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Proof step 8 from proof bytes2int-pos (previous step)

bytes = by + bytes0 ⊦ bytes = [], ¬ bytes2bits(bytes).first ↔ ¬ b2i(by) < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ ¬ (byte2bits(by) + bytes2bits(bytes0)).first ↔ ¬ b2i(by) < 0

Rule argument

Additional information

Used simplifier rules: bytes2bits-rec: ⊦ bytes2bits(by ' + bytes) = byte2bits(by) + bytes2bits(bytes);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' + bytes ≠ [];% ⊦ by + bytes = by ' + bytes;