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Proof step 6 from proof bytes2int-zero-02 (previous step)

⊦ bits2int(byte2bits(i2b(0)) + byte2bits(i2b(0)) + bytes2bits(bytes)) = bits2int(bytes2bits(i2b(0) ' + bytes))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ bits2int(byte2bits(i2b(0)) + byte2bits(i2b(0)) + bytes2bits(bytes)) = bits2int(byte2bits(i2b(0)) + bytes2bits(bytes))

Rule argument

Additional information

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