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Proof step 30 from proof min-skolemf (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: m0

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used terms: m0