Glass' Partially Ordered GroupsLemma 2.3.2 says:
Let G be a p.o. group and $g,h\in G$. If $g\vee h$ exists, then $g^{-1} \wedge h^{-1}=(g\vee h)^{-1}$
Proof: If $f\leq g^{-1},h^{-1}$ then $f^{-1}\geq g,h$. Thus $f^{-1}\geq g\vee h$. Hence $f\leq (g\vee h)^{-1}$.
I don't understand the second step. $g\vee h\geq g,h$, so the fact that $f^{-1}\geq g,h$ doesn't seem to imply that it's bigger than their join.
Recall that $g \vee h$ is the least upper bound (supremum) of $g , h$. This means that anything which is $\geq$ both $g$ and $h$ must also be $ \geq g \vee h$.