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Proof step 17 from proof maxinf-fin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
max∞(np) = ⌜n0⌝ , np(n), np(n0), n0 < n, ∀ m. np(m) → ¬ n0 < m ⊦ n0 = n, n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1