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Proof step 21 from proof byte-law-neg (previous step)

¬(~128 ≤ ~(i % 256) ∧ ~(i % 256) ≤ 127), ~128 ≤ i % 256, i < ~128, i < 0, i < i % 256 ⊦ b2i(i2b(~ i)) = b2i(i2b(~(i % 256)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
i % 256 < ~127, i < i % 256, i < 0, i < ~128, ~128 ≤ i % 256 ⊦ i2b(~ i) = i2b(~(i % 256)), i % 256 = 0

Rule argument

Additional information

Used simplifier rules: byte-equal: ⊦ b2i(byi) = b2i(byj) ↔ byi = byj;% ⊦ i < 0 ∧ 0 ≤ j → i < j;%0 < j ⊦ i % (* j) < 0 ↔ i < 0 ∧ i % j ≠ 0;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ 0 < j ∧ i < 0 → ¬ 0 < i % j;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i - j ≤ i0 ↔ i ≤ j + i0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;