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Proof step 5 from proof div-add1-negpos-neg-01 (previous step)

i1 < j, i1 < 0, j + i1 * j ≤ 0, 0 ≤ i0, 1 + i1 * j < j, 1 + i1 * j + i0 < j, 0 < j, i1 * j + i0 < j, i1 * j + i0 < 0, i1 + 1 < j ⊦ (i1 * j + i0) / j = i1 + i0 / j + 1, j = 1, i0 % j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)
Proof step 7:
i0 = i2 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i2, i1 < j, i1 < 0, j + i1 * j ≤ 0, 0 ≤ i0, 1 + i1 * j < j, 1 + i1 * j + i0 < j, 0 < j, i1 * j + i0 < j, i1 * j + i0 < 0, i1 + 1 < j ⊦ (i1 * j + i0) / j = i1 + i2 + 1, j = 1, j0 = 0

Rule argument

spec: int-div inst: name: div-elim-pospos seq: ⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0) subst: [i, j, i0, j0] → [i0, j, i, j0] flags: rewrite, current

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: i0, j, i, j0