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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: Min-left: ⊦ j < i → min(j, i) = j;