# Does proving $Q^{(1)}(\mathcal{N}\otimes\mathcal{N})=Q^{(1)}(\mathcal{N})+Q^{(1)}(\mathcal{N})$ imply additivity for arbitrary $n$?

I have been reading the proofs that are usually presented in order to proof the additivity of degradable and conjugate degradable channels, and they usually present that the coherent information is additive for two channel uses, i.e. $$Q^{(1)}(\mathcal{N}\otimes\mathcal{N})=Q^{(1)}(\mathcal{N})+Q^{(1)}(\mathcal{N})$$, and then immediately state that such thing implies that the quantum channel capacity is single-letter characterized.

My question is then, does proving that the channel coherent information to be additive for the two-channel use case imply that it is additive for larger $$n$$?

Cross-posted on physics.SE

After discussing this topic with several colleagues, the answer to this questions is that proving the two-shot additivity does not imply additivity for arbitrary $$n$$ channel uses. In fact, it has been studied that an unbounded number of channel uses may be required to be able to detect non-additive effects of the coherent information (Unbounded number of channel uses may be required to detect quantum capacity). It seems that the paper I was citing in the last update of the answer ("Quantum channel capacities") is referring to what it is called "strong additivity", which in fact means that a channel N is additive respect to an entropic function for any oteher channel M. Strong additivity is not being proved in the question posted, so one thing does not imply the other.