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Proof step 17 from proof fits-law-div-stronger (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
b2i(by1) < 0, b2i(by1) / i + b2i(by1) ≤ 0, b2i(by1) < ~127, 0 < b2i(by1) / i, 127 < b2i(by1) / i ⊦ b2i(by1) = 0, b2i(by1) / i = b2i(i2b(b2i(by1) / i)), i = 0, i = ~1

Rule argument

Additional information

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