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Proof step 20 from proof fits-law-div-stronger (previous step)

b2i(by1) < 0, b2i(by1) / i + b2i(by1) ≤ 0, b2i(by1) + 1 ≤ ~127, 0 < b2i(by1) / i, 127 < b2i(by1) / i ⊦ b2i(by1) = 0, b2i(by1) / i = b2i(i2b(b2i(by1) / i)), i = 0, i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
~128 / i = 128, b2i(by1) = ~128 ⊦ i = 0, i = ~1

Rule argument

Additional information

Used simplifier rules: byte-min: ⊦ b2i(by) ≤ ~128 ↔ b2i(by) = ~128;%bib-pos-high: ⊦ 0 ≤ i ∧ 127 < i % 256 → b2i(i2b(i)) = i % 256 - 256;%i + 1 = i0 ⊦ (* i) < j → (j ≤ (* i0) ↔ j = i0);%i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i - j ≤ i0 ↔ i ≤ j + i0;% ⊦ 0 < i → i1 < i1 + i;% ⊦ i - j < i0 ↔ i < j + i0;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;