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Proof step 10 from proof associativity (previous step)

i * (j * i0 + j) = (i * j) * (i0 + 1), ∀ i1, j0. i1 * j0 * (i0 + 1) = (i1 * j0) * (i0 + 1) ⊦ i * j * i0 = (i * j) * i0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ i * (j + j0) = i * j + i * j0
Proof step 12:
i * j * i0 + i * j = (i * j) * (i0 + 1), ∀ i1, j0. i1 * j0 * (i0 + 1) = (i1 * j0) * (i0 + 1) ⊦ i * j * i0 = (i * j) * i0

Rule argument

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

Additional information

Used terms: j, j * i0, i