Proof. Let $A,B\in\cali{A}$ then $A=\dcups{n}{i=1}A_{i},B=\dcups{m}{i=1}B_{i}$ for some $\curl{A_i}_{1}^{n},\curl{B_i}_{1}^{m}\suf\cali{E}$. If $A\cap B=\emp$, then

so $\cali{A}$ is closed under disjoint unions.
(A1): By (P1) and (E1) there exists $\curl{E_i}_{1}^{n}\suf\cali{E}$ such that $X=\emp^{c}=\dcups{n}{i=1}E_{i}\in\cali{A}$
(A2’): Let $A=\dcups{n}{i=1}A_{i},B=\dcups{m}{i=1}B_{i}\in\cali{A}$ with $A\suf B$ then

By (E) and (P2) $\caps{n}{j=1}B_{i}\del A_{j}$ is a finite disjoint union of elements of $\cali{E}$ hence $B\del A\in\cali{A}$.
(A3): Let $A=\dcups{n}{j=1}A_{j},B=\dcups{m}{i=1}B_{i}\in\cali{A}$ then

so $\cali{A}$ is closed under intersections. Thus by (A2’)

and hence

$\square$