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Proof step 89 from proof byte-mult-256 (previous step)

i0 + i = ~1, 0 ≤ i0, i < 0, ~256 ≤ i1, i1 < 0, i1 < ~128, ~256 < i1 ⊦ i1 + 256 = b2i(i2b(i1 + 256)), i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 90:
⊦ 0 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i

Rule argument

name: byte-in rotate: F subst: [i] → [i1 + 256]

Additional information

Used simplifier rules: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ ¬ i ≤ j ↔ j < i;%i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; Used terms: i1 + 256