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Proof step 7 from proof bytes2int-mone (previous step)

b2i(by) < 0 ⊦ bits2int(byte2bits(i2b(~1)) + byte2bits(by) + bytes2bits(bytes)) = bits2int(bytes2bits(by ' + bytes))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
b2i(by) < 0 ⊦ bits2int(byte2bits(i2b(~1)) + byte2bits(by) + bytes2bits(bytes)) = bits2int(byte2bits(by) + bytes2bits(bytes))

Rule argument

Additional information

Used simplifier rules: bytes2bits-rec: ⊦ bytes2bits(by ' + bytes) = byte2bits(by) + bytes2bits(bytes);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;