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

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
max∞(np) = ⌜n1⌝ , np(n1), ∀ m. np(m) → ¬ max∞(np) < ⌜m⌝ , (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n)) ⊦ max∞(np) = ∞

Rule argument

spec: nat-inf-basic inst: name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [max∞(np), m0] flags:

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(np), m0