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Proof step 58 from proof inv-get-inverse-between (previous step)

n2 = m1 + 1, nσ[n2] = nσ[n1] + 1, # nσ = ∞, n1 + m + 1 < n2, stutf(nσ), ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), nσ[m1] ≤ nσ[n2], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∀ m. nσ[m] = nσ[n2] → ¬ n2 < m, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n2], n2 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 59:
nσ[m1 + 1] = nσ[n1] + 1, # nσ = ∞, stutf(nσ), ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), nσ[m1] ≤ nσ[n1] + 1, n1 + m < m1, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∀ m. nσ[m] = nσ[n1] + 1 → ¬ m1 + 1 < m, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n1] + 1, m1 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2