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Proof step 20 from proof div-div (previous step)

(i / j) / abs(j0) = i / (j * abs(j0)), i < j, i < 0, 0 < j, i + 1 < j ⊦ j0 = 0, (i / j) / j0 = i / (j * j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ i * j = j * i
Proof step 22:
(i / j) / abs(j0) = i / (abs(j0) * j), i < j, i < 0, 0 < j, i + 1 < j ⊦ j0 = 0, (i / j) / j0 = i / (j * j0)

Rule argument

spec: int-mult inst: name: commutativity seq: ⊦ i * j = j * i subst: [j, i] → [abs(j0), j] flags: rewrite

Additional information

Used terms: abs(j0), j