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Proof step 7 from proof shiftright-byte-8-neg (previous step)

b2i(by) % 256 ≠ 0 ∧ b2i(by) < 0, b2i(by) < 0 ⊦ b2i(by) / 256 - 1 = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
b2i(by) < 0 ⊦ b2i(by) / 256 = 0, b2i(by) = 0

Rule argument

Additional information

Used simplifier rules: byte-mod-100x: ⊦ b2i(by) % 256 = b2i(by);% ⊦ i - j = i0 ↔ i = i0 + j;