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 $\sigma(C+D)=\tau(C+D)=\sigma(C)+\sigma(D)$.
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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)$. |
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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 $\sigma(C+D)=\sigma(C)+\sigma(D)$. |
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