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Proof step 4 from proof div-ls-zero (previous step)

0 < j, i < 0 ⊦ i / j < 0 ↔ i + j ≤ 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 6:
i = i0 * j + j0, i < 0, 0 < j, abs(j0) < j, j0 ≤ 0, i0 ≤ 0 ⊦ i0 < 0 ↔ i + j ≤ 0

Rule argument

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

Additional information

Used terms: i, j, i0, j0