Cardinality of a free Boolean algebra on countably many generators The cardinality of a free Boolean algebra on finitely many generators is $2^{2^n}$, where $n$ is the number of generators.
Why is the the free Boolean algebra on countably many generators countable? How to prove it?

Let $S$ be the set of generators. Define sets $A_n$ recursively:
$$A_0=S\cup\{0,1\}$$
$$A_{n+1}=A_n\cup\{x\vee y:x,y\in A_n\}\cup\{x\wedge y:x,y\in A_n\}\cup\{x':x\in A_n\}$$
(I would have gladly used the notation you are using for the Boolean operations, but you didn't tell me what they are.)
You can easily show by induction that all of the sets $A_n$ are countable, and so is their union $A=\bigcup_{n=1}^\infty A_n$. So $A$ is countable, and it's a Boolean algebra, and it's the Boolean algebra generated by $S.$
This trivial argument has nothing to do with "free" or "Boolean"; in an algebraic structure with countably many operations, each of finite "arity", a countable set generates a countable set, and a set of infinite cardinality $\aleph_\alpha$ generates a set of cardinality $\aleph_\alpha.$
On the other hand, if you allow infinitary operations (for example, complete Boolean algebras), then all hell breaks loose.
