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Proof step 17 from proof inv-get-inverse-between (previous step)

∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n, # nσ = ∞, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ max∞(λ n0. nσ[n0] = n + 1).n = 0, nσ[∞.n + m + 1] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# nσ = ∞, ⌜nσ[m0]⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ[n0] = nσ[m0] + 1).n, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ max∞(λ n0. nσ[n0] = nσ[m0] + 1).n = 0, nσ[∞.n + m + 1] = nσ[m0] + 1

Rule argument

Additional information

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