Proof, [Lemma 25.1, Zhu93]. Let $U, V \in A$ be partial isometries such that $P = U^{*}U$, $UU^{*} \le Q$, $Q = V^{*}V$, and $VV^{*} \le P$. Denote $Q_{0} = Q$ and $P_{0} = P$. For each $n \in \natz$, inductively define $P_{n+1}= VQ_{n}V^{*}$ and $Q_{n+1}= UP_{n}U^{*}$, then:

1. (i)

   For each $n \in \natz$, $P_{n}, Q_{n} \in \text{Proj}(A)$.

2. (ii)

   For each $n \in \natz$, $P_{n} \le P$ and $Q_{n} \le Q$.

3. (iii)

   For each $n \in \natz$, $P_{n+1}\le P_{n}$ and $Q_{n+1}\le Q_{n}$.

As $\seq{P_n}, \seq{Q_n}\subset \text{Proj}(A)$ are non-increasing sequences, by Theorem 37.3.1, there exists $P_{\infty}, Q_{\infty} \in \text{Proj}(A)$ such that $P_{n} \to P_{\infty}$ and $Q_{n} \to Q_{\infty}$ in the strong operator topology as $n \to \infty$.

For each $n \in \natz$, $U(P_{n} - P_{n+1})U^{*} = Q_{n+1}- Q_{n+2}$, so

and $P_{n} - P_{n+1}\sim Q_{n+1}- Q_{n+2}$. Similarly, $Q_{n} - Q_{n+1}\sim P_{n+1}- P_{n+2}$. As $P_{n+1}= VQ_{n}V^{*}$ for all $n \in \natz$, $P_{\infty} \sim Q_{\infty}$ after passing through a strong-operator limit.

For each $N \in \natz$, $\sum_{n = 0}^{N} (P_{n} - P_{n+1}) = P - P_{N+1}$, so $P = P_{\infty} + \sum_{n = 0}^{\infty} (P_{n} - P_{n+1})$. Similarly, $Q = Q_{\infty} + \sum_{n = 0}^{\infty} (Q_{n} - Q_{n+1})$. Therefore

because $\sim$ is preserved through direct sums.$\square$