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Proof step 200 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic2%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%less-succ-leq: ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; from specification nat-inf%eqls-succ: ⊦ n∞0 + ⌜1⌝ = n∞ → (⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0); from specification nat-inf%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%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%les: ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); from specification nat