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Proof step 2 from proof div-le-modzero (previous step)

i % j = 0, 0 < j ⊦ i0 ≤ i / j ↔ i0 * j ≤ i

User applied rule case distinction

And obtained the following new goals

Proof step 3:
i0 ≤ i / j, i % j = 0, 0 < j ⊦ i0 * j ≤ i
Proof step 6:
i0 * j ≤ i, i % j = 0, 0 < j ⊦ i0 ≤ i / j

Rule argument

pos: right 1

Additional information