Proof. Suppose $\delta $ is defined by the splitting
and $\delta '$ is defined by the splitting with $\pi ',s'$ in place of $\pi ,s$. Then
since $bs' = bs = 1_ z$ and $\pi s = 0$. Similarly,
Since $\delta = \pi d(s)$ and $\delta ' = \pi 'd(s')$ as constructed in Lemma 22.20.1, we may compute
using $\pi a = 1_ x$, $ba = 0$, and $\pi 'sbd(s') = \pi 'sba\pi d(s') = 0$ by formula (5) in Lemma 22.20.1. $\square$
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