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Proof step 2 from proof div-add1-negpos-neg-01 (previous step)

i % j = 0, j + i ≤ 0, (i + i0) + 1 < j, i < 0, i < j, i + 1 < j, 0 ≤ i0, 0 < j, i + i0 < 0, i + i0 < j ⊦ j = 1, (i + i0) / j = i / j + i0 / j + 1, i0 % j = 0

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:
i = i1 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i1 ≤ 0, j0 = 0, j + i ≤ 0, (i + i0) + 1 < j, i < 0, i < j, i + 1 < j, 0 ≤ i0, 0 < j, i + i0 < 0, i + i0 < j ⊦ j = 1, (i + i0) / j = i1 + i0 / j + 1, i0 % j = 0

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] → [i, j, i1, j0] flags: rewrite, current

Additional information

Used simplifier rules: zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: i, j, i1, j0