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Proof step 58 from proof byte-law-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 59:
0 < i % 256, 127 < i, 0 ≤ i, 0 < i, 128 < i % 256 ⊦ 256 = b2i(i2b(256 - i % 256)) + i % 256

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j - i0 ↔ i + i0 < j;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ ~ i < 0 ↔ 0 < i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;%j ≠ 0 ⊦ ~ i % j = ~(i % j);