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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
max∞(np) = ∞, np(n), (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n)), ∀ m. np(m) → ¬ n < m ⊦

Rule argument

Additional information

Used simplifier rules: disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic