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Proof step 4 from proof div-mult-three-nls (previous step)

j * j * j = i0 * 3 + j0 ∧ 0 ≤ j0 ∧ j0 < 3 ∧ 0 ≤ i0 ∧ 0 < 3 ∧ 0 ≤ j * j * j, i0 < j, 2 ≤ j, 0 ≤ j, 0 < j ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult