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Proof step 2 from proof max-a (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < max(n, n0) ⊦ max(n, n0) = max(max(m, n), n0), max(n, n0) = 0

Rule argument

Additional information

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