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Proof step 39 from proof bytes2int-rec (previous step)

¬ bytes2bits(bytes).first ∧ bytes2bits(bytes) ≠ [] ⊦ bytes2bits(bytes) + byte2bits(by) = [], bits2uint(bytes2bits(bytes) + byte2bits(by)) = bits2uint(bytes2bits(bytes)) * 256 + band(b2i(by), 255), bytes = [], (bytes2bits(bytes) + byte2bits(by)).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
⊦ bytes2bits(bytes) + byte2bits(by) = [], bits2uint(bytes2bits(bytes) + byte2bits(by)) = bits2uint(bytes2bits(bytes)) * 256 + band(b2i(by), 255), bytes = [], (bytes2bits(bytes) + byte2bits(by)).first, bytes2bits(bytes).first

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];