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Proof step 10 from proof dist-rsplit (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ i * (j + 1) = i * j + i

Rule argument

spec: inst: name: succ-rsplit seq: ⊦ i * (j + 1) = i * j + i subst: [j, i] → [j + i0, i] flags: rewrite, current

Additional information

Used simplifier rules: add-in: ⊦ i + j = i + i0 ↔ j = i0; 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 Used terms: j + i0, i