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Proof step 11 from proof div-mult-three-nls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: numlslsdel-01: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult