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Proof step 5 from proof div-add1-negpos-pos (previous step)

0 ≤ i1 * j + i0, i1 < 0, 0 ≤ i0, 0 < j ⊦ (i1 * j + i0) / j = i1 + i0 / j

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

Rule argument

spec: int-div inst: 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] → [i0, j, i, j0] flags: rewrite, current

Additional information

Used simplifier rules: div-mult-pos: ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ i0 < j → (j * i + i0) / j = i;%dist-rmerge: ⊦ i * j + i * j0 = i * (j + j0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-ge-add: ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); from specification int-mult%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2 Used terms: i0, j, i, j0