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Proof step 8 from proof div-est-num (previous step)

i1 * j + j1 < j, i1 * j + j1 < 0, i1 < i / j, i1 < j, i1 ≤ 0, j1 < j, j1 ≤ 0, i + 1 < j, i < j, i < 0, 0 < j, j0 ≤ i / j, i ≤ i1 * j + j1, i1 < j0, (i1 * j + j1) + 1 < j, abs(j1) < j ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 10:
i = i0 * j + j2 ∧ j2 ≤ 0 ∧ abs(j2) < j ∧ i0 ≤ 0, i1 * j + j1 < j, i1 * j + j1 < 0, i1 < i0, i1 < j, i1 ≤ 0, j1 < j, j1 ≤ 0, i + 1 < j, i < j, i < 0, 0 < j, j0 ≤ i0, i ≤ i1 * j + j1, i1 < j0, (i1 * j + j1) + 1 < j, abs(j1) < j ⊦

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, i0, j2] 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, i0, j2