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Proof step 14 from proof max-a (previous step)

n < m, n0 ≤ m, n < n0 ⊦ max(m, n) < n0, n0 = 0, m = max(max(m, n), n0), m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
n < max(m, n), n0 ≤ max(m, n), n < n0, n0 ≤ m, n < m ⊦ m = max(m, n), n0 = 0, m = 0, max(m, n) = 0

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat