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Proof step 8 from proof div-mult-two-nls-01 (previous step)

2 * (i0 * j + j0) = i1 * j + j1, j1 < j, j0 < j, 1 + 2 * i0 < i1, 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j1, 0 ≤ i1 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i * (j + j0) = i * j + i * j0
Proof step 10:
2 * i0 * j + 2 * j0 = i1 * j + j1, j1 < j, j0 < j, 1 + 2 * i0 < i1, 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j1, 0 ≤ i1 ⊦

Rule argument

spec: int-mult inst: name: dist-rsplit seq: ⊦ i * (j + j0) = i * j + i * j0 subst: [j0, j, i] → [j0, i0 * j, 2] flags: rewrite

Additional information

Used terms: j0, i0 * j, 2