I'm attempting to prove the following theorem left as an exercise from Bell & Slomson's Models & Ultraproducts (1969). I'd like to know whether my attempted proof is correct, and if not, I'd like some hints. I'm unclear how to go about this. I don't think I can use any facts about homomorphisms or order-preserving functions, since this comes before that material is presented (7 pages later), and that strikes me as the natural way to do the proof.
"Suppose $B$ = < B, $\lor$, $\land$, *, $\emptyset$, 1>. is a Boolean algebra. Let $\le$ be defined as follows: $\forall$ $x$, $y$ $\in$ B, $x$$\le$ $y$ iff $x$ $\lor$ $y$ = $y$. Then < B, $\le$ > is a complemented, distributive lattice."
Proof:
We note |B| = |$B$|. We suppose B is partially ordered by $\le$. Take $x$, $y$ $\in$ B. Since B is partially ordered by $\le$, we have $x$ $\le$ $y$. So by definition above, $x$ $\lor$ $y$ = $y$. By dual this means $x$ $\land$ $y$ = $x$. By definition of the structure $B$, there will be $\top$ = 1 and $\bot$ = $\emptyset$. Need to show each element of B has a unique complement, which will then yield that < B, $\le$ > is a complemented, distributive lattice. So suppose for contradiction that not every element of < B, $\le$ > has a unique complement. Then there exists $x$ $\in$ B such that $x$ has two complements, say $y$ , $z$. Because $B$ is a Boolean algebra, this means $B$ complemented and distributive, which means each element has a unique complement. Since B is in the structure $B$, given $x$ has two complements $y$, $z$ $\in$ B these two elements must exists for x $\in$ $B$, i.e.there's an $x$* $\in$ $B$ such that $x$* = $y$, $z$ (complement of x is y, z). But each element of $B$ has unique complement, since $B$ is a Boolean algebra. Contradiction. Hence it must be that every element of B has a unique complement.
END PROOF.
NOTE: Courtesy of @bof, I learned that I need to show that $B$ is partially ordered by $\le$, as defined above.
Reflexivity: Given definition above and $x$ $\in$ B, it follows by substitution of $x$ for $y$ that $x$ $\lor$ $x$ = $x$ $\in$ $B$; and by the definition, this is just $x$ $\le$ $x$.
Anti-symmetry: Suppose for $x$, $y$ $\in$ B we have $x$ $\lor$ $y$ = $x$ $\land$ $y$ $\in$ $B$. Given the definition for $\lor$ above, we have a dual: $x$ $\le$ $y$ iff $x$ $\land$ $y$ = $x$; so we can have $x$ $\lor$ $y$ = $y$ = $x$ $\land$ $y$ = $x$ or $x$ = $y$. If it happens that $y$ = $x$ $\land$ $y$ and $x$ = $x$ $\lor$ $y$ with $x$ $\lor$ $y$ = $x$ $\land$ $y$, we have $x$ = $y$. So either way, $x$ = $y$.
Transitivity: Suppose $x$ $\lor$ $y$ = $y$ & $y$ $\lor$ $z$ = $z$ for $x$, $y$, $z$ $\in$ B and $x$ $\lor$ $y$ & $y$ $\lor$ $z$ $\in$ $B$. By our definition, we have $x$ $\le$ $y$, $y$ $\le$ $z$, and so it must be that $z$ $\ge$ $x$, $y$, which by our definition means that $z$ = $x$ $\lor$ $z$, and again this is just $x$ $\le$ $z$.