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Proof step 2 from proof div-mult-02 (previous step)

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

User 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:
i * i0 = i1 * j * i0 + j0, 0 < j, 0 < i0, j0 < j * i0, 0 < j * i0, 0 ≤ i, 0 ≤ j0, 0 ≤ i * i0 ⊦ i1 = i / j

Rule argument

spec: <toplevel> 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] → [i * i0, j * i0, i1, j0] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s) Used terms: i * i0, j * i0, i1, j0