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Proof step 14 from proof fits-law-div-stronger (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
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

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)%i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)