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Proof step 14 from proof zero-04 (previous step)

i1 + j / i0 < j, i1 + j / i0 < i1 * i0 + j, 0 ≤ i1 * i0 + j, j0 < i0, 0 < j, i1 * i0 + j0 < i0, (i1 * i0 + j0) + 1 < i0, i1 + j / i0 < (i1 * i0 + j0) + j, i1 + j / i0 < i0, j0 ≤ 0, abs(j0) < i0, i1 * i0 + j0 < 0, 0 ≤ (i1 * i0 + j0) + j, i1 + j / i0 < 0, 0 < i0, (i1 + j / i0) + 1 < i0, j0 ≤ i1 * i0 + j, (i1 * i0 + j0) + 1 < j, i1 * i0 + j0 < j, j0 < j, (i1 + j / i0) + 1 < j ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
⊦ 0 < j ∧ 0 < i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j)

Rule argument

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

Additional information

Used simplifier rules: 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%modzero-03: ⊦ i < 0 ∧ j0 < j ∧ 0 < j → ¬ 0 ≤ i * j + j0; from specification int-mult Used terms: j, i0, i, j1