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Proof step 19 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 < j * 3 + 3, j = 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ i + j = j + i
Proof step 21:
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

Rule argument

spec: int-basic2 inst: name: add-c seq: ⊦ i + j = j + i subst: [j, i] → [3, j * 3] flags: rewrite

Additional information

Used terms: 3, j * 3