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Proof step 11 from proof min-skolemf (previous step)

n = n1 +1, mon<(nσ), ∀ n0, nσ. n0 < n → mon<(nσ) → nσ = [] ∨ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n0 ∨ (∃ m0. ⌜m0⌝ < # nσ ∧ n0 ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n0 ≤ nσ[m] → m0 ≤ m)) ⊦ nσ = [], # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n, ∃ m0. ⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → m0 ≤ m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
mon<(nσ), ∀ n0, nσ. n0 < n1 +1 → mon<(nσ) → nσ = [] ∨ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n0 ∨ (∃ m0. ⌜m0⌝ < # nσ ∧ n0 ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n0 ≤ nσ[m] → m0 ≤ m)) ⊦ nσ = [], # nσ ≠ ∞ ∧ ¬ n1 < nσ[(# nσ).n - 1], ∃ m0. ⌜m0⌝ < # nσ ∧ n1 < nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n1 < nσ[m] → m0 ≤ m)

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2