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Proof step 19 from proof less-mult-div-01 (previous step)

0 < i0 * j + j0, 0 < j, 0 < j1, i1 < j, j0 < j, 0 < (i1 * (i0 * j + j0)) / j, 0 ≤ (i1 * (i0 * j + j0)) / j, 0 ≤ i0 * j + j0, i0 * j + j0 ≤ (i1 * (i0 * j + j0)) / j, 0 ≤ i0, 0 ≤ j0 ⊦ (j0 = 0 ∨ 0 ≤ i1) ∧ (i0 = 0 ∨ 0 ≤ i1)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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