Showing that $S(\rho_{XB}||\sigma_{XB})=\sum_{x}p(x)D(\rho_{B}^{x}||\sigma_{B}^{x})$ for classical-quantum states

Question

Having some trouble showing that $S(\rho_{XB}||\sigma_{XB})=\sum_{x}p(x)D(\rho_{B}^{x}||\sigma_{B}^{x})$ for $\rho_{XB}=\sum_{x}p(x)|x\rangle\langle x|\otimes\rho_{B}^{x}$ and $\sigma_{XB}=\sum_{x}p(x)|x\rangle\langle x|\otimes\sigma_{B}^{x}$

I know that $$S(\rho_{XB}||\sigma_{XB})=\mathrm{Tr}(\rho_{XB}\log(\rho_{XB}))-\mathrm{Tr}(\rho_{XB}\log(\sigma_{XB}))$$ and $$\mathrm{Tr}(\rho_{XB}\log(\rho_{XB}))=-S(X)-\sum_{x}p(x)S(\rho_{B}^{x})$$ due to its classical-quantum nature. However, this implies that $$-\mathrm{Tr}(\rho_{XB}\log(\sigma_{XB}))=S(X)+\sum_{x}p(x)S(\rho_{B}^{x})+\sum_{x}p(x)D(\rho_{B}^{x}||\sigma_{B}^{x})$$.

Taking $$-\mathrm{Tr}(\rho_{XB}\log(\sigma_{XB}))=-\mathrm{Tr}(\sum_{x}p(x)|x\rangle\langle x|\otimes\rho_{B}^{x} \log(\sum_{x}p(x)|x\rangle\langle x|\otimes\sigma_{B}^{x}))=$$

$$-\mathrm{Tr}(\sum_{x}p(x)|x\rangle\langle x|\otimes\rho_{B}^{x}(\sum_{x}|x\rangle\langle x|\otimes \log(p(x)\sigma_{B}^{x})))=\mathrm{Tr}(\sum_{x}p(x)|x\rangle\langle x|\otimes\rho_{B}^{x}\log(p(x)\sigma_{B}^{x}))=$$

$$-\sum_{x}p(x)\mathrm{Tr}(\rho_{B}^{x}\log(p(x)\sigma_{B}^{x}))=-\sum_{x}p(x)\mathrm{Tr}(\rho_{B}^{x}(\log(p(x))+\log(\sigma_{B}^{x})))=$$

$$-\sum_{x}p(x)(\mathrm{Tr}(\rho_{B}^{x}(\log(p(x)))+\mathrm{Tr}(\rho_{B}^{x}\log(\sigma_{B}^{x})))=-\sum_{x}p(x)\log(p(x))+\sum_{x}p(x)D(\rho_{B}^{x}||\sigma_{B}^{x})$$

As can be seen, this is not enough to cancel the other terms, which can be seen when putting these two together, you get $$-S(X)-\sum_{x}p(x)S(\rho_{B}^{x})+S(X)+\sum_{x}p(x)D(\rho_{B}^{x}||\sigma_{B}^{x})=\sum_{x}p(x)D(\rho_{B}^{x}||\sigma_{B}^{x})-\sum_{x}p(x)S(\rho_{B}^{x})$$ This could only equal the desired exprerssion if $\rho_{B}^{x}$ were pure states, which is specified nowhere in the text I am reading.

Where have I gone wrong?

Edit: As Rammus stated in his answer, I went wrong with $$-\sum_{x}p(x)(\mathrm{Tr}(\rho_{B}^{x}(\log(p(x)))+\mathrm{Tr}(\rho_{B}^{x}\log(\sigma_{B}^{x})))=-\sum_{x}p(x)\log(p(x))+\sum_{x}p(x)D(\rho_{B}^{x}||\sigma_{B}^{x})$$ Instead, it should be $$-\sum_{x}p(x)(\mathrm{Tr}(\rho_{B}^{x}(\log(p(x)))+\mathrm{Tr}(\rho_{B}^{x}\log(\sigma_{B}^{x})))=-\sum_{x}p(x)\log(p(x))-\sum_{x}p(x)Tr(\rho_{B}^{x}log(\sigma_{B}^{x}))$$ Putting this back into the original equation gives the desired equality.

Answer 1

As you say, $$ \mathrm{Tr}[\rho_{XB} \log \rho_{XB}] = -S(X) + \sum_{x} p(x) \mathrm{Tr}[\rho_{B}^x \log \rho_B^x]. $$

But if you can prove the above statement, then the exact same derivation gives you $$ \mathrm{Tr}[\rho_{XB} \log \sigma_{XB}] = - S(X) + \sum_{x} p(x) \mathrm{Tr}[\rho_B^x \log \sigma_B^x]. $$

If you put both together then you get $$ \begin{aligned} D(\rho_{XB} \|\sigma_{XB}) &= \mathrm{Tr}[\rho_{XB} \log \rho_{XB}] - \mathrm{Tr}[\rho_{XB} \log \sigma_{XB}] \\ &= -S(X) + \sum_{x} p(x) \mathrm{Tr}[\rho_{B}^x \log \rho_B^x] + S(X) - \sum_{x} p(x) \mathrm{Tr}[\rho_B^x \log \sigma_B^x] \\ &= \sum_{x} p(x) \mathrm{Tr}[\rho_{B}^x \log \rho_B^x] - \sum_{x} p(x) \mathrm{Tr}[\rho_B^x \log \sigma_B^x] \\ &= \sum_{x} p(x) \left(\mathrm{Tr}[\rho_{B}^x \log \rho_B^x] - \mathrm{Tr}[\rho_B^x \log \sigma_B^x]\right) \\ &= \sum_{x} p(x) D(\rho_{B}^x \|\sigma_B^x) \end{aligned} $$

To comment on your derivation

You have several errors in the 7th and 8th equalities in your long derivation.