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Proof step 4 from proof shiftright-byte-8-neg (previous step)

b2i(by) < 0 ∧ 8 ≤ 8, b2i(by) < 0 ⊦ (b2i(by) - 255) / 256 >> (8 - 8) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
b2i(by) < 0 ⊦ (b2i(by) - 255) / 256 = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ i >> 0 = i;