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Proof step 4 from proof div-rec (previous step)

0 < j, 0 < i0 + j, 0 ≤ i0 ⊦ (i0 + j) / j = i0 / j + 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 6:
i0 = i1 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j, 0 < j, 0 < i0 + j, 0 ≤ i0 ⊦ (i0 + j) / j = i1 + 1

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, i, j0] flags: rewrite, current

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: i0, j, i, j0