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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
mon<(nσ), # 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)

Rule argument

Additional information

Used simplifier rules: ref-1: ⊦ ¬ n < n; from specification nat-basic1%ler: ⊦ m ≤ m; from specification nat