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Proof step 15 from proof div-mult-three-nls (previous step)

0 < j, 2 < j, i0 < j, j * j * j < 3 * (j + 1), 0 ≤ i0, 0 ≤ j0, 0 ≤ j, j0 < 3 ⊦ 3 * (j + 1) < j * (j + 1), j = 3, j = 2

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: ls-pos-01 seq: ⊦ 0 < j → (i * j < i0 * j ↔ i < i0) subst: [i0, j, i] → [j, j + 1, 3] flags: rewrite

Additional information

Used simplifier rules: lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%num-between: i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs Used terms: j, j + 1, 3