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Proof step 5 from proof maxinf-inf (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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