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Proof step 32 from proof min-skolemf (previous step)

⌜m⌝ < # nσ, nσ[m0] < nσ[m], ⌜m0⌝ + ⌜1⌝ < # nσ, mon<(nσ), m ≤ m0, # 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[m] = 0, # nσ ≠ ∞ ∧ ¬ nσ[m0] < nσ[(# nσ).n - 1], ∃ m. ⌜m⌝ < # nσ ∧ nσ[m0] < nσ[m] ∧ (∀ m1. ⌜m1⌝ < # nσ ∧ nσ[m0] < nσ[m1] → m ≤ m1)

User applied rule case distinction

And obtained the following new goals

Proof step 33:
⌜m0⌝ < # nσ ∧ (∀ m. ⌜m⌝ < # nσ ∧ nσ[m0] ≤ nσ[m] → m0 ≤ m), ⌜m⌝ < # nσ, nσ[m0] < nσ[m], ⌜m0⌝ + ⌜1⌝ < # nσ, mon<(nσ), m ≤ m0, ∀ 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[m] = 0, # 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

pos: left 6

Additional information

Used simplifier rules: ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1