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Proof step 3 from proof byte-law-mult (previous step)

⊦ i2b(b2i(i2b(i)) * j) = i2b(b2i(i2b(i)) * b2i(i2b(j)))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ i * j = j * i subst: [j, i] → [j, b2i(i2b(i))] rotate: F spec: int-mult inst: name: commutativity

Additional information

Used terms: j, b2i(i2b(i))