0P6L Proof. The lemma follows from the commutativity of the following diagram: E2iF1iE2(m)⊕E2iF1iE1(m)\textstyle{E_{2}^{i}F_{1}^{i}E_{2}(m)\oplus E_{2}^{i}F_{1}^{i}E_{1}(m)}E2(m)⊕E1(m)\textstyle{E_{2}(m)\oplus E_{1}(m)}E2iF1iE2(m~)⊕E2iF1iE1(m~)\textstyle{E_{2}^{i}F_{1}^{i}E_{2}(\tilde{m})\oplus E_{2}^{i}F_{1}^{i}E_{1}(\tilde{m})}E2(m~)⊕E1(m~)\textstyle{E_{2}(\tilde{m})\oplus E_{1}(\tilde{m})}E2ςi∘λ(1⋯2i+1)\scriptstyle{E_{2}\varsigma_{i}\circ\lambda_{(1\cdots 2i+1)}}E1ςi∘λ(1⋯2i+1)\scriptstyle{E_{1}\varsigma_{i}\circ\lambda_{(1\cdots 2i+1)}}∑r=1iE2ςi−1∘E2iF1i−1ε1∘λ(1⋯r)(2i⋯i+r)\scriptstyle{\sum_{r=1}^{i}E_{2}\varsigma_{i-1}\circ E_{2}^{i}F_{1}^{i-1}\varepsilon_{1}\circ\lambda_{(1\cdots r)(2i\cdots i+r)}}E2iF1iE2f\scriptstyle{E_{2}^{i}F_{1}^{i}E_{2}f}E2iF1iE1f\scriptstyle{E_{2}^{i}F_{1}^{i}E_{1}f}E2f\scriptstyle{E_{2}f}E1f\scriptstyle{E_{1}f}E2ς~i∘λ(1⋯2i+1)\scriptstyle{E_{2}\tilde{\varsigma}_{i}\circ\lambda_{(1\cdots 2i+1)}}E1ς~i∘λ(1⋯2i+1)\scriptstyle{E_{1}\tilde{\varsigma}_{i}\circ\lambda_{(1\cdots 2i+1)}}∑r=1iE2ς~i−1∘E2iF1i−1ε1∘λ(1⋯r)(2i⋯i+r)\scriptstyle{\sum_{r=1}^{i}E_{2}\tilde{\varsigma}_{i-1}\circ E_{2}^{i}F_{1}^{i-1}\varepsilon_{1}\circ\lambda_{(1\cdots r)(2i\cdots i+r)}} ∎