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Proof step 16 from proof bytes2int-append (previous step)

bytes2bits(bytes) ≠ [], bytes2bits(bytes).first ⊦ bytes2bits(bytes) + bytes2bits(bytes0) = [], bytes0 = [], bytes = [], bits2uint((bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)))+1) + bits2int(bytes2bits(bytes)) * 256 ^ n→i(# bytes0) + band(bits2int(bytes2bits(bytes0)), 256 ^ n→i(# bytes0) - 1) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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