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.
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.