$X$ on the state by performing the operation $O_x$ on $\rho$, resulting in a state $\rho^{(x)}=O_x(\rho)$ with discord $\delta^{(x)}({\rm A}|{\rm B})$.

Let us decompose the discord of $\rho$, $\delta({\rm A}|{\rm B})$ into three components: $$\label{eq_discord_decomp}
\delta({\rm A}|{\rm B}) = \delta_{\rm loss} + \bar\delta({\rm A}|{\rm B}) + \delta_{\rm con}({\rm A}|{\rm B})$$ The first component $\delta_{\rm loss}$ is the amount of discord lost to the environment during the encoding process. This can be evaluated by first defining $$\delta_{\rm loss}^{(x)} = \delta({\rm A}|{\rm B}) - \delta^{(x)}({\rm A}|{\rm B})$$ as the loss of discord for each possible value of $x$ that Alice can encode, and then taking the weighted average over the probability of encoding that $x$. This results in $$\delta_{\rm loss} = \sum_x p_x \delta_{\rm loss}^{(x)}$$ The second component $\bar\delta({\rm A}|{\rm B})$ is the discord of $\bar\rho=p_0\rho^{(0)}+p_1\rho^{(1)}$, the state after encoding. This is the state seen by Bob who is oblivious to the value of $X$.

We term the remaining component the [*consumed discord*]{} $\delta_{\rm con}({\rm A}|{\rm B})$, and represents the discord in $\rho$ that remain unaccounted for. In prior literature, it was proposed to capture the amount of discord consumed to encode the value of $X$ on the state $\rho$ [@weedbrook2016discord]. For 