How to prove a symmetric encryption scheme provides perfect secrecy? I learned in class that in order to achieve perfect secrecy, the source of the plaintext $\mathcal{P}$ needs to be independent from the source of the encryption key $\mathcal{K}$. We also learned that this is always the case for independent sources: $H(\mathcal{P}) \le H(\mathcal{K})$, but that it shouldn't be used as an exhaustive proof, because in the case where you have a high-entropy key but it is badly designed (such as having different lengths for each code word), you could still derive a lot of information from attempted decryption. 

We saw an example where someone used a coin to determine a random key composed of 16 bits to mask a plaintext composed of 4 digits coded into 4 bits each (using xor logic). In this case, $H(\mathcal{P}) = 13.2877$ (there are 10000 possible combinations of digits), and $H(\mathcal{K}) = 16$, so the inequality holds. 

Suppose now that 8 bits of the key are generated randomly, and then just copied over for the remaining 8 bits. The key source and the plaintext source now become dependent, however only if the "hacker" knows that the key was copied twice over (because it is technically a possible outcome). The inequality doesn't hold anymore either because $H(\mathcal{K}) = 8$. I'm assuming that this encryption scheme fails to provide perfect secrecy, but I remain unconvinced because of my previous argument of the conditional state of this "copied" key. In this case, is there a better way to provide a proof other than the two methods mentioned above?
Xor-ing the first half of the encrypted message with the second gives the same result as xor-ing the first and second half of the original message. (When the key is duplicated). This contradicts perfect secrecy as some information can be obtained from the cyphertext.