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Proof step 2 from proof maxinf-skolemf (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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