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

i < j, i < 0, 0 < j, i + 1 < j ⊦ i < i / j ↔ 1 < j

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 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0, i < j, i < 0, 0 < j, i + 1 < j ⊦ i < i0 ↔ 1 < 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, 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, i0, j0