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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information