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Proof step 9 from proof min-skolemf (previous step)

n = 0, 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 10:
mon<(nσ), ∀ n0, nσ. n0 < 0 → mon<(nσ) → nσ = [] ∨ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n0 ∨ (∃ m0. ⌜m0⌝ < # nσ ∧ n0 ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n0 ≤ nσ[m] → m0 ≤ m)) ⊦ nσ = [], ∃ m0. ⌜m0⌝ < # nσ ∧ (∀ m. ⌜m⌝ < # nσ → m0 ≤ m)

Rule argument

Additional information

Used simplifier rules: lez: ⊦ 0 ≤ m; from specification nat%zero-01: ⊦ ¬ m < 0; from specification nat-basic1