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Proof step 6 from proof min-skolemf (previous step)

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

User applied rule induction

And obtained the following new goals

Proof step 7:
mon<(nσ), ∀ n0, nσ. n0 < n → mon<(nσ) → nσ = [] ∨ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n0 ∨ (∃ m0. ⌜m0⌝ < # nσ ∧ n0 ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n0 ≤ nσ[m] → m0 ≤ m)) ⊦ nσ = [] ∨ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n ∨ (∃ m0. ⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → m0 ≤ m))

Rule argument

indvar: n wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: n