Theorem 37.7.8 (Kaplansky’s Formula).label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then
\[[(P \vee Q) - Q] \sim [(P - P \wedge Q)]\]
Proof. Using Theorem 37.3.1, assume without loss of generality that $I \in A$. In which case, by (1) of Lemma 37.7.4,
\[[(I - Q)P](H)^{\perp} = \ker(P(I - Q)) = \ker(Q)^{\perp} + [\ker(Q) \cap \ker(P)]\]
and since $P \vee Q$ and $Q$ commute,
\begin{align*}[(I - Q)P](H)&= [\ker(Q)^{\perp} + [\ker(Q) \cap \ker(P)]]^{\perp} \\&= \ker(Q) \cap [\ker(Q) \cap \ker(P)]^{\perp} \\&= \ker(Q) \cap [\ker(Q)^{\perp} + \ker(P)^{\perp}]\\&= (I - Q)(H) \cap [Q(H) + P(H)]\\&= (I - Q)(H) \cap [(Q \vee P)(H)] \\&= (P \vee Q)(I - Q)(H) = [(P \vee Q) - Q](H)\end{align*}
by (2) of Lemma 37.7.4. Similarly,
\[[P(I - Q)](H)^{\perp} = \ker((I - Q)P) = \ker(P) + [\ker(Q)^{\perp} \cap \ker(P)^{\perp}]\]
so
\begin{align*}[P(I - Q)](H)&= [\ker(P) + [\ker(Q)^{\perp} \cap \ker(P)^{\perp}]]^{\perp} \\&= \ker(P)^{\perp} \cap [\ker(Q) + \ker(P)] \\&= P(H) \cap [(I - Q)(H) + (I - P)(H)] \\&= P(H) \cap [(I - Q) \vee (I - P)](H) \\&= P(H) \cap [I - (P \wedge Q)](H) \\&= [P - (P \wedge Q)](H)\end{align*}
Therefore
\[[(P \vee Q) - Q](H) = [(I - Q)P](H) \sim P(I - Q)(H) = [P - (P \wedge Q)](H)\]
by Lemma 37.7.7.$\square$
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