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Proof step 21 from proof min-stutters (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
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 = []

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%lels-03: ⊦ m < n → m ≤ n; from specification nat