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Proof step 1 from proof byte-is-cast-01 (conclusion)

⊦ i2b(i) = i2b(cast(8, i))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ i2b(i) = i2b(j) ↔ b2i(i2b(i)) = b2i(i2b(j))
Proof step 3:
⊦ b2i(i2b(i)) = b2i(i2b(cast(8, i)))

Rule argument

name: i2b-equal rotate: F subst: [j, i] → [cast(8, i), i]

Additional information

Used terms: cast(8, i), i