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Proof step 21 from proof div-mult-three-nls (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-basic2 inst: name: plst seq: i < j, ¬ j0 < i0 ⊦ i + i0 < j + j0 subst: [j0, j, i0, i] → [j * 3, 3, j * 3, j0] flags: rewrite

Additional information

Used simplifier rules: lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%numlelsdel: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); from specification int-abs%ls-pos: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); from specification int-mult Used terms: j * 3, 3, j * 3, j0