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Proof step 12 from proof associativity (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ i * (j + j0) = i * j + i * j0

Rule argument

spec: <toplevel> name: dist-rsplit seq: ⊦ i * (j + j0) = i * j + i * j0 subst: [j0, j, i] → [1, i0, i * j] flags: <all paths>

Additional information

Used simplifier rules: one: ⊦ 1 * i = i;%commutativity: ⊦ i * j = j * i;% ⊦ i + j = i + i0 ↔ j = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: 1, i0, i * j