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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i < max(j, j0) ⊦ max(j, j0) = max(max(i, j), j0)

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ j < i → max(j, i) = i;