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Proof step 201 from proof inv-get-inverse-between (previous step)

# nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n] = nσ[n1] + 1, nσ[n1 + 1] < nσ[n], nσ[n1 + 1] < nσ[m0], n1 + m + 1 < m0, n1 + m + 1 < n, nσ[n1] < nσ[m0], n1 + m < n, nσ[n1 + m + 1] < nσ[n], n1 + m < m0, nσ[n1 + m + 1] < nσ[m0], stutf(nσ), nσ[n1 + 1] ≤ nσ[n1], nσ[n1 + m + 1] ≤ nσ[n1], nσ[n] ≤ nσ[m0], n ≤ m0, n1 ≤ m0, nσ[n1 + 1] ≤ nσ[n1 + m + 1], mon≤(nσ) ∧ m0 ≤ n1 + 1 ∧ ⌜n1 + 1⌝ < # nσ → nσ[m0] ≤ nσ[n1 + 1], ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n] → ¬ n < m ⊦ nσ[n1 + m + 1] = nσ[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 202:
# nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n] = nσ[n1] + 1, nσ[n1 + m + 1] < nσ[m0], n1 + m < m0, nσ[n1 + m + 1] < nσ[n], n1 + m < n, nσ[n1] < nσ[m0], n1 + m + 1 < n, n1 + m + 1 < m0, nσ[n1 + 1] < nσ[m0], nσ[n1 + 1] < nσ[n], stutf(nσ), nσ[n1 + 1] ≤ nσ[n1 + m + 1], n1 ≤ m0, n ≤ m0, nσ[n] ≤ nσ[m0], nσ[n1 + m + 1] ≤ nσ[n1], nσ[n1 + 1] ≤ nσ[n1], ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n] → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n]

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true);% ⊦ m < n → m ≤ n; (s)% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ n∞0 + ⌜1⌝ = n∞ → (⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0); (s)%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)