Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
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 $\sigma(C+D)=\tau(C+D)=\sigma(C)+\sigma(D)$.
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 $\sigma(C+D)=\tau(C+D)=\sigma(C)+\sigma(D)$.
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)$.
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 $\sigma(C+D)=\sigma(C)+\sigma(D)$.
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 $\sigma(C+D)=\tau(C+D)=\sigma(C)+\sigma(D)$.
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 $\sigma(C+D)=\sigma(C)+\sigma(D)$.
