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Proof step 28 from proof bytes2int-append (previous step)

¬ # bitsinvert(bytes2bits(bytes0)) ≠ 0, bytes2bits(bytes).first ⊦ bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)) = [], (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), bytes = [], bytes0 = [], bytes2bits(bytes) + bytes2bits(bytes0) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: bytes2bits-length: ⊦ # bytes2bits(bytes) = i→n(8) * # bytes;% ⊦ # bitsinvert(bits) = # bits;