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Proof step 4 from proof min-skolemf (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 5:
mon<(nσ) ⊦ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n, nσ = [], ∃ m0. (# nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n ⊃ m0 = (# nσ).n;⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → m0 ≤ m))

Rule argument

fma: # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n

Additional information

Used formulas: # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n