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Proof step 2 from proof byte-law-and-one (previous step)

⊦ i2b(band(i, j)) = i2b(band(b2i(i2b(i)), j)), band(i, j) = band(b2i(i2b(i)), j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i2b(i) = i2b(j) ↔ b2i(i2b(i)) = b2i(i2b(j))
Proof step 4:
⊦ b2i(i2b(band(i, j))) = b2i(i2b(band(b2i(i2b(i)), j))), band(i, j) = band(b2i(i2b(i)), j)

Rule argument

name: i2b-equal rotate: F subst: [j, i] → [band(b2i(i2b(i)), j), band(i, j)]

Additional information

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