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Proof step 5 from proof byte-law-mult (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i2b(i * j) = i2b(b2i(i2b(i)) * j)

Rule argument

name: byte-law-mult-one rotate: F subst: [j, i] → [b2i(i2b(i)), j]

Additional information

Used simplifier rules: ⊦ i * j = j * i; Used terms: b2i(i2b(i)), j