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Proof step 9 from proof byte-law-add-one (previous step)

~256 < j0, i0 ≤ 0, j0 < 0, i0 * 256 + j0 < 0 ⊦ i2b(j0 + j) = i2b(b2i(i2b(j0)) + j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ ~256 < i ∧ i < ~128 → b2i(i2b(i)) = i + 256
Proof step 11:
¬(~256 < j0 ∧ j0 < ~128), ~256 < j0, i0 ≤ 0, j0 < 0, i0 * 256 + j0 < 0 ⊦ i2b(j0 + j) = i2b(b2i(i2b(j0)) + j)

Rule argument

name: bib-lower rotate: F subst: [i] → [j0]

Additional information

Used simplifier rules: byte-add-256: ⊦ i2b(i + 256) = i2b(i);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0; Used terms: j0