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Proof step 4 from proof fits-law-div-stronger (previous step)

~ abs(b2i(by1)) ≤ b2i(by1) / i ∧ b2i(by1) / i ≤ abs(b2i(by1)) ⊦ 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 5:
b2i(by1) / i ≤ abs(b2i(by1)), 0 ≤ abs(b2i(by1)) + b2i(by1) / i ⊦ i = ~1, i = 0, b2i(by1) / i = b2i(i2b(b2i(by1) / i))

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;