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Proof step 3 from proof nummult-ls (previous step)

0 < i, 0 ≤ j ⊦ i * i0 < j ↔ i0 < (i + j - 1) / i

User applied rule case distinction

And obtained the following new goals

Proof step 4:
i * i0 < j, 0 < i, 0 ≤ j ⊦ i0 < (i + j - 1) / i
Proof step 15:
i0 < (i + j - 1) / i, 0 < i, 0 ≤ j ⊦ i * i0 < j

Rule argument

pos: right 1

Additional information