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Proof step 38 from proof min-skolemf (previous step)

⌜m0⌝ < # nσ, mon<(nσ), ∀ m. ⌜m⌝ < # nσ ∧ nσ[m0] ≤ nσ[m] → m0 ≤ m, ∀ n0, nσ0. n0 < nσ[m0] +1 → mon<(nσ0) → nσ0 = [] ∨ (∃ m0. ⌜m0⌝ < # nσ0 ∧ n0 ≤ nσ0[m0] ∧ (∀ m. ⌜m⌝ < # nσ0 ∧ n0 ≤ nσ0[m] → m0 ≤ m)) ∨ # nσ0 ≠ ∞ ∧ nσ0[(# nσ0).n - 1] < n0 ⊦ nσ = [], ⌜m0⌝ + ⌜1⌝ < # nσ, ∃ m. ⌜m⌝ < # nσ ∧ nσ[m0] < nσ[m] ∧ (∀ m1. ⌜m1⌝ < # nσ ∧ nσ[m0] < nσ[m1] → m ≤ m1), # nσ ≠ ∞ ∧ ¬ nσ[m0] < nσ[(# nσ).n - 1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
⌜m0⌝ + ⌜1⌝ = # nσ, nσ[m0] < nσ[(# nσ).n - 1], mon<(nσ), ∀ m. ¬ m0 < m ∧ nσ[m0] ≤ nσ[m] → m0 ≤ m, ∀ n0, nσ0. n0 < nσ[m0] +1 → mon<(nσ0) → nσ0 = [] ∨ (∃ m0. ⌜m0⌝ < # nσ0 ∧ n0 ≤ nσ0[m0] ∧ (∀ m. ⌜m⌝ < # nσ0 ∧ n0 ≤ nσ0[m] → m0 ≤ m)) ∨ # nσ0 ≠ ∞ ∧ nσ0[(# nσ0).n - 1] < n0 ⊦ nσ = [], ∃ m. ¬ m0 < m ∧ nσ[m0] < nσ[m] ∧ (∀ m1. ¬ m0 < m1 ∧ nσ[m0] < nσ[m1] → m ≤ m1)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ n∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (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% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ n∞ < n∞0 → (n∞ + ⌜1⌝ < n∞0 ↔ ¬(n∞ + ⌜1⌝ = n∞0 ∧ n∞ ≠ ∞)); (ss)