0PBK Proof. Let s′∈I′s^{\prime}\in I^{\prime}. Since |θθ′(s′)|+|θs′′|=|θθ′(s′)∘θs′′||\theta_{\theta^{\prime}(s^{\prime})}|+|\theta^{\prime}_{s^{\prime}}|=|\theta_{\theta^{\prime}(s^{\prime})}\circ\theta^{\prime}_{s^{\prime}}|, it follows that i(θθ′(s′),idi)+i(θs′′,idi)=i(θθ′(s′)∘θs′′,idi)i(\theta_{\theta^{\prime}(s^{\prime})},\operatorname{id}\nolimits_{i})+i(\theta^{\prime}_{s^{\prime}},\operatorname{id}\nolimits_{i})=i(\theta_{\theta^{\prime}(s^{\prime})}\circ\theta^{\prime}_{s^{\prime}},\operatorname{id}\nolimits_{i}) for all i∈I0i\in I_{0}. As a consequence, i((θ∘θ′)⊠idI0)−i(θ⊠idI0)−i(θ′⊠idI0)=i(θ∘θ′)−i(θ)−i(θ′).i((\theta\circ\theta^{\prime})\boxtimes\operatorname{id}\nolimits_{I_{0}})-i(\theta\boxtimes\operatorname{id}\nolimits_{I_{0}})-i(\theta^{\prime}\boxtimes\operatorname{id}\nolimits_{I_{0}})=i(\theta\circ\theta^{\prime})-i(\theta)-i(\theta^{\prime}). The lemma follows now from Lemma 7.4.9. ∎