I want to prove the Kaplansky's formula
For two projections $p, q$, $$(p\lor q)-p\sim q-(p\land q).$$
A step where I'm stuck is in trying to prove that the projection onto the kernel of $(1-p)q$ is $(1-q)+(q\land p)$.
Moreover, how does one find the right support of $(1-p)q$ from above?
If anyone can provide a reference where proof of Kaplansky's formula is explained properly, that'd be useful too!
Suppose that $(1-p)qx=0$. Then $$ qx=pqx+(1-p)qx=pqx. $$ So $qx\in (p\land q)\,H$, and then $$ x=(1-q)x+qx\in \big[(1-q)+(p\land q)\big]\,H. $$
Right support: if $xp=x$, then $px^*=x^*$. The least such projection is the propjection onto the range. So the right support is $$ P_{[\ker (1-p)q]^\perp}=1-(1-q)-p\land q=q-p\land q. $$
The proof I know is the one in Kadison Ringrose II, Theorem 6.1.7.