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Proof step 10 from proof min-skolemf (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
mon<(nσ), ∀ n0, nσ. n0 < 0 → mon<(nσ) → nσ = [] ∨ (∃ m0. ⌜m0⌝ < # nσ ∧ n0 ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n0 ≤ nσ[m] → m0 ≤ m)) ∨ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n0 ⊦ nσ = [], ∃ m0. ⌜m0⌝ < # nσ ∧ (∀ m. ⌜m⌝ < # nσ → m0 ≤ m)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ ¬ m < 0; (s)