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

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
j < j0, i < max(j, j0) ⊦ max(j, j0) = max(max(i, j), j0)

Rule argument

fma: j < j0

Additional information

Used simplifier rules: Max-left: ⊦ ¬ j < i → max(j, i) = j;%Max-right: ⊦ j < i → max(j, i) = i; Used formulas: j < j0