Boolean algebras without atoms Why is the theory of Boolean algebras without atoms $\omega$-categoric?

An answer using back-and-forth is as follows:
Let $A=\{a_n:n\in \mathbb N \}$ and $B=\{b_n:n\in \mathbb N \}$ with $a_0=0=b_0$ and $a_1=1=b_1$.
Put $\pi(a_i)=b_i$ for $i=0,1$.
Suppose we have defined $\pi$ for the subalgebra generated by $a_0,\ldots,a_n$, in a way that $\pi$ restricted to this subalgebra is an embedding into $B$. Let $x_0,\ldots,x_m$ be the atoms of $\langle a_0,\ldots,a_n \rangle$.
If $a_{n+1}\in \langle a_0,\ldots,a_n \rangle$, we are done.
Now suppose $a_{n+1}\notin \langle a_0,\ldots,a_n \rangle$, let $y_0,\ldots,y_k$ be the atoms of $\langle a_0,\ldots,a_{n+1} \rangle$, then for each $i\leq m$ we can find $y_{i,0},\ldots,y_{i,n_i}\in \{y_0,\ldots,y_k\}$  with $y_{i,0}+\cdots+y_{i,n_i}=x_i$. Then $\{y_{i,j}:i\leq m, j\leq n_i\}=\{y_0,\ldots,y_k\};$ as $\Sigma\{x_0,\ldots,x_n\}=1$.
Since $B$ is atomless, for each $i\leq m$ we can find $w_{i,0},\ldots,w_{i,n_i}\in B$ pairwise disjoint such that $w_{i,0}+\cdots+w_{i,n_i}=\pi(x_i)$, then $\{w_{i,j}:i\leq m, j\leq n_i\}$ is a set of pairwise disjoint elements of $B$ whose sum is $1$, hence if we define $\pi(y_{i,j})=w_{i,j}$ for all $i\leq m$ and $j\leq m_i$, $\pi$ is an embedding into $B$ and $\pi$ is an extension of our previous definition of $\pi$.
For the "forth" step interchange the roles of $a$ and $b$.
