I'm looking through the book Models and Ultraproducts by Bell & Slomson, and I'm checking that my proof is correct (exercise 1.13). I'm pretty sure it is, but I thought I'd post it anyway.
It is stated as follows on page 10 of my Dover Thrift edition:
"Show that in a complemented lattice $L$, for all x $\in$ $L$,
(1) x $\land$ 1 = x,
(2) x $\lor$ 1 = 1,
(3) x $\land$ $\emptyset$ = $\emptyset$,
(4) x $\lor$ $\emptyset$ = x."
(This is a nice intuitive-sounding property.)
Proof: Divides into cases.
Case (1): We have that $L$ is complemented. So we have the existence of a 1 in $L$ such that 1 $\geq$ x for every x in L. This includes every x $\neq$ 1. $L$ is partially ordered, so that every x $\neq$ 1 is also x $\lt$ 1. Hence Infimum(x,1) or greatest lower bound of (x,1) = x, for any x. But this is just x $\land$ 1 = x for all x $\in$ $L$. END CASE 1.
Case (2): Again, given $L$ is complemented, 1 exists. Thus 1 is the maximum of $L$. As such, 1 is unique. Hence Supremum(x, 1) = 1 or Least upper bound (x,1) = 1 for any x in $L$. But this is just x $\lor$ 1 = 1 for all x $\in$ $L$. END CASE 2.
Case (3): $L$ is complemented, so $\emptyset$ exists, where $\emptyset$ is the minimum of $L$. Thus $\emptyset$ $\leq$ x for every x $\in$ $L$. Hence Infimum(x, $\emptyset$) = $\emptyset$ for all x $\in$ $L$. But this is just x $\land$ $\emptyset$ = $\emptyset$ for all x $\in$ $L$. END CASE 3.
Case (4): $L$ is complemented, so $\emptyset$ exists, where again $\emptyset$ is the minimum of $L$. Hence Supremum(x, $\emptyset$) = x for every x $\in$ $L$. But this is just x $\lor$ $\emptyset$ = x for every x $\in$ $L$. END CASE 4.
END PROOF.