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Proof step 3 from proof div-mult-three-nls (previous step)

(j * j * j) / 3 < j, 0 < j, 2 ≤ j, 0 ≤ j ⊦

User 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 ∧ 0 ≤ i0)
Proof step 5:
j * j * j = i0 * 3 + j0, 0 < j, i0 < j, j0 < 3, 0 < 3, 0 ≤ j, 2 ≤ j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j * j * j ⊦

Rule argument

spec: <toplevel> 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] → [j * j * j, 3, i, j0]

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 Used terms: j * j * j, 3, i, j0