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Proof step 4 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 5:
⊦ min(j, j0) = min(min(i, j), j0), i < min(j, j0)

Rule argument

Additional information

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