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Proof step 2 from proof byte-mult (previous step)

⊦ i2b(i * 256) = i2b(0)

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(i * 256)) = b2i(i2b(0))

Rule argument

name: i2b-equal rotate: F subst: [j, i] → [0, i * 256]

Additional information

Used terms: 0, i * 256