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Proof step 3 from proof max-a (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
n < n0, m < max(n, n0) ⊦ max(n, n0) = max(max(m, n), n0), max(n, n0) = 0

Rule argument

fma: n < n0

Additional information

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