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Proof step 9 from proof bytes2int-rec (previous step)

¬(¬ (bytes2bits(bytes) + byte2bits(by)).first ∧ bytes2bits(bytes) + byte2bits(by) ≠ []) ⊦ bytes = [], bits2int(bytes2bits(bytes) + byte2bits(by)) = bits2int(bytes2bits(bytes)) * 256 + band(b2i(by), 255)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(bytes2bits(bytes) + byte2bits(by)).first ⊦ bytes = [], bits2int(bytes2bits(bytes) + byte2bits(by)) = bits2int(bytes2bits(bytes)) * 256 + band(b2i(by), 255)

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];