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Proof step 10 from proof maxinf-inf (previous step)

∃ n0. n1 < n0 ∧ np(n0), max∞(np) = ⌜n1⌝ , np(n1), ∀ m. np(m) → ¬ n1 < m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
max∞(np) = ⌜n1⌝ , n1 < n, np(n), np(n1), ∀ m. np(m) → ¬ n1 < m ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1