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Proof step 5 from proof max-a (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
n < n0, m < n0 ⊦ max(m, n) < n0, n0 = max(max(m, n), n0), n0 = 0

Rule argument

fma: max(m, n) < n0

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; Used formulas: max(m, n) < n0