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Proof step 3 from proof div-elim-posneg (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
⊦ j < 0 ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < abs(j))
Proof step 5:
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

Rule argument

spec: <toplevel> name: div-elim1 seq: ⊦ j < 0 ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < abs(j)) subst: [i, j, i0, j0] → [i, j, i1, j1] flags: <rotated> <all paths>

Additional information

Used terms: i, j, i1, j1