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Proof step 5 from proof div-add-neg (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j)
Proof step 7:
i0 = i2 * j + j1 ∧ j1 ≤ 0 ∧ abs(j1) < j, j1 = 0, i1 * j + j0 < j, j0 < j, i0 < 0, i0 < j, i0 + 1 < j, i1 * j + j0 < 0, 0 < j, abs(j0) < j, j0 ≤ 0, (i1 * j + j0) + 1 < j ⊦ ((i1 * j + j0) + i0) / j = i1 + i2

Rule argument

spec: int-div inst: name: div-elim2 seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j) subst: [i, j, i0, j0] → [i0, j, i, j1] flags: rewrite, current

Additional information

Used terms: i0, j, i, j1