0NNI Proof. First, since we have a natural DK-equivalence 𝒞→ℳ(𝒞){\mathcal{C}}\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{M}}({\mathcal{C}}), there is a natural equivalence K(𝒞)→K(ℳ(𝒞))K({\mathcal{C}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}K({\mathcal{M}}({\mathcal{C}})) [13, 19, 81]. Next, since the category ℳ(𝒞){\mathcal{M}}({\mathcal{C}}) satisfies the hypothesis of Theorem 7.8, the second equivalence holds. ∎