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Proof step 5 from proof div-elim-posneg (previous step)

i = i1 * j + j1, j < 0, j1 < abs(j), 0 ≤ i, 0 ≤ j1 ⊦ i0 = i1 ∧ j0 = j1 ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < abs(j) ∧ i0 ≤ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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

Rule argument

Additional information

Used simplifier rules: ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)