I was able to complete the proof, but I'm afraid I may have used a circular argument or made too many assumptions somewhere. A better/simpler proof would be nice.
Consider two boolean algebras $(B,\leq, \lor, \land,^c,0,1)$ and $(B',\leq, \lor, \land,^c,0,1)$. A map $h:B\to B'$ is a homomorphism, if for all $x,y\in B$ we have:
- $h(x\land y) = h(x) \land h(y)$
- $h(x^c) = (h(x))^c$
Then show that $h(x\lor y) = h(x) \lor h(y)$ for all $x,y \in B$, $h(0) = 0$ and $h(1) = 1$.
My work:
I started with $h(x\land y) = h(x) \land h(y)$ and got $(h(x\land y))^c = (h(x) \land h(y))^c$. Then, using De-morgan's laws (is it okay to use these?) and $h(x^c) = (h(x))^c$, I ended up with $h(x^c\lor y^c) = h(x^c) \lor h(y^c)$. Now, we know that a Boolean algebra is a complemented lattice, i.e. every element $x \in B$ has a complement $x^c \in B$. So $h(x^c\lor y^c) = h(x^c) \lor h(y^c)$ essentially means $h(p\lor q) = h(p) \lor h(q)$ for all $p,q \in B$, which proves what is desired. Is this last line of reasoning okay?
Lastly, putting $y = x^c$ in $h(x\land y) = h(x) \land h(y)$ and $h(x\lor y) = h(x) \lor h(y)$ gives $h(0) = 0$ and $h(1) = 1$.
Thanks!
Everything seems fine. Double negation makes your life easier when combined with De Morgan: $$h(x\vee y) = h(x^{cc}\vee y^{cc}) = h((x^c \wedge y^c)^c) \stackrel{(\ast)}{=} (h(x)^c\wedge h(y)^c)^c = h(x)\vee h(y),$$where in $(\ast)$ we used the two homomorphism axioms in one shot.
Then for $0$, you have $$h(0) = h(x \wedge x^c) = h(x)\wedge h(x)^c = 0,$$and for $1$ we have $$h(1)= h(x \vee x^c) = h(x)\vee h(x)^c = 1.$$