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Proof step 2 from proof mod-add-01 (previous step)

i + j ≤ 0, j < 0, 0 < i ⊦ (i + j) % i = j % i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 4:
j = i0 * i + j0 ∧ j0 ≤ 0 ∧ abs(j0) < i ∧ i0 ≤ 0, i + j ≤ 0, j < 0, 0 < i ⊦ (i + j) % i = j0

Rule argument

spec: int-div inst: 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] → [j, i, i0, j0] flags: rewrite, current

Additional information

Used terms: j, i, i0, j0