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Proof step 10 from proof max-a (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n < m, n0 ≤ m, n < n0 ⊦ n0 = 0, m = max(max(m, n), n0), m = 0

Rule argument

Additional information

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