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Proof step 1 from proof min-skolemf (conclusion)

⊦ mon<(nσ) → (∃ m0. (nσ = [] ⊃ m0 = 0 ;(# nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n ⊃ m0 = (# nσ).n;⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → m0 ≤ m)) ))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
mon<(nσ) ⊦ ∃ m0. (nσ = [] ⊃ m0 = 0;(# nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n ⊃ m0 = (# nσ).n;⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → m0 ≤ m)))

Proof step 1 from proof min-skolemf (conclusion)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information