Proving both distributive laws for Boolean algebra given by certain axioms.

Question

In the book "Introduction to mathematical logic", Elliott Mendelson gives the following axiomatization of Boolean algebra:

We call the triple $(B,\cap,')$ a Boolean algebra whenever $B$ has at least two elements, $\cap$ (meet) is two-argument operator and $'$ (complement) is one-argument operator and the following axioms are satisfied:

1. $x\cap y=y\cap x$
2. $x\cap (y\cap z)=(x\cap y)\cap z$
3. $x\cap y'=z\cap z' \iff x\cap y=x$

I managed to obtain a few results like:

1. $x\cap x=x$
2. $x\cap x'=y\cap y'$ (hence we can define $\mathbb{0}$)
3. $x\cap\mathbb{0}=\mathbb{0}$
4. $x''=x$

Next we define $x\cup y:=(x'\cap y')'$ and $\mathbb{1}:=\mathbb{0}'$ and I proved some more properties:

1. $x\cup y=y\cup x$
2. $(x\cup y)\cup z= x\cup(y\cup z)$
3. $x\cup(x\cap y)=x$ (absorption)
4. $x\cap(x\cup y)=x$ (absorption)
5. $x\cup x'=\mathbb{1}$

What I can't prove are both distributive laws:

1. $(x\cap y)\cup z=(x\cup z)\cap(y\cup z)$
2. $(x\cup y)\cap z=(x\cap z)\cup(y\cap z)$

Can you help me?

Answer 1

Having defined $0$, rewrite your third axiom as \begin{equation}\label{eq:0} x \wedge y' = 0 \Leftrightarrow x \wedge y = x,\tag{0} \end{equation} and since you have already proven that the structure is a complemented lattice (and only need distributivity to conclude it's a Boolean algebra), then $x \leq y$ iff $x \wedge y = x$, which is notationally convenient. Thus \begin{equation}\label{eq:1} x \wedge y' = 0 \Leftrightarrow x \leq y\tag{1} \end{equation} Now, with the equations that you claim you proved, $$x \wedge (x \wedge y)' \wedge y = (x \wedge y) \wedge (x \wedge y)' = 0,$$ whence, by \eqref{eq:0} $$x \wedge (x \wedge y)' \wedge y' = x \wedge (x \wedge y)',$$ and analogously, $$x \wedge (x \wedge z)' \wedge z' = x \wedge (x \wedge z)',$$ yielding $$x \wedge (x \wedge y)' \wedge (x \wedge z)' \wedge (y' \wedge z') = x \wedge (x \wedge y)' \wedge (x \wedge z)',$$ or $$x \wedge (x \wedge y)' \wedge (x \wedge z)' \leq y' \wedge z',$$ and, by \eqref{eq:1}, $$x \wedge (x \wedge y)' \wedge (x \wedge z)' \wedge (y' \wedge z')' = 0.$$ From your definition of join and some identities you proved, this is $$x \wedge ((x \wedge y) \vee (x \wedge z))' \wedge (y\vee z) = 0,$$ or, by \eqref{eq:1}, $$x \wedge (y \vee z) \leq (x \wedge y) \vee (x \wedge z).$$ Since $$x \wedge (y \vee z) \geq (x \wedge y) \vee (x \wedge z)$$ holds in every lattice, we conclude that $$x \wedge (y \vee z) = (x \wedge y) \vee (x \wedge z),$$ and therefore, $B$ is a Boolean algebra.
(I suppose you know that in a lattice each distributive law follows from the other.)