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Proof step 1 from proof max-a (conclusion)

⊦ max(i, max(j, j0)) = max(max(i, j), j0)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 2:
i < max(j, j0) ⊦ max(i, max(j, j0)) = max(max(i, j), j0)
Proof step 6:
⊦ i < max(j, j0), max(i, max(j, j0)) = max(max(i, j), j0)

Rule argument

fma: i < max(j, j0)

Additional information

Used formulas: i < max(j, j0)