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Proof step 68 from proof byte-mult-256 (previous step)

i0 + i < 0, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i0 + i = 0, i = 0, i2b(256 * (i0 + i) + j0) = i2b(j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 69:
⊦ i2b(i) = i2b(j) ↔ b2i(i2b(i)) = b2i(i2b(j))
Proof step 70:
i0 + i < 0, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i0 + i = 0, i = 0, b2i(i2b(256 * (i0 + i) + j0)) = b2i(i2b(j0))

Rule argument

name: i2b-equal rotate: F subst: [j, i] → [j0, 256 * (i0 + i) + j0]

Additional information

Used terms: j0, 256 * (i0 + i) + j0