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Proof step 21 from proof min-skolemf (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: m

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: m