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Proof step 2 from proof div-eq-zero (previous step)

0 < i, 0 < j ⊦ i / j = 0 ↔ i < j

User applied rule case distinction

And obtained the following new goals

Proof step 3:
i / j = 0, 0 < i, 0 < j ⊦ i < j
Proof step 6:
i < j, 0 < i, 0 < j ⊦ i / j = 0

Rule argument

pos: right 1

Additional information