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Proof step 72 from proof bytes2int-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
8 * # bytes0 = 0 ⊦ bytes2bits(bytes) + bytes2bits(bytes0) = [], bits2uint(bytes2bits(bytes) + bytes2bits(bytes0)) = bits2uint(bytes2bits(bytes)) * 256 ^ n→i(# bytes0) + band(bits2int(bytes2bits(bytes0)), 256 ^ n→i(# bytes0) - 1), bytes = [], bytes0 = [], bytes2bits(bytes).first

Rule argument

Additional information

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