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Proof step 3 from proof maxinf-skolemf (previous step)

np(n0) ⊦ ∃ n0. n < n0 ∧ np(n0), ∃ n∞. (n∞ = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));np(n∞.n) ∧ (∀ m. np(m) → ¬ n∞ < ⌜m⌝ ))

User applied rule cut formula

And obtained the following new goals

Proof step 4:
∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0)), np(n0) ⊦ ∃ n0. n < n0 ∧ np(n0), ∃ n∞. (n∞ = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));np(n∞.n) ∧ (∀ m. np(m) → ¬ n∞ < ⌜m⌝ ))
Proof step 8:
np(n0) ⊦ ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0)), ∃ n0. n < n0 ∧ np(n0), ∃ n∞. (n∞ = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));np(n∞.n) ∧ (∀ m. np(m) → ¬ n∞ < ⌜m⌝ ))

Rule argument

fma: ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))

Additional information

Used formulas: ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0))