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Proof step 33 from proof inv-get-inverse-between (previous step)

max∞(λ n0. nσ[n0] = n + 1) = ∞, # 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 ⊦ nσ[n1 + m + 1] = n + 1, max∞(λ n0. nσ[n0] = n + 1).n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
max∞(λ n0. nσ[n0] = n + 1) = ∞, # nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , n1 + m + 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 ⊦ nσ[n1 + m + 1] = n + 1, ∞.n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1