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Proof step 3 from proof byte2bits-val (previous step)

¬(¬ byte2bits(by).first ∧ byte2bits(by) ≠ []) ⊦ bits2uint(byte2bits(by)) = band(b2i(by), 255)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
b2i(by) < 0 ⊦ bits2uint(byte2bits(by)) = b2i(by) + 256

Rule argument

Additional information

Used simplifier rules: byte2bits-first: ⊦ byte2bits(by).first ↔ b2i(by) < 0;%byte2bits-empty: ⊦ byte2bits(by) ≠ [];% ⊦ band(i, j) = band(j, i);% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);% ⊦ b2i(by) < 0 → band(b2i(by), 255) = b2i(by) + 256;