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Proof step 51 from proof byte-law-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 52:
i % 256 = 128, 127 < i, 0 < i ⊦ 0 = i, i2b(~ i) = i2b(128)

Rule argument

Additional information

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