞(λ n0. nσ[n0] = n + 1) = ⌜n2⌝ , # nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , n1 + m + 1 < n2, stutf(nσ), ⌜n⌝ + ⌜1⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦ max∞(λ n0. nσ[n0] = n + 1) = ∞, nσ[n1 + m + 1] = n + 1, n2 = 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 + 1), m2] flags:

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. nσ[n0] = n + 1), m2