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Proof step 9 from proof byte-law-neg (previous step)

¬(i < 0 ∧ ~128 ≤ i % 256), i < ~128, i < 0 ⊦ i2b(~ i) = i2b(~ b2i(i2b(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i % 256 < ~128, i < 0, i < ~128 ⊦ i2b(~ i) = i2b(~(i % 256)), i % 256 = 0

Rule argument

Additional information

Used simplifier rules: bib-neg-low: ⊦ i < 0 ∧ i % 256 < ~128 → b2i(i2b(i)) = i % 256 + 256;%byte-add-256-01: ⊦ i2b(~(i + 256)) = i2b(~ i);% ⊦ i < 0 ∧ 0 ≤ j → i < j;%0 < j ⊦ i % (* j) < 0 ↔ i < 0 ∧ i % j ≠ 0;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;