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Proof step 32 from proof min-stutters (previous step)

mon<(nσ) ∧ min(nσ, n) < min(nσ, n) + 1 ∧ ⌜min(nσ, n) + 1⌝ < # nσ → nσ[min(nσ, n)] < nσ[min(nσ, n) + 1], min(nσ, n) + 1 < min(nσ, n + 1), min(nσ, n) < min(nσ, n + 1), ⌜min(nσ, n + 1)⌝ < # nσ, n < nσ[min(nσ, n + 1)], mon<(nσ), n ≤ nσ[min(nσ, n)], ⌜min(nσ, n)⌝ < # nσ, # nσ = ∞ ∨ ¬ nσ[(# nσ).n - 1] < n, # nσ = ∞ ∨ n < nσ[(# nσ).n - 1], ∀ m. ⌜m⌝ < # nσ ∧ n < nσ[m] → min(nσ, n + 1) ≤ m, ∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m ⊦ min(nσ, n + 1) = 1 + min(nσ, n), min(nσ, n + 1) = min(nσ, n), nσ = [], n < nσ[min(nσ, n) + 1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%sum-notinf: ⊦ n∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); 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%ls-notlssucc: ⊦ ⌜m⌝ < n∞ → (¬ ⌜m⌝ + ⌜1⌝ < n∞ ↔ ⌜m⌝ + ⌜1⌝ = n∞); from specification nat-inf%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%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%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2