are $$\rho^{(x)}_{{\rm A}|b}=\frac{\Tr_{\rm B}(\Pi_b\rho_{{\rm AB}}^{(x)})}{p_b}.$$

Our goal is to prove $\delta_{\rm con}({\rm A}|{\rm B})=\chi_q-\chi_c$. Because of condition 2, $\delta^{(1)}({\rm A}|{\rm B})=0$, as so the consumed discord is $$\begin{aligned}
\delta_{\rm con}({\rm A}|{\rm B}) &= p_0\delta^{(0)}({\rm A}|{\rm B})-\bar\delta({\rm A}|{\rm B}) \\
&= p_0(S(\rho_{\rm B}^{(0)})-S(\rho_{{\rm AB}}^{(0)})+\sum_b p_bS(\rho^{(0)}_{{\rm A}|b})) \\
&-S(\bar\rho_{\rm B})+S(\bar\rho_{{\rm AB}})-\sum_b p_b S(\bar\rho_{{\rm A}|b}).\end{aligned}$$ We also have that: $$\begin{aligned}
\chi_q-\chi_c&=S(\bar\rho_{{\rm AB}})-p_0S(\rho_{{\rm AB}}^{(0)})-p_1S(\rho_{{\rm AB}}^{(1)}) \\
& +\sum_bp_b (-S(\bar\rho_{{\rm A}|b})+p_0S(\rho_{{\rm A}|b}^{(0)})+p_1S(\rho_{{\rm A}|b}^{(1)})).\end{aligned}$$ This leads to $$\begin{aligned}
&\delta_{\rm con}({\rm A}|{\rm B})-(\chi_q-\chi_c)\\
&=p_0 S(\rho_{\rm B}^{(0)}) -S(\bar\rho_{\rm B})+p_1S(\rho_{{\rm AB}}^{(1)}) -\sum_bp_bp_1S(\rho_{{\rm A}|b}^{(1)}).\end{aligned}$$ From condition 1, we have that $\rho_{\rm B}^{(0)}=\rho_{\rm B}^{(1)}=\bar\rho_{\rm B}$. From condition 2, $\rho_{{\rm AB}}^{(1)}$ is a product state, so $S(\rho_{{\rm AB}}^{(1)})=S(\rho_{\rm A}^{(1)})+S(\rho_{{\rm B}}^{(1)})$ and $\rho^{(1)}_{{\rm A}|b}=\rho^{(1)}_{\rm A}$. So this becomes $$\begin{aligned}
&\delta_{\rm con}({\rm A}|{\rm B})-(\chi_q-\chi_c)\\
&=S(\rho_{\rm B}^{(0)})(p_0-1+p_1)+S(\rho_{\rm A}^{(1)})(p_1-p_1)\\
&=0.\end{aligned}$$

In continuous-variable quantum illumination, condition 1 is satisfied since the idler is not interacting with the illumination object. Condition 2 is met by the fact that the swap operation decorrelates mode A and mode B. By restricting ourselves to Gaussian quantum discord, together with the assumption that a Gaussian heterodyne measurement is the optimal measurement for the quantities in condition 3, we have $\delta^{\rm G}_{\rm con}({\rm A}|{\rm B})=\chi_q-\chi_c$. This assumption is justified by numerical results in 