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Proof step 20 from proof min-skolemf (previous step)

⌜m0⌝ < # nσ, n1 < nσ[m0], mon<(nσ), ⌜m⌝ < # nσ, n1 < nσ[m], m < m0, ∀ m. ⌜m⌝ < # nσ ∧ n1 ≤ nσ[m] → m0 ≤ m, ∀ 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 = [], n1 = nσ[m0], nσ[m0] = 0, nσ[m] = 0, m0 = 0, # nσ ≠ ∞ ∧ ¬ n1 < nσ[(# nσ).n - 1], ∃ m0. ⌜m0⌝ < # nσ ∧ n1 < nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n1 < nσ[m] → m0 ≤ m)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: m

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used terms: m