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Proof step 9 from proof div-addthree-rew (previous step)

j0 < j, 0 < j, j1 < j, 0 ≤ i2, 0 ≤ j0, 0 ≤ i1, 0 ≤ j1, 0 ≤ i3 ⊦ (j * (i3 + i2) + j0 + j1 + i1) / j = i2 + i3 + i1 / j + (j0 + j1 + i1 % j) / j

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)
Proof step 11:
i1 = i0 * j + j2, j1 < j, 0 < j, j0 < j, j2 < j, 0 ≤ i3, 0 ≤ j1, 0 ≤ i1, 0 ≤ j0, 0 ≤ i2, 0 ≤ j2, 0 ≤ i0 ⊦ (j * (i3 + i2) + j0 + j1 + i1) / j = i2 + i3 + i0 + (j0 + j1 + j2) / j

Rule argument

spec: <toplevel> 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] → [i1, j, i0, j2] flags: <rotated> <all paths>

Additional information

Used terms: i1, j, i0, j2