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Proof step 33 from proof byte-law-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
~127 ≤ i % 256, i % 256 ≤ 128, i < i % 256, i < 0, i < ~128 ⊦ i % 256 + b2i(i2b(~ i)) = 0

Rule argument

Additional information

Used simplifier rules: j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i < ~ j ↔ i + j < 0;%j ≠ 0 ⊦ ~ i % j = ~(i % j);% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i + 0 = i;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ 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;% ⊦ ~ i = j ↔ i + j = 0;