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Proof step 3 from proof nummult-modle (previous step)

j % i = 0, i < j, 0 < i, 0 < j, 1 < j ⊦ i * i0 ≤ j * j0 ↔ i0 ≤ j / i * j0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
⊦ 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, i, i1, j1] flags: rewrite, current

Additional information

Used simplifier rules: add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%le-pos: ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j, i, i1, j1