Every BA is isomorphic to a Lindenbaum-Tarski algebra I have read in many places that every boolean algebra is isomorphic to a Lindenbaum-Tarski algebra, but can't seem to find the proof. Could someone please enlighten me? Thank you very much
 A: Here is the proof I was shown on a course on Boolean Algebras.
Let $\mathbf{A}$ be a Boolean Algebra, and let $X$ be a set of variables with the same cardinality as $A$. Fix any surjective map $h : X \rightarrow A$.
Denote by $\mathbf{Fm}(X)$ the absolutely free algebra generated by $X$ (known as the term algebra over $X$ in Universal Algebra, and as the formula algebra with variables in $X$ in Abstract Algebraic Logic). So $h$ extends to a surjective homomorphism $h : \mathbf{Fm}(X) \rightarrow \mathbf{A}$.
Let $\Gamma := \{\varphi \in \mathbf{Fm}(X) : h(\varphi) = 1\}$, where $1$ is the top element of $\mathbf{A}$.
Define the relation $\Omega(\Gamma)$ in $\mathbf{Fm}(X)$ as follows:
$$\langle \varphi,\psi \rangle \in \Omega(\Gamma) \iff \Gamma \models \varphi \leftrightarrow \psi.$$
You can check that $\Omega(\Gamma)$ is a congruence of $\mathbf{Fm}(X)$.
Claim 1. $\Gamma \models \varphi$ implies $h(\varphi) = 1$ for all $\varphi \in \mathbf{Fm}(X)$.
Proof. Suppose $\Gamma \models \varphi$ but $h(\varphi) \neq 1$. Then, $h(\neg\varphi) \neq 0$, and therefore there is some ultrafilter $\mathcal{U} \subseteq A$ with $h(\neg\varphi) \in \mathcal{U}$. Define the valuation $v : X \rightarrow \{T,F\}$ by setting $v(x) := T$ iff $h(x) \in \mathcal{U}$. You can check by induction that $v(\psi) = T$ iff $h(\psi) \in \mathcal{U}$ for every formula $\psi$. Since $h(\gamma) = 1 \in \mathcal{U}$ for all $\gamma \in \Gamma$, $v[\Gamma] \subseteq \{T\}$, and thus you get $v(\varphi) = T$ from $\Gamma \models \varphi$. This means $h(\varphi) \in \mathcal{U}$, which is impossible because $h(\neg\varphi) \in \mathcal{U}$. $\blacksquare$
Claim 2. For every formulas $\varphi,\psi \in \mathbf{Fm}(X)$,
$$\langle \varphi,\psi \rangle \in \Omega(\Gamma) \iff h(\varphi) = h(\psi).$$
Proof. Assume $\langle \varphi,\psi \rangle \in \Omega(\Gamma)$. Then, $\Gamma \models \varphi \leftrightarrow \psi$, so $h(\varphi \leftrightarrow \psi) = 1$ by Claim 1, whence $h(\varphi) = h(\psi)$. Conversely, assume $h(\varphi) = h(\psi)$. Then, $h(\varphi \leftrightarrow \psi) = h(\varphi) \leftrightarrow^\mathbf{A} h(\psi) = 1$, so $\varphi \leftrightarrow \psi \in \Gamma$. $\blacksquare$
Finally, define the map $g : \mathbf{Fm}(X) / \Omega(\Gamma) \rightarrow \mathbf{A}$ by setting $g([\varphi]) := h(\varphi)$ for every $\varphi \in \mathbf{Fm}(X)$, where $[\varphi]$ denotes the equivalence class of $\varphi$. By Claim 2, this map is well-defined. If $a \in A$, then there is some $x \in X$ such that $h(x) = a$ because $h$ is surjective, and thus $g([x]) = h(x) = a$, so $g$ is also surjective. And it is injective as well by Claim 2. You only have to check that $g$ is a homomorphism, which follows easily from the fact that $h$ is a homomorphism; for example,
$$g([\varphi] \land [\psi]) = g([\varphi \land \psi]) = h(\varphi \land \psi) = h(\varphi) \land^\mathbf{A} h(\psi) = g([\varphi]) \land^\mathbf{A} g([\psi]).$$
Conclusion: $\mathbf{A} \cong \mathbf{Fm}(X) / \Omega(\Gamma)$.
