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Proof step 13 from proof min-skolemf (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 14:
mon<(nσ), nσ = [] ∨ (∃ m0. ⌜m0⌝ < # nσ ∧ n1 ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n1 ≤ nσ[m] → m0 ≤ m)) ∨ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n1, ∀ 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 ⊦ nσ = [], ∃ m0. ⌜m0⌝ < # nσ ∧ n1 < nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n1 < nσ[m] → m0 ≤ m), # nσ ≠ ∞ ∧ ¬ n1 < nσ[(# nσ).n - 1]

Rule argument

subst: [n0, nσ] → [n1, nσ]

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1, nσ