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Proof step 22 from proof bytes2int-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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