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Proof step 14 from proof min-skolemf (previous step)

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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
mon<(nσ), ⌜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]

Proof step 14 from proof min-skolemf (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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