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Proof step 35 from proof min-skolemf (previous step)

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