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Proof step 2 from proof mult-02 (previous step)

0 ≤ i0, i0 < j, 0 ≤ i, 0 < j0, 0 < j ⊦ (i * j + i0) / (j0 * j) = i / j0

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)
Proof step 4:
i = i1 * j0 + j1 ∧ 0 ≤ j1 ∧ j1 < j0, 0 ≤ i0, i0 < j, 0 ≤ i, 0 < j0, 0 < j ⊦ (i * j + i0) / (j0 * j) = i1

Rule argument

spec: int-div inst: name: div-elim-pos seq: ⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j) subst: [i, j, i0, j0] → [i, j0, i1, j1] flags: rewrite, current

Additional information

Used terms: i, j0, i1, j1