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Proof step 23 from proof byte-mult-256 (previous step)

~256 ≤ j0, 0 < i, ~256 < j0, j0 < 0 ⊦ i = 0, i2b(256 + j0) = i2b(j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ i2b(i) = i2b(j) ↔ b2i(i2b(i)) = b2i(i2b(j))
Proof step 25:
~256 ≤ j0, 0 < i, ~256 < j0, j0 < 0 ⊦ i = 0, b2i(i2b(256 + j0)) = b2i(i2b(j0))

Rule argument

name: i2b-equal rotate: F subst: [j, i] → [j0, 256 + j0]

Additional information

Used terms: j0, 256 + j0