Problem: Let $M$ be a (properly) infinite factor and $e$ and $f$ be projections of $M$. Then $$ e\lor f \sim 1 \iff e\sim1\ \text{ or } f\sim1 .$$
Idea: As $M$ is a factor, its center is trivial. Being properly infinite means that $1$ is properly infinite, therefore, by the halving lemma, there is a projection $g$ such that $g\sim 1\sim1-g$. But I don't know how to proceed. Moreover, we have the relations below: $$ e \lor f-f \sim e-e\land f\ $$ and: $$e-e\land\ (1-f)\sim\ f-(1-e)\land\ f.$$
Any help would be highly appreciated.
$\Leftarrow$) If $e=1$, then $e\lor f=1$.
$\Rightarrow$) Because $M$ is a factor, $e$ and $f$ are comparable. So we assume, without loss of generality, that $f\preceq e$ (otherwise, we switch roles). We also have that $e$ is infinite, because if both $e$ and $f$ are finite, so is $e\lor f$. With $M$ a factor, $e$ is properly infinite.
We can halve $e$, that is there exists a projection $e_1\leq e$ with $ e_1\sim e\sim e-e_1$. Using Kaplanski's formula, $e\lor f-e\sim f-f\land e\leq f$. That is $$\tag1 e\lor f-e\preceq f\preceq e\sim e-e_1. $$ We combine this with $$\tag2 e\sim e_1, $$ Because the projections on the left of $(1)$ and $(2)$ are orthogonal to each other, and the same happens with both projections on the right, $$ e\lor f=e+(e\lor f -e)\preceq e_1+(e-e_1)=e. $$ As $e\leq e\lor f$, we have shown that $e\sim e\lor f\sim 1$.