I have a pretty simple proof. As Tom has already observed, we have a bounded lattice. Now one characterization of Boolean algebra is a (bounded) lattice such that for every element $a$ there is an element $-a$ such that

$$a \wedge b \leq c \qquad iff \qquad a \leq -b \vee c$$

In category-speak, notice that the poset map $(-) \wedge b$ has a right adjoint $-b \vee (-)$. This immediately implies that $(-) \wedge b$ distributes over any joins that exist.

So let's check that this iff condition holds. Under the hypotheses, the left side is equivalent to saying

$$a \wedge b \wedge -c = 0.$$

Similarly, the right-hand side says $a \wedge -(-b \vee c) = 0$. But according to the characterization of joins, we can rewrite this as $a \wedge b \wedge -c = 0$, and we are done.

I'll see if I can dig up a reference in just a moment. It's pretty well-known in categorical circles.

I have a pretty simple proof. As Tom has already observed, we have a bounded lattice. Now one characterization of Boolean algebra is a (bounded) lattice such that for every element $a$ there is an element $-a$ such that

$$a \wedge b \leq c \qquad iff \qquad a \leq -b \vee c$$

In category-speak, notice that the poset map $(-) \wedge b$ has a right adjoint $-b \vee (-)$. This immediately implies that $(-) \wedge b$ distributes over any joins that exist.

So let's check that this iff condition holds. Under the hypotheses, the left side is equivalent to saying

$$a \wedge b \wedge -c = 0.$$

Similarly, the right-hand side says $a \wedge -(-b \vee c) = 0$. But according to the characterization of joins, we can rewrite this as $a \wedge b \wedge -c = 0$, and we are done.

Edit: One reference is the nLab. Alternatively, notice that combining Tom's observation with mine shows that we have a complemented distributive lattice.