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Proof step 1 from proof min-a (conclusion)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 2:
i < min(j, j0) ⊦ min(i, min(j, j0)) = min(min(i, j), j0)
Proof step 4:
⊦ min(i, min(j, j0)) = min(min(i, j), j0), i < min(j, j0)

Rule argument

fma: i < min(j, j0)

Additional information

Used formulas: i < min(j, j0)