Define $\tau(C+D):=\sigma(C)+\sigma(D)$. Then $\tau$ is a summation method for $C+D$, and by your assumption of uniqueness it follows that $\sigma(C+D)=\tau(C+D)=\sigma(C)+\sigma(D)$.
I think, you have already answered your own question. Define $\tau(C+D):=\sigma(C)+\sigma(D)$. Then $\tau$ is a summation method for $C+D$, and by your assumption of uniqueness it follows that $\sigma(C+D)=\sigma(C)+\sigma(D)$.\sigma(C+D)=\tau(C+D)=\sigma(C)+\sigma(D)$. 1 I think, you have already answered your own question. Define$\tau(C+D):=\sigma(C)+\sigma(D)$. Then$\tau$is a summation method for$C+D$, and by your assumption of uniqueness it follows that$\sigma(C+D)=\sigma(C)+\sigma(D)\$.