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Proof step 6 from proof div-elim-posneg (previous step)

j < 0, j1 + j < 0, 0 ≤ j1, 0 ≤ i1 * j + j1 ⊦ i0 = i1 ∧ j0 = j1 ↔ i1 * j + j1 = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 + j < 0 ∧ i0 ≤ 0

User applied rule case distinction

And obtained the following new goals

Proof step 7:
i0 = i1, j0 = j1, j1 + j < 0, j < 0, 0 ≤ i1 * j + j1, 0 ≤ j1 ⊦ i1 * j + j1 = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 + j < 0 ∧ i0 ≤ 0
Proof step 10:
i1 * j + j1 = i0 * j + j0, j1 + j < 0, j < 0, j0 + j < 0, 0 ≤ i1 * j + j1, 0 ≤ j1, 0 ≤ j0, i0 ≤ 0 ⊦ i0 = i1 ∧ j0 = j1

Rule argument

pos: right 1

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