Just out of curiousity. If we view a formal system as an algebra, what do 'prime' sentences look like? $\varphi$ is prime iff $(\psi\wedge\eta\to \varphi)\to(\psi\to \varphi\vee \eta\to \varphi)$ is true for every $\psi,\eta$.
For instance, in language of set theory(assuming no set axioms), $(\exists x)(x=x)$ should be prime. For if $\psi\wedge\eta\to (\exists x)(x=x)$, the empty set is not a model of $\psi \wedge \eta$. Therefore empty set can't be model of $\psi,\eta$ both.
Another example. $\varphi=(\exists x)(x=\{x\})$ is not prime. $((\exists x)(x=\{\{x\}\})\wedge (\forall x)(x=\{\{x\}\}\to x=\{x\}))\to \varphi$. Denote the two sentences by $\psi,\eta$. It is obvious that $\eta\not\to \varphi$, since $\eta$ has an empty model. To show $\psi\not\to\varphi$, consider $M=\{0,1\}$ and let $\{0\}=1$ and $\{1\}=0$. Then $\{\{0\}\}=0$ in $M$ but no $x\in M$ satisfies $\{x\}=x$.
Is there some interesting facts about such algebras?
 A: The algebras you refer to in your question are called Lindenbaum-Tarski algebras. For any language $L$, the set of $L$-sentences up to logical equivalence naturally forms a Boolean algebra. More generally, we can fix an $L$-theory $T$ and a context of free variables $(x_1,\dots,x_n)$, and the set of $L$-formulas with free variables from $(x_1,\dots,x_n)$ up to equivalence modulo $T$ forms a Boolean algebra (when the context is empty, we get the case of sentences). These Boolean algebras are studied extensively in the field of model theory.
Your question is about prime elements of the algebra of sentences modulo logical equivalence. In the comments, halrankard2 suggested that if $\lnot \varphi$ is complete, then $\varphi$ is prime. This is true, and in fact this is a characterization of the prime sentences. [Caveat: I think you definition of "prime" should also include the condition that $\varphi$ is not a tautology, i.e. $\varphi$ generates a prime (and, in particular, proper) ideal in the Boolean algebra of sentences. If we do this, then we should take "complete" to mean "complete and consistent", i.e. $\varphi$ is complete if $\varphi$ is not a contradiction and for any sentence $\psi$, either $\varphi\rightarrow \psi$ or $\varphi\rightarrow \lnot \psi$. Alternatively, if you want to consider a tautology to be prime, then you can allow contradictions to be complete and you'll still get the equivalence.]
Here's a proof:
Suppose $\lnot \varphi$ is complete. Suppose $\psi\land \eta\rightarrow\varphi$. Then $\lnot \varphi\rightarrow \lnot (\psi \land \eta)$. Now since $\lnot \varphi$ is complete, $\lnot \varphi\rightarrow \psi$ or $\lnot \varphi\rightarrow \lnot \psi$. Similarly, $\lnot \varphi\rightarrow \eta$ or $\lnot \varphi\rightarrow \lnot \eta$. But if $\lnot \varphi\rightarrow \psi$ and $\lnot \varphi\rightarrow \eta$, then $\lnot \varphi\rightarrow (\psi\land \eta)$. Since $\lnot \varphi$ does not imply a contradiction, we conclude that $\lnot \varphi\rightarrow \lnot \psi$ or $\lnot \varphi\rightarrow \lnot \eta$. So $\psi\rightarrow \varphi$ or $\eta\rightarrow \varphi$. Hence $\varphi$ is prime.
Conversely, suppose $\varphi$ is prime. Since $\varphi$ is not a tautology, $\lnot \varphi$ is not a contradiction. Now for any sentence $\psi$, we have $\psi\land \lnot \psi \rightarrow \varphi$, so $\psi\rightarrow \varphi$ or $\lnot \psi \rightarrow \varphi$. Thus $\lnot \varphi\rightarrow \lnot \psi$ or $\lnot\varphi\rightarrow \psi$. So $\lnot \varphi$ is complete.

Similarly, fixing a theory $T$, a sentence $\varphi$ is prime in the algebra of sentences modulo $T$ if and only if $\lnot \varphi$ is complete modulo $T$, i.e., for every sentence $\varphi$, $T\vdash \lnot\varphi\rightarrow \psi$ or $T\vdash \lnot\varphi\rightarrow \lnot\psi$.
The presence of the negation in this characterization is a little bit annoying - it's often more convenient in logic to think about filters, rather than ideals, on Boolean algebras. We get that $\varphi$ is complete if and only if $\varphi$ generates a prime filter: If $\varphi\rightarrow \psi\lor \eta$, then $\varphi\rightarrow \psi$ or $\varphi\rightarrow \eta$.
And that last fact is really just a rephrasing of the statement that every prime filter on a Boolean algebra is a maximal filter (ultrafilter). Equivalently, every prime ideal in a Boolean ring is maximal.
