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Proof step 2 from proof mod-add-pos (previous step)

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

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] → [i0, j, i1, j0] flags: rewrite, current

Additional information

Used terms: i0, j, i1, j0