Let $(M,\tau)$ be a finite von Neumann factor (in my case $M=R^\omega$, but I don't think this additional hypothesis might be useful for this particular problem) and fix a projection $p$. Let $\tau_p$ (resp. $\tau_{1-p}$) denote the unique normalized trace on $pMp$ (resp. $(1-p)M(1-p)$.

**Convention.** Given two projections $q_1,q_2$ in some finite factor, I can do several operations between them; for instance, $q_1\vee q_2$, but also $q_1\wedge q_2^{\perp}$. I will describe these operations using *formulas*. So when I will say that a certain property of two projections is true for all formula $f$, will mean that this property is true for any projection obtained by $q_1,q_2$ doing operations using $\vee, \wedge$ and $\perp$. I will use the formulas as functions; namely, $f(q_1,q_2)$ is the projection obtained by $q_1,q_2$ when the formula $f$ is applied. For instance, $f(q_1,q_2)=q_1\vee q_2$.

Suppose that we have the following data:

- projections $p_1,p_2,q_1,q_2\in pMp$
- projections $p_1',p_2',q_1',q_2'\in (1-p)M(1-p)$

such that for all formula $f$ one has

$$ \tau_p(f(p_1,p_2))\sim\tau_p(f(q_1,q_2)) \qquad\text{and}\qquad\tau_{1-p}(f(p_1',p_2'))\sim\tau_{1-p}(f(q_1',q_2')) $$

Question:Is it true that for all formula $f$, one has $$ \tau(f(p_1+p_1',p_2+p_2')\sim\tau(f(q_1+q_1',q_2+q_2')) ? $$

It should be reasonable. Basically I have two pairs of component-wise orthogonal projections $(p_1,p_2)$ and $(p_1',p_2')$ and then I have another two pairse of component-wise orthogonal projections that have approximately the same geometry as the first two pairs. I want to conclude that the sums still have approximately the same geometry... I'd been trying a bit but I am in trouble, basically because there is no way to distribute the sum of projections with $f$. One needs maybe some clever idea (or a counterexample), but I'm stuck now.

Thanks in advance for any help,

Valerio