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Proof step 2 from proof div-addthree-rew (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)
Proof step 4:
i = i2 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i2, 0 ≤ i, 0 ≤ i0, 0 ≤ i1, 0 < j ⊦ (i + i0 + i1) / j = i2 + i0 / j + i1 / j + (j0 + i0 % j + i1 % j) / j

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] → [i, j, i2, j0] flags: rewrite, current

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: i, j, i2, j0