I am studying Ebbinghaus book "Mathematical Logic". In the section "Theories and Decidability" there is the following exercise:
Let $T = \Phi^{\vDash}$ be a theory, where $\Phi$ is $R$-enumerable. Show that $T$ is $R$-axiomatizable. (Hint: Starting with an enumeration $\phi_0, \phi_1, \ldots$ of $\Phi$, consider the set $ \{ \phi_{0}, \phi_{0} \land \phi_1, \ldots \}$.)
Here $\Phi$ is a set of sentences and $\Phi^{\vDash} := \{ \phi \in L_{0}^S \ | \ \Phi \vDash \phi \}$.
How to solve this exercise?
EDIT - please don't downvote for lack of dedication: I have put my attempt in this question as an answer. Of course, I welcome feedback and other answers too.
According to Ebbinghaus definition, $T$ will be $R$-axiomatizable if there is an $R$-decidable set $\Psi$ of sentences such that T = $\Psi^{\vDash}$. Maybe the most immediate idea is to take $\Psi = \Phi$. However, although $\Phi$ is $R$-enumerable, we can't guarantee that $\Phi$ will be $R$-decidable, so this idea fails.
Let's follow the hint in the exercise and take $\Psi = \{\phi_0, \phi_0 \land \phi_1, \ldots \}$. Notice that $\Psi^\vDash = \Phi^\vDash = T$, which means that all we need to do is to prove that $\Psi$ is $R$-decidable. The sentences in $\Psi$ always increase by size, so it's easy to decide whether a formula $\psi$ is in $\Psi$ or not: check all formulas of $\Psi$ until you either find $\psi$ or you find a formula with size greater than $\psi$ (in this latter case, we know for sure $\psi \notin \Psi$).