Can't find a proof to a basic boolean algebra principle. I really have searched for an answer to my issue but it's so specific that I couldn't find it. The problem is very easy, and it consists of the exercise B. in this page: The Joy of Sets, p. 25. I'm trying to figure out the proof by treating zero and unity as ordinals $0$ and $1$, and probably my mistake is precisely that. 
I'm a newbie and I hope you'll excuse me if my question seems too stupid.
Thank you very much.
 A: It's no surprise that you could not solve the problem, because the axioms given are not strong enough to prove this. In fact, you can see that axiom (B5) is just a special case of axiom (B3), and can therefore be omitted; but then there are no axioms for $\neg$ (complement) at all, and in fact the axioms just define a distributive lattice together with any unary operation $\neg$ (without any restrictions on it). A simple example of this is the set of integers $\mathbb Z = \{\ldots,-2,-1,0,1,2,\ldots\}$, where $a \land b = \min(a,b)$ and $a \lor b = \max(a,b)$. Then complementation $\neg$ can be any function -- say, we could define $\neg a = a + 1$. This set is absolutely not a Boolean algebra, and in particular you can see that it is not true that $a \lor \neg a = b \lor \neg b$ for all $a,b$.
A correct axiomatisation of Boolean algebras replaces (B5) with
$$
(a \land \neg a) \lor b = b \qquad (a \lor \neg a) \land b = b.
$$
Now you can give the proof that you are asked, and it is not very difficult anymore. But do note that the names $0,1$ here have nothing to do with ordinals/natural numbers.
