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


*

*$x\cap y=y\cap x$

*$x\cap (y\cap z)=(x\cap y)\cap z$

*$x\cap y'=z\cap z' \iff x\cap y=x$
I managed to obtain a few results like:


*

*$x\cap x=x$

*$x\cap x'=y\cap y'$ (hence we can define $\mathbb{0}$)

*$x\cap\mathbb{0}=\mathbb{0}$

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


*

*$x\cup y=y\cup x$

*$(x\cup y)\cup z= x\cup(y\cup z)$

*$x\cup(x\cap y)=x$ (absorption)

*$x\cap(x\cup y)=x$ (absorption)

*$x\cup x'=\mathbb{1}$
What I can't prove are both distributive laws:


*

*$(x\cap y)\cup z=(x\cup z)\cap(y\cup z)$

*$(x\cup y)\cap z=(x\cap z)\cup(y\cap z)$
Can you help me?
 A: 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.)
