Characterizing sub-algebras

Question

Let $B$ be a boolean algebra and $A$ and $A^{'}$ two sub-algebras. Is it true that the sub algebra generated by $A \cup A^{'}$ inside $B$ is the same as the algebra $C=\{ (a_{1}\wedge b_{1}) \vee \dots \vee (a_{k} \wedge b_{k})| a_{1},\dots, a_{k} \in A, b_{1},\dots, b_{k} \in A^{'} $ and $k \in \mathbb{N}\}$. If so, how can I prove such statement?

Answer 1

Yes. The subalgebra generated by $A\cup A'$ consists of all terms in the language $\{\land,\lor,\lnot,\top,\bot\}$ of Boolean algebras, evaluated at elements of $A\cup A'$. By disjunctive normal form, any such term is equivalent to a disjunction of conjunctions of elements and their negations.

We can simplify any such conjunction (like $c_1\land \lnot c_2\land \lnot c_3\land c_4$, for example) to one of the form $a\land b$ with $a\in A$ and $b\in A'$, by replacing all the terms in $A$ by their conjunction (or $\top$ if there are none) and similarly for $A'$. So we are left with a disjunction of the form stated in the question.