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Proof step 6 from proof maxinf-fin (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 7:
max∞(np) = ∞, np(n), ∀ m. np(m) → ¬ n < m, ∀ n. ∃ n0. n < n0 ∧ np(n0) ⊦
Proof step 10:
max∞(np) = ∞, np(n), ∀ m. np(m) → ¬ n < m, ∀ n. ¬ np(n) ⊦

Rule argument

pos: left 4

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